// Numbas version: finer_feedback_settings {"name": "Skills Audit for Maths and Stats Catalogue", "metadata": {"description": "

The full list of questions to choose from to build a bespoke Skills Audit for Maths and Stats

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Simplifying expressions from $\\frac{x^mx^n}{x^p}$ to $x^{m+n-p}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Simplify the following expression:

\\[x^{\\var{m}}x^{\\var{n}}\\]

", "advice": "

To simplify $x^{\\var{m}}x^{\\var{n}}$, we want to make use of the following rule:

\\[a^n \\times a^m = a^{n+m}\\]

Applying this rule,

\\[\\begin{split}x^{\\var{m}}x^{\\var{n}} &\\,=x^{\\simplify[!collectNumbers]{{m}+{n}}}\\\\ \\\\&\\,=x^{\\var{m+n}}. \\end{split}\\]

Use this link to find some resources which will help you revise this topic.

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Find the missing whole number power in an equation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the value of $n$ if

\\[\\frac{x^n}{x^\\var{p}}=x^\\var{m}\\]

", "advice": "

To find $n$ we need to re-write the expression such that we have $x^n$ on the left. We can multiply through by $x^\\var{p}$ to get

\\[x^n=x^\\var{m}{x^\\var{p}}\\]

Then applying the rule $x^p \\times x^q = x^{p+q}$ we get

\\[x^n=x^{\\var{m}+\\var{p}}=x^\\var{m+p}\\]

Hence, $n =\\var{m+p}$

Use this link to find some resources which will help you revise this topic.

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Calculate an answer involving a fractional index.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate the following expression:

\\[\\var{a^n}^{\\frac{1}{\\var{n}}}\\]

", "advice": "

To find $\\var{a^n}^{\\frac{1}{\\var{n}}}$, we want to make use of the fact that a power of $\\frac{1}{n}$ is the same as the $n$th root. Since

\\[\\var{a^n}=\\var{a}^\\var{n},\\]

we have,

\\[ \\var{a^n}^{\\frac{1}{\\var{n}}} =\\left(\\var{a}^\\var{n}\\right)^{\\frac{1}{\\var{n}}}=\\var{a}. \\]

Use this link to find some resources which will help you revise this topic.

Using indices rules to rewrite an expression from $a^\\frac{m}{n}$ to $b$, for integers $a$, $b$, $m$ and $n$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate the following expression:

\\[\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}\\]

", "advice": "

To find $\\var{a^n}^{\\frac{\\var{m}}{\\var{n}}}$, we want to make use of the following rule:

\\[\\left(a^n\\right)^m = a^{n\\times m}\\]

By rewriting the power $\\frac{\\var{m}}{\\var{n}}$ as a product of $\\var{m} \\times \\frac{1}{\\var{n}}$, we can apply this rule:

\\[ \\begin{split} \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} &\\,= \\var{a^n}^{\\left(\\var{m} \\times \\frac{1}{\\var{n}}\\right)} \\\\ &\\,= \\left(\\var{a^n}^\\frac{1}{\\var{n}}\\right)^\\var{m} \\\\ &\\,= \\var{a}^\\var{m}\\end{split} \\]

Then calculating what is left:

\\[ \\begin{split} \\var{a}^\\var{m} &\\,=\\var{a^(m)} \\end{split} \\]

Therefore,

\\[ \\var{a^n}^{\\frac{\\var{m}}{\\var{n}}} =\\var{a^(m)}. \\]

Use this link to find some resources which will help you revise this topic.

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perform a calculation involving negative indices.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate and simplify the following expression:

\\[\\frac{\\var{x}^{\\var{n}}}{\\var{y}^{\\var{m}}}\\]

", "advice": "

To simplify this expression we use the rule $a^{-n}=\\frac1{a^n}$.

\\[\\frac{\\var{x}^{\\var{n}}}{\\var{y}^{\\var{m}}}=\\frac{\\var{y}^{\\var{-m}}}{\\var{x}^{\\var{-n}}}=\\frac{\\var{y^-m}}{\\var{x^-n}}=\\simplify{{y^-m}/{x^-n}}\\]

Use this link to find some resources which will help you revise this topic.

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Simple exercise in collecting terms in different powers of \$x\$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}$ and $\\var{f}$ are ordinary numbers. We can combine them to get $\\var{b+f}$

We can combine $\\var{a}x$ and $\\var{d}x$ to get $\\var{a+d}x$.

We combine $\\var{c}y$ and $\\var{e}y$ to get $\\var{c+e}y$. So our answer is:

$\\simplify{{c+e}y+{a+d}x+{b+f}}$

Use this link to find some resources which will help you revise this topic.

$\\simplify[!collectNumbers]{{a}x+{b}+{c}y+{d}x+{f}+{e}y}$

", "answer": "({c}+{e})y+({a}+{d})x+({b}+{f})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "`+-$n`?*y+`+-$n`?*x+`+-$n`?", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC07 Collect terms (including squares)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Simple exercise in collecting terms in different powers of \$x\$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}$ and $\\var{f}$ are ordinary numbers. We can combine them to get $\\var{b+f}$

We can combine $\\var{a}x$ and $\\var{d}x$ to get $\\var{a+d}x$.

There are also $\\var{c}$ times $x^2$. So our answer is:

$\\simplify{{c}x^2+{a+d}x+{b+f}}$

Use this link to find some resources that will help you revise how to collect like terms.

$\\simplify[!collectNumbers]{{a}x+{b}+{c}x^2+{d}x+{f}}$

", "answer": "{c}x^2+({a}+{d})x+({b}+{f})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "`+-$n`?*x^2+`+-$n`?*x+`+-$n`?", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": [{"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC08 Collecting terms (higher powers)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Simple exercise in collecting terms in different powers of \$x\$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following expression by combining \"like\" terms.

", "advice": "

First we expand the minus sign in the bracket.

\\[\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}=\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4+{-f}x+{-e}x^3}\\]

The idea is to collect together and combine any terms that are the same kind of term so:

$\\var{b}x$ and $\\var{-f}x$ both have an $x$ term. We can combine them to get $\\var{b-f}x$

We can combine $\\var{a}x^4$ and $\\var{d}x^4$ to get $\\var{a+d}x^4$.

We combine $\\var{c}x^3$ and $\\var{-e}x^3$ to get $\\var{c-e}x^3$. So our answer is:

$\\simplify{{a+d}x^4+{c+e}x^3+{b+f}}$

Use this link to find some resources which will help you revise this topic.

$\\simplify[!collectNumbers]{{a}x^4+{b}x+{c}x^3+{d}x^4-({f}x+{e}x^3)}$

", "answer": "({a}+{d})x^4+({c}-{e})x^3+({b}-{f})x", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "mustmatchpattern": {"pattern": "`+-$n`?*x^4+`+-$n`?*x^3+`+-$n`?*x", "partialCredit": 0, "message": "", "nameToCompare": "", "warningTime": "submission"}, "valuegenerators": [{"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC09 Multiply algebraic terms", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Filling in the blanks from the answer to a simplified expression involving indices.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the missing factor in the following statement:

\\[ \\var{3*a}x^{\\var{b}}y^{\\var{c}} = \\var{a}x^{\\var{d}}(?)\\]

", "advice": "

We can divide the left handside of the expression by the factor given on the right hand side of the expression to work out the missing factor:

\\[\\begin{split}
\\var{3*a}x^{\\var{b}}y^{\\var{c}}&=\\var{a}x^{\\var{d}}(?)\\\\
\\Rightarrow \\frac{\\var{3*a}x^{\\var{b}}y^{\\var{c}}}{\\var{a}x^{\\var{d}}}&=(?)\\\\
\\Rightarrow 3x^{\\var{b}-\\var{d}}y^{\\var{c}} &=(?),
\\end{split}\\]

which after simplifying gives the answer:

\\[(?) = 3x^{\\var{b-d}}y^{\\var{c}}\\]

Use this link to find some resources which will help you revise this topic.

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This question is made up of 10 exercises to practice the multiplication of brackets by a single term.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Expand the expression below by multiplying each of the terms inside the brackets by the term outside. Give the answer in its simplest form.

", "advice": "

Expand brackets using the general formula $\\displaystyle a(x+c)=ax+ac$. This means we multiply each term inside the brackets by the term outside the brackets.

It is easy to forget that the sign outside the brackets also needs to be involved in the multiplication so remember that when two of the same sign are multiplied, the resultant term is positive and when opposite signs are multiplied, the result is negative.

\\[
\\begin{align}
\\simplify[terms]{{a[7]}x({a[8]}x^2+{a[9]}x)}&=
\\simplify[!collectNumbers]{{a[7]}x{a[8]}x^2+{a[7]}x{a[9]}x}\\\\&=
\\simplify{{a[7]}*{a[8]}x^3+{a[7]}*{a[9]}x^2}\\text{.}
\\end{align}
\\]

Use this link to find resources to help you revise how to expand single brackets

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$\\simplify{{a[7]}x({a[8]}x^2+{a[9]}x)}=$ [[0]]

Factorise an expression of 2 or 3 terms where the gcd is a letter times a number. Part of HELM Book 1.3.4

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Factorise $\\var{q2expr}$

", "advice": "

The two terms share a common factor of $\\var{q2gcd}\\var{latex(q2v[0])}$ which can be factored out.

So $\\var{q2expr} = \\var{q2ans}$

Use this link to find some resources which will help you revise this topic.

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Expanding two linear brackets multiplied together.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Expand the brackets and simplify

", "advice": "

To expand the brackets $\\simplify{({a[1]}x+{a[2]})({a[3]}x+{a[4]})}$ We first multiply all the terms in the left bracket by all the terms in the right bracket. This gives us

\\[\\var{a[1]}\\times\\var{a[3]}x^2+\\var{a[1]}x\\times\\var{a[4]}+\\var{a[2]}\\times\\var{a[3]}x+\\var{a[2]}\\times\\var{a[4]}=\\var{a[1]*a[3]}x^2+\\var{a[1]*a[4]}x+\\var{a[2]*a[3]}x+\\var{a[2]*a[4]}.\\]

We can then collect the terms to give us the final answer of

\\[\\var{a[1]*a[3]}x^2+\\var{a[1]*a[4]+a[2]*a[3]}x+\\var{a[2]*a[4]}.\\]

Use this link to find some resources which will help you revise this topic.

$\\simplify{({a[1]}x+{a[2]})({a[3]}x+{a[4]})}=$[[0]]

Expand two brackets involving powers of $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Expand the brackets and simplify

", "advice": "

To expand the brackets $\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}$ We first multiply all the terms in the left bracket by all the terms in the right bracket. This gives us

\\[\\var{a[1]}x^\\var{b[1]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[1]}x^\\var{b[1]}\\times\\var{c[1]}x^\\var{b[4]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{a[3]}x^\\var{b[3]}+\\var{a[2]}x^\\var{b[2]}\\times\\var{c[1]}x^\\var{b[4]}\\]

We can then simplify to give us the final answer of

$\\simplify{{a[1]*a[3]}*x^{b[1]+b[3]}+{a[1]*c[1]}*x^{b[1]+b[4]}+{a[2]*a[3]}*x^{b[2]+b[3]}+{a[2]*c[1]}*x^{b[2]+b[4]}}.$

Use this link to find some resources which will help you revise this topic.

$\\simplify{({a[1]}x^{b[1]}+{a[2]}x^{b[2]})({a[3]}x^{b[3]}+{c[1]}x^{b[4]})}=$[[0]]

Fiind the Highest Common Factor of two algebraic expressions involving a coefficient and powers of $x$ and $y$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the highest common factor of $\\var{c[0]}x^\\var{xp[0]}y^\\var{yp[0]}$ and $\\var{c[1]}x^\\var{xp[1]}y^\\var{yp[1]}$?

", "advice": "

In order to find the highest common factor of two single term algebraic expressions you can first find the highest common factor of the coefficients.

In this case the Highest common factor of $\\var{c[0]}$ and $\\var{c[1]}$ is $\\var{cans}$.

Then work through each of the variables (letters) in turn and see what powers of each appear. In the first expression there is $x^\\var{xp[0]}$ and the second expression there is $x^\\var{xp[1]}$. So they both have at least $x^\\var{xpans}$ in them. Similarly, the first expression there is $y^\\var{yp[0]}$ and the second expression there is $y^\\var{yp[1]}$. So they both have at least $y^\\var{ypans}$ in them.

Hence, the Highest Common Factor (HCF) of the two expressions is:

\\[\\var{cans}x^\\var{xpans}y^\\var{ypans}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"c": {"name": "c", "group": "Ungrouped variables", "definition": "repeat(2*random(6..25),2)", "description": "", "templateType": "anything", "can_override": false}, "xp": {"name": "xp", "group": "Ungrouped variables", "definition": "repeat(random(1..8),2)", "description": "", "templateType": "anything", "can_override": false}, "yp": {"name": "yp", "group": "Ungrouped variables", "definition": "repeat(random(1..8 except [xp[0],xp[1]]),2)", "description": "", "templateType": "anything", "can_override": false}, "cans": {"name": "cans", "group": "Ungrouped variables", "definition": "GCD(c[0],c[1])", "description": "", "templateType": "anything", "can_override": false}, "xpans": {"name": "xpans", "group": "Ungrouped variables", "definition": "min(xp[0],xp[1])", "description": "", "templateType": "anything", "can_override": false}, "ypans": {"name": "ypans", "group": "Ungrouped variables", "definition": "min(yp[0],yp[1])", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "xp[0]<>xp[1] AND c[0]<>c[1]", "maxRuns": 100}, "ungrouped_variables": ["c", "xp", "yp", "cans", "xpans", "ypans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{cans}*x^{xpans}*y^{ypans}", "answerSimplification": "basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}, {"name": "y", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC15 Algebraic substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Substitute values into an algebraic expression and calculate the result.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate the following expression,

\\[\\simplify{p^{n}+{a}*r*t+{c}},\\]

when $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$.

", "advice": "

In order to evaluate $\\simplify{p^{n}+{a}*r*t+{c}},$ with the given values, $p = \\var{pval}$, $r = \\var{rval}$, and $t = \\var{tval}$, we replace each instance of that letter with its corresponding value and then apply the rules of BIDMAS:

\\[\\var{pval}^\\var{n}+\\var{a}\\times \\var{rval} \\times \\var{tval} + \\var{c}\\]

Which gives the answer $\\var{ans}$.

Follow this link for more help on tackling these kind of questions.

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Rearrange a specific formula. No randomisation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rearrange the following equation, to make $y$ the subject:

\\[{cy -b = 3x}\\]

", "advice": "

In order to rearrange the equation so that it is in terms of $y$, we must first add $b$ to both sides, and then divide both sides of the equation by $c$:

\\begin{split} cy-b &= 3x \\\\ cy &= 3x + b \\\\ y &=\\frac{3x+b}{c} \\end{split}

Use this link to find some resources which will help you revise this topic.

$y=$ [[0]]

Simplify (qx+a)/(rx+b) +/- (sx+c)/(tx+d)

x is a randomised variable. a,b,c,d,q,r,s,t are randomised integers. a,b,c,d run from -5 to 5, including 0. q,r,s,t run from -3 to 3, and can be 0 if the constant term is nonzero, but are mostly 1.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express $\\displaystyle{\\var{te[0]}\\var{sgnl}\\var{te[1]}}$ as a single fraction.

", "advice": "

\\[\\begin{align*} \\var{te[0]}\\var{sgnl}\\var{te[1]} &= \\frac{\\var{ndnde[3]}}{\\var{ndnde[3]}}\\times\\var{te[0]}\\var{sgnl}\\frac{\\var{ndnde[1]}}{\\var{ndnde[1]}}\\times\\var{te[1]}\\\\&=\\frac{\\var{cnd[0]}\\var{sgnl}\\var{cnd[1]}}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\frac{(\\var{cnd[2]})\\var{sgnl}(\\var{cnd[3]})}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\frac{\\var{cnd[4]}}{(\\var{ndnde[1]})(\\var{ndnde[3]})}\\\\&=\\var{ans} \\end{align*}\\]

There is no benefit in expanding the denominator. In fact, it is best to leave the denominator factorised, because then it is easier to see if the fraction can be simplified.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"te": {"name": "te", "group": "Ungrouped variables", "definition": "[simplify(expression(\"(\"+ndnd[0]+\")/(\"+ndnd[1]+\")\"),\"all\"),\n simplify(expression(\"(\"+ndnd[2]+\")/(\"+ndnd[3]+\")\"),\"all\")\n]", "description": "", "templateType": "anything", "can_override": false}, "v": {"name": "v", "group": "Ungrouped variables", "definition": "random(\"a\",\"b\",\"c\",\"d\",\"f\",\"g\",\"h\",\"k\",\"m\",\"n\",\"p\",\"q\",\"r\",\"s\",\"t\",\"u\",\"v\",\"w\",\"x\",\"y\",\"z\")", "description": "

the variable to use

", "templateType": "anything", "can_override": false}, "xc": {"name": "xc", "group": "Ungrouped variables", "definition": "repeat(weighted_random([ [1,0.7] , [random(1..3),0.3] ])\n ,4)", "description": "

the x-coefficients

", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "map(\nswitch(xc[i]=0, random(2..5),\n xc[i]=1, random(-5..5 except 0),\n xc[i]=2, random(-5..5 except [-4,-2,2,4]),\n random(-5..5 except [-xc[i],xc[i]])\n),i,[0,1,2,3])", "description": "

the constants. Make sure that the constants are coprime with their x-coefficient, and if their x-coefficient is 0, that they are positive. There is a condition in variable testing to ensure that no fraction = 1.

", "templateType": "anything", "can_override": false}, "ndnd": {"name": "ndnd", "group": "Ungrouped variables", "definition": "[\"+\"+c[0],\n xc[1]+\"*\"+v+\"+\"+c[1],\n \"+\"+c[2],\n xc[3]+\"*\"+v+\"+\"+c[3]\n ]", "description": "

numerator 1, denominator 1, numerator 2, denominator 2, as strings.

", "templateType": "anything", "can_override": false}, "sgn": {"name": "sgn", "group": "Ungrouped variables", "definition": "random(\"+\",\"-\")", "description": "", "templateType": "anything", "can_override": false}, "sgnl": {"name": "sgnl", "group": "Ungrouped variables", "definition": "latex(sgn)", "description": "

for display purposes

", "templateType": "anything", "can_override": false}, "ndnde": {"name": "ndnde", "group": "Ungrouped variables", "definition": "map(simplify(expression(x),\"all\"),x,ndnd)", "description": "", "templateType": "anything", "can_override": false}, "cnd": {"name": "cnd", "group": "Ungrouped variables", "definition": "[simplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\"all\"),\nsimplify(expression(\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\"all\"),\nsimplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),[\"expandBrackets\",\"all\"]),\nsimplify(expression(\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),[\"expandBrackets\",\"all\"]),\nsimplify(expression(\n string(simplify(\n expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"])\n )+\"+\"+\n string(simplify(\n expression(sgn+\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"])\n )\n),[\"expandBrackets\",\"basic\"]),\n \nsimplify(expression(\n string(simplify(expression(\"(\"+ndnd[3]+\")*(\"+ndnd[0]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))+ \"+\" +\n string(simplify(expression(sgn+\"(\"+ndnd[1]+\")*(\"+ndnd[2]+\")\"),\n [\"expandBrackets\",\"all\",\"!noLeadingMinus\"]))\n ),[\"all\",\"!noLeadingMinus\"])\n ]", "description": "

The combined numerator and denominator terms:

0) numerator term 1, 1) numerator term 2,

2) brackets expanded num t1, 3) brackets expanded num t2

4) numerator, no brackets

5) numerator simplified

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Simplify the sum of two algebraic fractions where spotting factorising of both numerators and denominators can reduce the work massively.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ simplifying as much as possible. Your answer should be in the form $\\frac{\\alpha\\var{v}+\\beta}{\\delta\\var{v}^2-\\gamma}.$

", "advice": "

To write the following as a single fraction $\\frac{\\var{num1}}{\\var{den1}}+\\frac{\\var{num2}}{\\var{den2}}$ first factorise as much as possible and look for any cancellations:

\\[\\begin{split}
&\\frac{\\var{a}\\times\\var{b}}{\\var{den1fact}} + \\frac{\\var{num2}}{\\var{den2fact}}\\\\
& = \\frac{\\var{b}}{\\var{den1simp}} + \\frac{1}{\\var{f1c}}.
\\end{split}\\]

Then get a common denominator for the two fractions and combine into a single fraction:

\\[\\begin{split}
&\\frac{\\var{b}}{\\var{den1simp}} + \\frac{\\var{f1}}{\\var{den1simp}}\\\\
& = \\frac{\\var{b}+\\var{f1}}{\\var{den1simp}}\\\\
& = \\var{ans}.
\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

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{"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Luke Park", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/826/"}, {"name": "Anna Strzelecka", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2945/"}, {"name": "heike hoffmann", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2960/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": [], "metadata": {"description": "

A question to practice simplifying fractions with the use of factorisation (for binomial and quadratic expressions).

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simplify the following algebraic expression.

", "advice": "

\\[\\frac{{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}}{{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}}\\]

In this question there is a quadratic expression which needs to be factorised into the products of binomials in both the numerator and denominator.

\\[\\frac{({\\simplify{n+{e1}}})({\\simplify{n+{e2}}})}{({\\simplify{n+{e1}}})({\\simplify{n+{e3}}})}\\]

The repeated binomials in the numerator and denominator cancel, leaving:

\\[\\frac{({\\simplify{n+{e2}}})}{({\\simplify{n+{e3}}})}\\]

Use this link to find some resources which will help you revise this topic.

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\\[\\frac{\\simplify{(n^2+({e1}+{e2})n+{e1}{e2})}}{\\simplify{(n^2+({e1}+{e3})n+{e1}{e3})}}\\]

", "answer": "(n+{e2})/(n+{e3})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "notallowed": {"strings": ["^2", "^"], "showStrings": false, "partialCredit": 0, "message": ""}, "valuegenerators": [{"name": "n", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AC20 Multiplication of algebraic fractions 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["adding and subtracting fractions", "adding fractions", "converting between decimals and fractions", "converting integers to fractions", "Fractions", "fractions", "integers", "manipulation of fractions", "subtracting fractions", "taxonomy"], "metadata": {"description": "

Manipulate fractions in order to add and subtract them. The difficulty escalates through the inclusion of a whole integer and a decimal, which both need to be converted into a fraction before the addition/subtraction can take place.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following multiplication, giving the fraction in its simplest form.

\\[\\frac{\\var{a}}{y^\\var{n2}}\\times\\frac{y^\\var{n1}}{\\var{b}}\\]

", "advice": "

To multiply two fractions you just multiply the numerators and multiply the denominators. This means we have,

\\[\\frac{\\var{a}}{y^\\var{n2}}\\times\\frac{y^\\var{n1}}{\\var{b}}=\\frac{\\var{a}\\times{y^\\var{n1}}}{y^\\var{n2}\\times\\var{b}}=\\frac{\\var{a/gcd_ab}\\times{y^\\var{n1-n2}}}{\\var{b/gcd_ab}}\\]

Use this link to find some resources which will help you revise this topic.

[[0]] [[1]]

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Simplifying first is essential in terms of managing expressions that might need factorising.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Expand and simplify $\\displaystyle{\\var{LeftMul}\\times\\var{RightMul}}.$

", "advice": "

Before we multiply the fractions together first lets check if we can do any cancellation. Notice that $\\var{RightMulBottom}$ has a factor of $\\var{Num}$ so we can cancel this straight away.

We also have a factor of $x$ in both $\\var{QuadCoeff[0]}x^2+\\var{QuadCoeff[1]}x$ and $\\var{RightMulTop}$ so we're now left with multiplying

\\[\\frac1{\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]}}\\times\\frac{\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}}{\\var{Lin2Coeff[0]}x+\\var{Lin2Coeff[1]}}.\\]

We're not necesserily done with cancellation though! To make sure that a fraction with a quadratic is simplified we have to factorise it to make sure there are no linear factors we can cancel. In this case we have
\\[\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}={(x+\\var{Lin1Coeff})(\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]})}.\\]

This gives us one last factor to cancel and then we can finally multiply whats left of each fraction to give us a final answer of

\\[\\var{ans}.\\]

Use this link to find some resources which will help you revise this topic.

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Rewrite the expression $\\frac{mx^2+nx+k}{(x+a)(x^2+bx+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{Bx+C}{x^2+bx+c}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Rewrite the following expression as partial fractions:

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))}. \\]

", "advice": "

To express \\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} \\] as partial fractions, we want to set this equal to the sum of two fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x^2+{b}x+{c}}$. Since we have a linear factor and a quadratic factor, this tells us that the form of the partial fractions will be

\\[ \\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\simplify{A/(x+{a}) + (B*x+C)/(x^2+{b}x+{c})},\\]

where $A$, $B$, and $C$ are constants.

To find the values of $A$, $B$, and $C$, we want to first multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{m}x^2+{n}x+{k}=A(x^2+{b}x+{c})+B*x(x+{a}) + C(x+{a})}.\\]

(Note: To find $A$, $B$, and $C$, we will use a combination of choosing suitable values of $x$ to eliminate terms, and equating coefficients. It can be solved by only equating coefficients, but this is a more efficient process.)

To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:

\\[ \\simplify{{m*a^2-n*a+k}=A{(a^2-b*a+c)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]

To find $C$, we can eliminate $B$ by setting $x=0$ and substituting in the result of $A$:

\\[ \\simplify{{k}={c}A+{a}C} \\implies C=\\simplify[all,fractionNumbers]{({k}-{c}A)/{a}}.\\]

Hence,

\\[ C = \\simplify[fractionNumbers]{{Csol}}.\\]

Finally, by equating coefficients of the $x^2$-terms we can find $B$:

\\[ (x^2): \\quad \\var{m} = \\simplify{A+B} \\implies B=\\var{m}-A. \\]

Therefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]

and

{check}

Use this link to find some resources which will help you revise this topic.

\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify{{Asol}/(x+{a})+({Bsol}x+{Csol})/(x^2+{b}x+{c})}.\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sol2": {"name": "sol2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({m*c-m*b*a+n*a-k}x+{k*(a-b)-m*a*c+n*c})/({a^2-a*b+c}(x^2+{b}x+{c}))}.\\\\]

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\\\\[ \\\\simplify{({m}x^2+{n}x+{k})/((x+{a})(x^2+{b}x+{c}))} = \\\\simplify[all,fractionNumbers]{{m*a^2-n*a+k}/({a^2-a*b+c}(x+{a}))+({(m*c-m*b*a+n*a-k)/simp2}x+{(k*(a-b)-m*a*c+n*c)/simp2})/({(a^2-a*b+c)/simp2}(x^2+{b}x+{c}))}.\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random([1,2,3,5,7])", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "1", "description": "", "templateType": "anything", "can_override": false}, "pairs": {"name": "pairs", "group": "Ungrouped variables", "definition": "[[1,random(-1,1)*random([1,3,4,5])],[2,random(-1,1)*random([1,2,4,5])],[3,random(-1,1)*random([1,2,3,5])],[5,random(-1,1)*random([1,2,3,4,5,7])],[7,random(-1,1)*random([1,2,3,4,5,7])]]", "description": "", "templateType": "anything", "can_override": false}, "index": {"name": "index", "group": "Ungrouped variables", "definition": "random(0..4)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "a^2-a*b+c>0 or a^2-a*b+c<0", "maxRuns": 100}, "ungrouped_variables": ["a", "pairs", "index", "b", "c", "m", "k", "n", "Asol", "Bsol", "Csol", "check", "sol1", "sol2", "sol3", "simp1", "simp2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "\n

[[0]]

Basic calculation from a sum given in Sigma notation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate:

\\[\\displaystyle{\\Sigma_{n=1}^3} \\var{b}n.\\]

", "advice": "

The sigma notation $\\displaystyle\\sum_{n=1}^{3}\\var{b}n$ is asking us to find the sum of the first three terms of the sequence $\\var{b}n$.

\\[\\begin{split}\\Sigma_{n=1}^3 \\var{b}n &\\, = (\\var{b}\\times 1) + (\\var{b}\\times 2) + (\\var{b}\\times 3) \\\\ &\\, = \\var{b1} + \\var{b2} + \\var{b3} \\\\ &\\, = \\var{sum}.\\end{split}\\]

Use this link to find resources to help you revise sigma notation.

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Calculating gradient and finding intercept from a geogebra graph.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{app}
Find the gradient of the line.

", "advice": "

Firstly draw a right angled 'step' from left to right. This triangle can be anywhere, but it is more helpful for it to have corners on the vertices (whole number points) of the graph and it is easier to calculate with postive numbers.

{app_advice}

Before we start to calculate, notice that the line is {uod}, so the gradient will be {pon} and the line is {sos}, so the absolute value of the number will be {mol}.

Now find the coordinates of the places your triangle meets the line

$(x_1,y_1)=(\\var{ax},\\var{ay})$ and $(x_2,y_2)=(\\var{bx},\\var{by})$

We need to compare the 'rise on the y-axis' to the 'run across the x-axis', we can say that:

$\\text{gradient} = \\frac{\\text{rise}}{\\text{run}}$

This is equivalent to using the formula:

$ m = \\frac{y_2 - y_1}{x_2 - x_1} $

and substitute the coordinates of the vertices of the triangle:

$\\begin{split} &\\, m = \\frac{\\var{by} - \\var{ay}}{\\var{bx} - \\var{ax}} \\\\
&\\, = \\frac{\\var{by-ay}}{\\var{bx-ax}} \\\\
&\\, = \\var[fractionNumbers]{m} \\\\
\\end{split} $

Use this link to find resources to help you revise straight line graphs and how to find the gradient of them.

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if(m=abs(m),'positive','negative')

", "templateType": "anything", "can_override": false}, "ax": {"name": "ax", "group": "Ungrouped variables", "definition": "random(0,1)", "description": "", "templateType": "anything", "can_override": false}, "ay": {"name": "ay", "group": "Ungrouped variables", "definition": "random(0,1,2,3)", "description": "", "templateType": "anything", "can_override": false}, "bx": {"name": "bx", "group": "Ungrouped variables", "definition": "random(ax+1..3) \n", "description": "", "templateType": "anything", "can_override": false}, "by": {"name": "by", "group": "Ungrouped variables", "definition": "random(0..4 except ay)\n", "description": "", "templateType": "anything", "can_override": false}, "app_advice": {"name": "app_advice", "group": "Ungrouped variables", "definition": "geogebra_applet(\n 800,500,\n [\n A: [\n definition: p1,\n label_visible: false,\n visible: true\n ],\n B: [\n definition: p2,\n label_visible: false,\n visible: true \n ],\n \n C: [\n definition: p3,\n label_visible: false,\n visible: false \n ],\n \n line1: [\n definition: \"Line(A,B)\",\n label_visible: false,\n visible: true\n ],\n \n line2: [\n definition: \"Segment(A,C)\",\n label_visible: false,\n visible: true\n ],\n \n \n \n line3: [\n definition: \"Segment(C,B)\",\n label_visible: false,\n visible: true\n ]\n ]\n)", "description": "", "templateType": "anything", "can_override": false}, "p3": {"name": "p3", "group": "Ungrouped variables", "definition": "vector(bx,ay)", "description": "", "templateType": "anything", "can_override": false}, "pon": {"name": "pon", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(m=abs(m),'a positive number','a negative number'))", "description": "", "templateType": "anything", "can_override": false}, "sos": {"name": "sos", "group": "Ungrouped variables", "definition": "if(m=0,'horizontal',if(abs(m)<1,'shallow','steep'))", "description": "", "templateType": "anything", "can_override": false}, "mol": {"name": "mol", "group": "Ungrouped variables", "definition": "if(m=0,'zero',if(abs(m)<1,'less than 1','greater than or equal to 1'))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "m<>1", "maxRuns": 100}, "ungrouped_variables": ["app", "m", "c", "P1", "P2", "uod", "ax", "ay", "bx", "by", "app_advice", "p3", "pon", "sos", "mol"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you have incorrectly rounded this answer. You might want to look at some resources on rounded decimals. You can also leave your answer in fraction form as
$\\var[fractionNumbers]{m}$

", "useAlternativeFeedback": false, "answer": "{m}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "{m}", "showPreview": true, "checkingType": "dp", "checkingAccuracy": "1", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF03 Shapes of quadratics", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Multiple choice - select the quadratic graph.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of the following is the graph $y=x^2$.

", "advice": "

Use this link to find some resources to help you familiarise yourself with these graphs.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["{geogebra_applet('https://www.geogebra.org/m/tpfzv3w7')}", "{geogebra_applet('https://www.geogebra.org/m/zftpwq64')}", "{geogebra_applet('https://www.geogebra.org/m/we3gngqa')}", "{geogebra_applet('https://www.geogebra.org/m/cadkup6r')}"], "matrix": ["1", 0, 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF04 Graphs of trig functions (sin, cos, tan)", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Match the relevant graph (sin, cos, tan) with its equation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

This is about core knowledge of graphs. You should know the shapes of the fundamental trig graphs, if you don't familiarize yourself with them from the resources linked below. In this setting the $x$-axis is given with a scale in radians but you might also find some where it is given in degrees. You should also be aware of the difference between those two different units of angles.

Use this link to find some resources to help you familiarise yourself with these graphs.

Match the graph to its function.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["$\\sin(x)$", "$\\cos(x)$", "$\\tan(x)$"], "matrix": [["1", 0, 0], [0, "1", 0], [0, 0, "1"]], "layout": {"type": "all", "expression": ""}, "answers": ["{geogebra_applet('https://www.geogebra.org/m/ntqvuwqr')}", "{geogebra_applet('https://www.geogebra.org/m/fsqmnhsc')}", "{geogebra_applet('https://www.geogebra.org/m/yg6f9eqz')}"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AF05 Function notation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Evaluating a linear function for a given value of $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given $f(x)=\\simplify{{m}x+{c}}$, find $f(\\var{n})$.

", "advice": "

If $f(x)=\\simplify{{m}x+{c}}$, to find $f(\\var{n})$ we need to evaluate $f(x)$ when $x=\\var{n}$:

\\[ \\begin{split} f(\\var{n}) &\\,= \\simplify[alwaysTimes]{{m}({n})+{c}} \\\\ &\\,= \\simplify[!collectNumbers]{{m*n}+{c}} \\\\ &\\,= \\simplify{{m*n+c}}. \\end{split} \\]

Use this link to find resources to help you revise function notation.

$f(\\var{n})=$[[0]]

Determining the range of a function of the form $f = m|x| + a$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

The range is the set of values that can be taken by $f(x)$, i.e. the values on the $y$-axis.

{geogebra_applet('https://www.geogebra.org/m/aqcgkurg',[a: a, m: m])}

Therefore, for $f(x)=\\simplify{{m}x^2+{a}}$, the range is $[\\var{a}, \\infty)$.

Use this link to find some resources to help you revise how to find the domain and range of a function.

Given $f(x)=\\simplify{{m}x^2+{a}}$

What is the range of $f(x)$?

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Finding the inverse of a function of the form $f(x)=\\frac{mx+c}{x+a},\\,x\\neq-a$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If $f(x)=\\simplify{({m}x+{c})/(x+{a})},\\,x\\neq \\simplify{{-a}}$, find the inverse function, $f^{-1}(x)$.

", "advice": "

To find $f^{-1}x$, it can help to first set $f(x)$ to a different variable, which we will call $y$:

\\[ y = f(x) = \\simplify{({m}x+{c})/(x+{a})}\\]

Since the function $f(x)$ takes us from $x$ to $y$, the inverse function $f^{-1}$ will take us from $y$ to $x$. So to obtain $f^{-1}$, we want to rearrange $y=\\simplify{({m}x+{c})/(x+{a})}$ so that it is $x$ as a function of $y$:

\\[ \\begin{split} y &\\,= \\simplify{({m}x+{c})/(x+{a})} \\\\\\\\ \\simplify{(x+{a})y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y+{a}y} &\\,= \\simplify{{m}x+{c}} \\\\\\\\ \\simplify{x*y - {m}x} &\\,= \\simplify{{c}- {a}y} \\\\ \\\\ \\simplify{x(y-{m})} &\\,= \\simplify{{c}-{a}y} \\\\\\\\ x&\\,= \\simplify{({c}-{a}y)/(y-{m})}. \\end{split} \\]

Hence, $f^{-1}(y) =\\simplify{({c}-{a}y)/(y-{m})}$, and therefore \\[ f^{-1}(x) =\\simplify{({c}-{a}x)/(x-{m})}.\\]

(Note: The inverse is valid for all values of $x$ except $x=\\var{m}$, since this would make the denominator equal to 0.)

Use this link to find resources to help you revise how to find the inverse of functions.

$f^{-1}(x)=$[[0]]

Finding composite functions of 2 linear functions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$, find expressions for $f\\circ g(x)$ and $g \\circ f(x)$.

Recall: $f \\circ g(x) \\equiv f(g(x))$ and $g \\circ f(x) \\equiv g(f(x))$.

", "advice": "

To find the composition $f \\circ g(x)$ we are substituting the expression for $g(x)$ into the function $f(x)$, replacing the $x$-terms with the function $g(x)$. Similarly, to find the composition $g \\circ f(x)$ we are substituting the expression for $f(x)$ into the function $g(x)$, replacing the $x$-terms with the function $f(x)$.

So, for $f(x)=\\simplify{{m}x+{c}}$ and $g(x)=\\simplify{{n}x+{d}}$,

\\[ \\begin{split} f \\circ g(x) \\equiv f(g(x)) &\\,= \\simplify{{m}({n}x+{d})+{c}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{m*d}+{c}} \\\\ &\\,=\\simplify{{m*n}x+{m*d+c}}, \\end{split} \\]

and

\\[ \\begin{split} g \\circ f(x) \\equiv g(f(x)) &\\,= \\simplify{{n}({m}x+{c})+{d}} \\\\ &\\,=\\simplify[!collectNumbers,unitFactor]{{m*n}x+{n*c}+{d}} \\\\ &\\,=\\simplify{{m*n}x+{n*c+d}}. \\end{split} \\]

Use this link to find resources to help you revise how to find composite functions.

$f \\circ g(x)=$[[0]]

$g \\circ f(x)=$[[1]]

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Finding $x$ from a logarithmic equation of the form $\\log_ax = b$, where $a$ and $b$ are positive integers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the value of $x$:

\\[ \\log_\\var{a}x = \\var{n} \\]

", "advice": "

To find the value of $x$, recall that $\\log_a(x)=b$ is equivalent to $x=a^b$.

Therefore, \\[\\log_\\var{a}(x) = \\var{n} \\implies \\simplify[!collectNumbers]{x={a}^{n}}.\\]

Hence, \\[x=\\var{a^n}\\,.\\]

Use this link to find resources to help you revise logarithms.

$x=$ [[0]]

Solving $a\\log(x)+\\log(b)=\\log(c)$ for $x$, where $a$, $b$ and $c$ are positive integers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve for $x$:

\\[ \\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c}). \\]

", "advice": "

To solve $\\var{a}\\log(x)+\\log(\\var{b})=\\log(\\var{c})$ for $x$, we want to use the following logarithm rules:

$\\log(a^n) = n\\log(a)$;
$\\log(a)+\\log(b) = \\log(ab)$.

Hence,

\\[ \\begin{split} \\var{a}\\log(x)+\\log(\\var{b}) &\\,=\\log(\\var{c}) \\\\ \\log(x^\\var{a})+\\log(\\var{b}) &\\,= \\log(\\var{c}) \\\\ \\log(\\var{b}x^\\var{a}) &\\,= \\log(\\var{c}). \\end{split} \\]

If $\\log(a)=\\log(b)$ then this implies $a=b$. Therefore,

\\[ \\begin{split} \\var{b}x^\\var{a} &\\,=\\var{c} \\\\ x^\\var{a} &\\,= \\simplify[fractionNumbers]{{c/b}} \\\\ x &\\,= \\simplify[fractionNumbers]{({c/b})^(1/{a})} \\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}\\end{split} \\]

Use this link to find rsources to help you revise how the rules of logarithms to help you solve logarithmic equations.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Solving an equation of the form $a^x=b$ using logarithms to find $x$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve for $x$:

\\[ \\var{a}^x = \\var{b} \\,. \\]

", "advice": "

To solve $\\var{a}^x = \\var{b}$ for $x$, since $x$ is the exponent we want to make use of the following logarithm rule:

$\\log(a^n) = n \\log(a)$.

By taking the logarithm of each side and applying the above rule:

\\[ \\begin{split}\\var{a}^x &\\,= \\var{b} \\\\ \\log_{10}(\\var{a}^x) & \\,= \\log_{10}(\\var{b})\\\\ x \\log_{10}(\\var{a}) &\\,= \\log_{10}(\\var{b}) \\\\\\\\ x&\\,=\\simplify{log({b})/log({a})} \\\\\\\\ x &\\,= \\var{sol} \\text{ (2 d.p.)}. \\end{split} \\]

Use this link to find resources to help you revise how logarithms.

$x=$ [[0]] (Give you answer to 2 decimal places where necessary)

Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

Given $\\simplify{{m}w-{n} = {p}w+{q}}$, we can get all the $w$'s on the left hand side and all the numbers on the right hand side, and then divide both sides by the coefficient of $w$ to get $w$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n



$\\simplify{{m}w+{n}}$	$=$	$\\simplify{{p}w+{q}}$

$\\simplify[!cancelTerms,unitFactor]{{m}w-{n}-{p}w}$	$=$	$\\simplify[!cancelTerms,unitFactor]{{p}w+{q}-{p}w}$

$\\simplify{{m-p}w-{n}}$	$=$	$\\var{q}$

$\\var{m-p}w-\\var{n}+\\var{n}$	$=$	$\\var{q}+\\var{n}$

$\\var{m-p}w$	$=$	$\\var{q+n}$

$\\displaystyle{\\frac{\\var{m-p}w}{\\var{m-p}}}$	$=$	$\\displaystyle{\\frac{\\var{q+n}}{\\var{m-p}}}$

$w$	$=$	$\\displaystyle{\\simplify{{q+n}/{m-p}}} = \\var{precround(ansA,1)} \\text{ to 1 dp}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\simplify{({m}w-{n}) = {p}w+{q}}$

$w=$ [[0]]

Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To solve an equation like

$\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}},$

the first thing to deal with is the denominators of the fractions. In order to do that you multiply both sides of the equation by both denominators $\\var{num2}$ and $\\var{num3}$ (or their lowest common multiple to be slightly more efficient). This will give something equivalent to:

$\\displaystyle{\\var{num3 + num2} x+\\var{num3*num1} = \\var{num2*num3*num4}.}$

Then proceeding by subtracting $\\var{num3*num1} from both sides:

$\\displaystyle{\\var{num3 + num2} x = \\var{num2*num3*num4-num3*num1}.}$

And finally dividing by $\\var{num2+num3}$:

$\\displaystyle{x = \\frac{\\var{num2*num3*num4-num3*num1}}{\\var{num2+num3}}.}$

Use this link to find resources to help you revise how to solve linear equations

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Solve $\\displaystyle{\\frac{x+\\var{num1}}{\\var{num2}}+\\frac{x}{\\var{num3}}=\\var{num4}}$.

$x=$ [[0]]

Solving a pair of linear simultaneous equations, giving answers as integers or fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the simultaneous equations for x and y, giving your answers as integers or fractions, but not decimals.

\\[ \\begin{split} \\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\end{split}\\]

", "advice": "

\\[\\begin{split}\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\,=\\var{c} \\qquad\\qquad&(1)\\\\ \\simplify[!noLeadingminus,unitFactor]{{a1}x +{b1}y} &\\,=\\var{c1} \\qquad\\qquad&(2)\\end{split}\\]

{advice1}

Use this link to find some resources which will help you revise this topic.

For these equations, it is easiest to get a solution for $y$ first, due to the $x$-terms having {eqoroppa} coefficients.

\\n

If we {aorsa} equation (2) {torfa} equation (1) this eliminates the $x$-terms leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1)})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgna*(c1)}}\\\\\\\\ \\\\simplify{{b+sgna*(b1)}y} &\\\\,= \\\\simplify{{c+sgna*(c1)}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}x + \\\\var{b} \\\\times \\\\simplify[all, fractionNumbers]{{c+sgna*(c1)}/{b+sgna*(b1)}} &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{a}x &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c} - {c*b+b*sgna*(c1)}/{b+sgna*(b1)}} \\\\\\\\ x &\\\\,= \\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "eqoroppb": {"name": "eqoroppb", "group": "Ungrouped variables", "definition": "if(abs(b)*b1=abs(b1)*b,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "eqoroppa": {"name": "eqoroppa", "group": "Ungrouped variables", "definition": "if(abs(a)*a1=abs(a1)*a,'equal','equal and opposite')", "description": "", "templateType": "anything", "can_override": false}, "samey": {"name": "samey", "group": "Ungrouped variables", "definition": "\"

For these equations, it is easiest to get a solution for $x$ first, due to the $y$-terms having {eqoroppb} coefficients.

\\n

If we {aorsb} equation (2) {torfb} equation (1) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1)})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c}+{sgn*(c1)}}\\\\\\\\ \\\\simplify{{a+sgn*(a1)}x} &\\\\,= \\\\simplify{{c+sgn*(c1)}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a} \\\\times\\\\simplify[fractionNumbers]{{c+sgn*(c1)}/{a+sgn*(a1)}} + \\\\var{b}y &\\\\,= \\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{c} - {c*a+a*sgn*(c1)}/{a+sgn*(a1)}} \\\\\\\\ y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb": {"name": "lcmb", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (2) by $\\\\simplify{{abs(b/b1)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b/b1)}x +{b1*abs(b/b1)}y} &\\\\,=\\\\var{c1*abs(b/b1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({a}+{sgn*(a1*abs(b/b1))})x} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgn*(c1*abs(b/b1))}}\\\\\\\\ \\\\simplify{{a+sgn*(a1*abs(b/b1))}x} &\\\\,= \\\\simplify{{c+sgn*(c1*abs(b/b1))}} \\\\\\\\ x &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgn*(c1*abs(b/b1))}/{a+sgn*(a1*abs(b/b1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c)+a*sgn*c1*abs(b/b1)}/{(a)+sgn*a1*abs(b/b1)})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c+a*sgn*c1*abs(b/b1))/(a+sgn*a1*abs(b/b1))}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcmb1": {"name": "lcmb1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\simplify{{abs(b1/b)}}$ we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1/b)}x +{b*abs(b1/b)}y} &\\\\,=\\\\var{c*abs(b1/b)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4)\\\\\\\\ \\\\end{split} \\\\]

\\n

If we {aorsb} equation (4) {torfb} equation (3) this eliminates the $y$-terms, leaving us with one equation in terms of $x$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(a*abs(b1/b))}+{sgn*a1})x} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(b1/b))}+{sgn*c1}}\\\\\\\\ \\\\simplify{{(a*abs(b1/b))+sgn*a1}x} &\\\\,= \\\\simplify{{(c*abs(b1/b))+sgn*c1}} \\\\\\\\ x &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $y$ we can substitute this $x$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\var{a}\\\\times\\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({(c*abs(b1/b))+sgn*c1}/{(a*abs(b1/b))+sgn*a1}) + {b}y} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(a*c*abs(b1/b))+a*sgn*c1}/{(a*abs(b1/b))+sgn*a1})} \\\\\\\\ \\\\simplify{{b}y} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(a*c*abs(b1/b)+a*sgn*c1)/(a*abs(b1/b)+sgn*a1)}} \\\\\\\\ y &\\\\,=\\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "full": {"name": "full", "group": "Ungrouped variables", "definition": "\"

To get a solution for $x$, if we multiply equation (1) by $\\\\var{abs(b1)}$ and equation (2) by $\\\\var{abs(b)}$, we will have two equations with {eqoroppb} $y$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(b1)}x+{b*abs(b1)}y} &\\\\,=\\\\var{c*abs(b1)} \\\\qquad\\\\qquad&(3)\\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(b)}x +{b1*abs(b)}y} &\\\\,=\\\\var{c1*abs(b)} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

Now, {aorsb} equation (4) {torfb} equation (3) to eliminate the $y$ terms:

\\n

\\\\[ \\\\begin{split} (\\\\simplify[!collectNumbers]{{a*abs(b1)} +{sgn*a1*abs(b)}}) x &\\\\,= \\\\simplify[!collectNumbers]{{c*abs(b1)}+{sgn*c1*abs(b)}} \\\\\\\\ \\\\simplify{{a*abs(b1)+sgn*a1*abs(b)}} x &\\\\,= \\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}} .\\\\end{split} \\\\]

\\n

So the solution for $x$ is \\\\[ x=\\\\simplify{{c*abs(b1)+sgn*c1*abs(b)}/{a*abs(b1)+sgn*a1*abs(b)}}.\\\\]

\\n

To obtain a solution for $y$ we can substitute this value of $x$ into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[noLeadingminus,fractionNumbers,unitFactor]{{a} {xsimp} + {b}y} &\\\\,=\\\\var{c} \\\\\\\\ \\\\var{b}y &\\\\,= \\\\simplify[!collectNumbers,fractionNumbers]{{c}-{a*xsimp}} \\\\\\\\\\\\var{b}y &\\\\,= \\\\simplify[fractionNumbers]{{c-a*xsimp}} \\\\\\\\y &\\\\,= \\\\simplify[fractionNumbers]{{(c-a*xsimp)/b}} \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "aorsa": {"name": "aorsa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'subtract','add')", "description": "", "templateType": "anything", "can_override": false}, "torfa": {"name": "torfa", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,'from','to')", "description": "", "templateType": "anything", "can_override": false}, "sgna": {"name": "sgna", "group": "Ungrouped variables", "definition": "if(a*abs(a1)=abs(a)*a1,-1,1)", "description": "", "templateType": "anything", "can_override": false}, "lcma": {"name": "lcma", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (2) by $\\\\simplify{{abs(a/a1)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a}x+{b}y} &\\\\,=\\\\var{c} \\\\qquad\\\\qquad&(3)\\\\\\\\ \\\\simplify[!noLeadingminus,unitFactor]{{a1*abs(a/a1)}x +{b1*abs(a/a1)}y} &\\\\,=\\\\var{c1*abs(a/a1)} \\\\qquad\\\\qquad&(4)\\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({b}+{sgna*(b1*abs(a/a1))})y} &\\\\,= \\\\simplify[all, !collectNumbers, !noLeadingminus]{{c}+{sgna*(c1*abs(a/a1))}}\\\\\\\\ \\\\simplify{{b+sgna*(b1*abs(a/a1))}y} &\\\\,= \\\\simplify{{c+sgna*(c1*abs(a/a1))}} \\\\\\\\ y &\\\\,= \\\\simplify[all,fractionNumbers]{{c+sgna*(c1*abs(a/a1))}/{b+sgna*(b1*abs(a/a1))}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c+sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c)+b*sgna*c1*abs(a/a1)}/{(b)+sgna*b1*abs(a/a1)})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c+b*sgna*c1*abs(a/a1))/(b+sgna*b1*abs(a/a1))}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "lcma1": {"name": "lcma1", "group": "Ungrouped variables", "definition": "\"

To get a solution for $y$, if we multiply equation (1) by $\\\\simplify{{abs(a1/a)}}$ we will have two equations with {eqoroppa} $x$-coefficients:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!noLeadingminus,unitFactor]{{a*abs(a1/a)}x +{b*abs(a1/a)}y} &\\\\,=\\\\var{c*abs(a1/a)} \\\\qquad\\\\qquad&(3) \\\\\\\\\\\\simplify[!noLeadingminus,unitFactor]{{a1}x+{b1}y} &\\\\,=\\\\var{c1} \\\\qquad\\\\qquad&(4) \\\\end{split}\\\\]

\\n

If we {aorsa} equation (4) {torfa} equation (3) this eliminates the $x$-terms, leaving us with one equation in terms of $y$:

\\n

\\\\[ \\\\begin{split} \\\\simplify[!collectNumbers, !noLeadingminus]{({(b*abs(a1/a))}+{sgna*b1})y} &\\\\,= \\\\simplify[!collectNumbers, !noLeadingminus]{{(c*abs(a1/a))}+{sgna*c1}}\\\\\\\\ \\\\simplify{{(b*abs(a1/a))+sgna*b1}y} &\\\\,= \\\\simplify{{(c*abs(a1/a))+sgna*c1}} \\\\\\\\ y &\\\\,= \\\\simplify[all, fractionNumbers]{{(c*abs(a1/a))+sgna*c1}/{(b*abs(a1/a))+sgna*b1}}. \\\\end{split} \\\\]

\\n

To obtain a solution for $x$ we can substitute this $y$-value into either of our initial equations. Using equation (1), we obtain

\\n

\\\\[ \\\\begin{split} \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{{a}x + {b}}\\\\times \\\\simplify[all, !noLeadingminus, !expandBrackets, fractionNumbers]{({c*abs(a1/a)+sgna*c1}/{(b*abs(a1/a))+sgna*b1})} &\\\\,= \\\\var{c} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c} -({(b*c*abs(a1/a))+b*sgna*c1}/{(b*abs(a1/a))+sgna*b1})} \\\\\\\\ \\\\simplify{{a}x} &\\\\,= \\\\simplify[all, !noLeadingminus, fractionNumbers]{{c -(b*c*abs(a1/a)+b*sgna*c1)/(b*abs(a1/a)+sgna*b1)}} \\\\\\\\ x &\\\\,=\\\\simplify[fractionNumbers]{{(c*abs(b1)+sgn*c1*abs(b))/(a*abs(b1)+sgn*a1*abs(b))}}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "advice1": {"name": "advice1", "group": "Ungrouped variables", "definition": "if(abs(b)=abs(b1), {samey},if(abs(a)=abs(a1),{samex},if(lcm(abs(b),abs(b1))=abs(b),{lcmb},if(lcm(abs(b),abs(b1))=abs(b1),{lcmb1},if(lcm(abs(a),abs(a1))=abs(a),{lcma},if(lcm(abs(a),abs(a1))=abs(a1),{lcma1},{full}))))))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "abs(b-b1)>1 and\nabs(a-a1)>1 and\ngcd(a,c)=1 and\ngcd(a1,c1)=1", "maxRuns": 100}, "ungrouped_variables": ["a", "b", "a1", "b1", "c", "c1", "aorsa", "torfa", "aorsb", "torfb", "sgna", "sgn", "xn", "xd", "xsimp", "eqoroppa", "eqoroppb", "advice1", "samey", "samex", "lcmb", "lcmb1", "lcma", "lcma1", "full"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$x=$ [[0]]

$y=$ [[1]]

Solving a pair of simultaneous equations of the form $a_1x+y=c_1$ and $a_2x^2+b_2xy=c_2$ by forming a quadratic equation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the following simultaneous equations:

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\\\ \\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\end{split} \\]

Give your answers to 2 decimal places where necessary.

", "advice": "

To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that $y$ is the subject, which we can then substitute into the equation with the quadratic $x$-term. This will result in a quadratic equation in terms of $x$ only.

For the equations

\\[ \\begin{split} \\simplify{{a1}x+y} &\\,= \\var{c1} \\qquad \\qquad &(1) \\\\\\simplify{{a2}x^2+{b2}x*y} &\\,= \\var{c2} \\qquad \\qquad &(2) \\end{split} \\]

we can rearrange equation (1) to make $y$ the subject:

\\[ y = \\simplify{{c1}-{a1}x}. \\qquad\\qquad (3)\\]

Substituting this into equation (2):

\\[ \\begin{split}\\simplify{{a2}x^2+{b2}x({c1}-{a1}x)} &\\,=\\var{c2} \\\\ \\simplify[!cancelTerms,unitFactor]{{a2}x^2+{b2*c1}x-{b2*a1}x^2} &\\,=\\var{c2}. \\end{split} \\]

Collecting similar terms:

\\[ \\simplify{({a2}-{b2*a1})x^2+{b2*c1}x-{c2}} =0. \\qquad\\qquad (4) \\]

Using the quadratic formula, we find two solutions for $x$:

{check}

Therefore, the 2 pairs of solutions for these simultaneous equations are

\\[ (x_1,y_1) = (\\var{x1dp},\\var{y1dp}) \\] and \\[ (x_2,y_2) = (\\var{x2dp},\\var{y2dp}). \\]

Use this link to find some resources which will help you revise this topic.

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\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{ (2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text2": {"name": "text2", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text3": {"name": "text3", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\text{ (2 d.p.)}\\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\text{(2 d.p.)} \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\, \\\\text{(2 d.p.)} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "text": {"name": "text", "group": "Ungrouped variables", "definition": "\"

\\\\[ x_1 = \\\\var{x1dp} \\\\, \\\\quad \\\\text{and} \\\\quad x_2=\\\\var{x2dp}\\\\]

\\n

To find the corresponding $y$-values, we can plug these solutions for $x$ back into equation (3), which gives:

\\n

\\\\[ y_1 = \\\\var{y1dp} \\\\, \\\\quad \\\\text{and} \\\\quad y_2=\\\\var{y2dp} \\\\]

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$(x_1,y_1)=$[[0]]

$(x_2,y_2)=$[[1]]

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Solve linear equations with unkowns on both sides. Including brackets and fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To solve an equation like

$\\displaystyle{\\frac{\\var{a}}{y}=\\frac{\\var{b}}{y+\\var{c}}},$

the first thing to deal with is that the unknown ($y$) that you are trying to find is in the denominator (on the bottom) of the fractions. In order to do that you first times by $y$ on both sides and $(y+\\var{c})$ on both sides leading to

\\[\\var{a}(y+\\var{c}) = \\var{b}y.\\]

From here, multiply out the brackets,

\\[\\var{a}y +\\var{a*c} = \\var{b}y.\\]

Now collect the $y$-terms on one side and the numbers on the other,

\\[\\var{a-b}y=\\var{-a*c}.\\]

Finally divide by the coefficient of $y$,

\\[y=\\frac{\\var{-a*c}}{\\var{a-b}}.\\]

Use this link to find resources to help you revise how to solve linear equations

Solve $\\displaystyle{\\frac{\\var{a}}{y}=\\frac{\\var{b}}{y+\\var{c}}}$.

$y=$ [[0]] (Give your answer as a fraction)

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Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Factorise the following quadratic equation.

", "advice": "

Quadratic equations of the form

\\[x^2+bx+c=0\\]

can be factorised to create an equation of the form

\\[(x+m)(x+n)=0\\text{.}\\]

When we expand a factorised quadratic expression we obtain

\\[(x+m)(x+n)=x^2+(m+n)x+(m \\times n)\\text{.}\\]

To factorise an equation of the form $x^2+bx+c$, we need to find two numbers which add together to make $b$, and multiply together to make $c$.

We need to find two values that add together to make $\\var{v3+v4}$ and multiply together to make $\\var{v3*v4}$.

\\[\\begin{align}
\\var{v3} \\times \\var{v4}&=\\var{v3*v4}\\\\
\\var{v3}+\\var{v4}&=\\var{v3+v4}\\\\
\\end{align} \\]

So the factorised form of the equation is

\\[\\simplify{(x+{v3})(x+{v4})}=0\\text{.}\\]

Use this link to find some resources which will help you revise this topic

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$\\simplify{x^2+{v3+v4}x+{v3*v4}}=0$

[[0]] $=0$

The answer is a comma-separated list of numbers.

The list is marked correct if each number occurs the same number of times as in the expected answer, and no extra numbers are present.

You can optionally treat the answer as a set, so the number of occurrences doesn't matter, only whether each number is included or not.

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Is every number in the student's list valid?

", "definition": "if(all(map(not isnan(x),x,interpreted_answer)),\n true,\n let(index,filter(isnan(interpreted_answer[x]),x,0..len(interpreted_answer)-1)[0], wrong, bits[index],\n warn(wrong+\" is not a valid number\");\n fail(wrong+\" is not a valid number.\")\n )\n )"}, {"name": "is_sorted", "description": "

Are the student's answers in ascending order?

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Is each number in the expected answer present in the student's list the correct number of times?

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True if the student's list doesn't contain any numbers that aren't in the expected answer.

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Should the answer be considered as a set, so the number of times an element occurs doesn't matter?

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Numbers included in the student's answer that are not in the expected list.

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Solving a quadratic equation via factorisation (or otherwise) with the $x^2$-term having a coefficient of 1.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve the following quadratic equation by factorisation or otherwise:

\\[ \\simplify[unitFactor]{x^2+{b}x+{c}=0} \\]

", "advice": "

To solve a quadratic equation of the form \\[ x^2+bx+c=0\\] by factorisation, we want to factorise the equation into the form \\[(x+p)(x+q)=0,\\] where $p+q=b$ and $p \\times q = c$.

Hence, for the equation \\[\\simplify{x^2+{b}x+{c}=0}, \\]

this can be factorised to \\[\\simplify{(x+{p})(x+{q})=0}.\\] This equation is satisfied when either \\[\\simplify{x+{p}=0} \\quad \\text{or} \\quad \\simplify{x+{q}=0}, \\] which implies the solutions to this quadratic equation are \\[ \\simplify{x={-p}} \\quad \\text{and} \\quad \\simplify{x={-q}} .\\]

Use this link to find resources to help you revise how to solve quadratic equations by factorising the expression.

$x= $[[0]]

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Rearrange expressions in the form $ax^2+bx+c$ to $a(x+b)^2+c$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

We can rewrite quadratic equations given in the form $ax^2+bx+c$ as a square plus another term - this is called \"completing the square\".

This can be useful when it isn't obvious how to fully factorise a quadratic equation.

Rewrite the following expressions in the form \\[(x+b)^2-c\\]

", "advice": "

Completing the square works by noticing that

\\[ (x+a)^2 = x^2 + 2ax + a^2 \\]

So when we see an expression of the form $x^2 + 2ax$, we can rewrite it as $(x+a)^2-a^2$.

Replace $x^2+\\var{evens2}x$ with $(x+\\var{evens2/2})^2 - \\var{evens2/2}^2$. Remember to keep the $\\var{evens2-evens1}$ term on the end!

\\begin{align}
\\simplify[basic]{ x^2 + {evens2}x + {evens2-evens1}} &= \\simplify[basic]{ (x+{evens2/2})^2 - {evens2/2}^2 + {evens2-evens1} } \\\\
&= \\simplify[basic]{ (x+{evens2/2})^2 + {evens2-evens1 - evens2^2/4} }
\\end{align}

Use this link to find some resources which will help you revise this topic.

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$\\simplify {x^2+ {evens2}x +{evens2-evens1}} =$ [[0]]

It doesn't look like you've completed the square.

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "AS09 Quadratics - factorise (a not 1)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": ["coefficient of x^2 greater than 1", "factorisation", "Factorisation", "factorising", "factorising quadratic equations", "Factorising quadratic equations", "factorising quadratic equations with x^2 coefficients greater than 1", "taxonomy"], "metadata": {"description": "

Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\\var{a*c}$.

You may have to test a a few different options before you find one that works. In this case $a$ and $c$ are $\\var{a}$ and $\\var{c}$.

This means our factorised equation must take the form

\\[(\\var{a}x+b)(\\var{c}x+d)=0\\text{.}\\]

This expands to

\\[ \\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b} \\]

So we must find two numbers which add together to make $\\var{a*d+b*c}$, and multiply together to make $\\var{b*d}$.

Therefore $b$ and $d$ must satisfy

\\begin{align}
b \\times d &=\\var{b*d}\\\\
\\simplify{{a}d+{c}b} &= \\var{a*d+b*c}\\text{.}
\\end{align}

$b = \\var{b}$ and $d = \\var{d}$ satisfy these equations:

\\begin{align}
\\var{b} \\times \\var{d} &=\\var{b*d}\\\\
\\simplify[]{ {a}*{d} + {b}*{c} } &= \\var{a*d+b*c}
\\end{align}

So the factorised form of the equation is

\\[ \\simplify{({a}x+{b})({c}x+{d}) = 0} \\text{.}\\]

$\\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\\var{a}x+\\var{b} = 0$ or $\\var{c}x+ \\var{d} = 0$.

So the roots of the equation are $\\var[fractionnumbers]{-b/a}$ and $\\var[fractionnumbers]{-d/c}$.

Use this link to find some resources which will help you revise this topic.

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$b$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "last q", "definition": "random(2..8 except a)", "description": "

$c$ in $(ax+b)(cx+d)$

", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "last q", "definition": "random(2..3)", "description": "

$a$ in $(ax+b)(cx+d)$

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The roots of the equation

", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "last q", "definition": "random(-8..8 except 0)", "description": "

$d$ in $(ax+b)(cx+d)$

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Solve the following equation by factorisation to find $x$.

$\\simplify{{a*c}x^2+{a*d+b*c}x+{b*d}=0}\\text{.}$

Input your answers in ascending order.

$x=$ [[0]]

$x=$ [[1]]

Factorising a quadratic expression of the form $a^2x^2-b^2$ to $(ax+b)(ax-b)$, using the difference of two squares formula.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Factorise the following quadratic expression:

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2}} \\]

", "advice": "

For a quadratic expression of this form we can make use of the Difference of Squares formula, which states that \\[a^2-b^2 = (a+b)(a-b).\\]

Therefore,

\\[ \\simplify[unitFactor]{{a^2}x^2-{c^2} = ({a}x+{c})({a}x-{c})}. \\]

Use this link to find resources to help you revise how to factorise a quadratic equation using the difference of two squares formula.

Find the derivative of a function of the form $y=ax^b$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify{{a}x^{b}}$.

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=kx^n \\] has a derivative \\[ \\frac{df}{dx} = knx^{n-1}. \\]

So, for the function \\[ y=\\simplify{{a}x^{b}}, \\] the derivative is \\begin{split}\\frac{dy}{dx} &= (\\var{a}\\times\\var{b})x^{\\var{b}-1},\\\\ \\\\&= \\simplify{{a*b}x^{{b}-1}}.\\end{split}

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Differentiate a polynomial expression involving coefficients and, negative and fractional indices.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of $y=\\simplify[unitFactor, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}}$.

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=kx^n \\] has a derivative \\[ \\frac{df}{dx} = knx^{n-1}. \\]

Additionally, the derivative of the sum or difference of two or more functions is equal to the sum or difference of the derivatives of each function: \\[ \\frac{d}{dx}(f(x)\\pm g(x)) = \\frac{df}{dx} \\pm \\frac{dg}{dx}.\\]

{advice}

Use this link to find some resources which will help you revise this topic.

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionb": {"name": "solutionb", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionc": {"name": "solutionc", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutiond": {"name": "solutiond", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

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$\\frac{dy}{dx}=$[[0]]

Calculating the derivative of a function of the form $a \\ln(bx)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[unitFactor]{{a}*ln({a_1}*x^2+{a_2}*x+{a_3})}.$

", "advice": "

From the Table of Derivatives and the chain rule we see that a function of the form \\[ f(x)=a \\ln(g(x)) \\] has a derivative \\[\\frac{df}{dx}=\\frac{g'(x)}{g(x)}.\\]

In this case $g(x)=\\var{a_1}x^2+\\var{a_2}x+\\var{a_3}$ so

\\[g'(x)=\\var{2*a_1}x+\\var{a_2}\\]

Therefore, the function \\[ \\simplify[unitFactor]{y={a}ln({a_1}*x^2+{a_2}*x+{a_3})}\\] has a derivative \\[(\\var{a*a_1*2}x+\\var{a*a_2})/(\\var{a_1}x^2+\\var{a_2}x+\\var{a_3})\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Calculating the derivative of an exponential function of the form $ae^{bx}$, using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{{a}*e^({b}x)}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a e^{kx} \\] has a derivative \\[ak e^{kx}.\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*e^({b}x)}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})e^{\\simplify[unitFactor]{{b}x}}\\\\ &= \\simplify[unitFactor]{{a*b}e^({b}x)}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\cos(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of $y=\\simplify[unitFactor]{{a}cos({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\cos(kx+c) \\] has a derivative \\[-ak \\sin (kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*cos({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=-(\\var{a}\\times \\var{b})\\sin(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{-a*b}sin({b}x+{c})}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\sin(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}sin({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\sin(kx+c) \\] has a derivative \\[ak \\cos (kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*sin({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\cos(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}cos({b}x+{c})}.\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Find the derivative of a function of the form $y=a \\tan(bx+c)$ using a table of derivatives.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Using the Table of Derivatives, calculate the derivative of $y=\\simplify[unitFactor]{{a}tan({b}x+{c})}.$

", "advice": "

From the Table of Derivatives we see that a function of the form \\[ f(x)=a \\tan(kx+c) \\] has a derivative \\[ak \\sec^2(kx+c).\\]

Therefore, the function \\[y=\\simplify[unitFactor]{{a}*tan({b}x+{c})}\\] has a derivative\\[ \\begin{split} \\frac{dy}{dx} &=(\\var{a}\\times \\var{b})\\sec^2(\\simplify[unitFactor]{{b}x+{c}})\\\\ &= \\simplify[unitFactor]{{a*b}}\\sec^2(\\simplify[unitFactor]{{b}x+{c}}).\\end{split}\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

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Finding the stationary points of a cubic equation and determining their nature.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given the function \\[ \\simplify{y={a}x^3+{b}x^2+{c}x+{d}} ,\\] find its stationary points and determine their nature.

", "advice": "

To find the stationary points of the function, we must solve $\\tfrac{dy}{dx}=0$ for $x$. For the function $\\simplify{y={a}x^3+{b}x^2+{c}x+{d}}$,

\\[ \\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}. \\]

Setting $\\frac{dy}{dx}=0$ and solving for $x$:

\\[ \\simplify{{3a}x^2+{2b}x+{c}} =0 \\\\ \\\\ \\implies x=\\var{solx1dp} \\var{x1} \\text{ and } x=\\var{solx2dp} \\var{x2}. \\]

Hence, the function has two stationary points at $x=\\var{solx1dp}$ and $x=\\var{solx2dp}$. To find the corresponding $y$-coordinates, we want to plug these values back into the initial equation.

When $x=\\var{solx1dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx1dp})^3+{b}*({solx1dp})^2+{c}*({solx1dp})+{d}} \\\\ &\\,=\\simplify{{soly1dp}} \\var{y1}. \\end{split} \\]

When $x=\\var{solx2dp}$,

\\[ \\begin{split} y &\\,= \\simplify[unitFactor,!cancelTerms]{{a}*({solx2dp})^3+{b}*({solx2dp})^2+{c}*({solx2dp})+{d}} \\\\ &\\,=\\simplify{{soly2dp}} \\var{y2}. \\end{split} \\]

Therefore, the stationary points of $y=\\simplify{{a}x^3+{b}x^2+{c}x+{d}}$ are

\\[ (\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}}) \\, , \\,(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}}). \\]

Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$-value of the stationary points.

Recall:

If $\\tfrac{d^2y}{dx^2}<0$ the stationary point is a maximum;
If $\\tfrac{d^2y}{dx^2}>0$ the stationary point is a minimum;
If $\\tfrac{d^2y}{dx^2}=0$ the stationary point is a point of inflection.

To calculate $\\tfrac{d^2y}{dx^2}$, we want to differentiate $\\tfrac{dy}{dx}$ again with respect to $x$:

\\[ \\begin{split} &\\frac{dy}{dx} = \\simplify{{3a}x^2+{2b}x+{c}}, \\\\ \\\\\\implies &\\frac{d^2y}{dx^2} = \\simplify{{6a}x+{2b}}. \\end{split}\\]

For $(\\simplify{{solx1dp}},\\, \\simplify{{soly1dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check}}$, so it is a minimum.

For $(\\simplify{{solx2dp}},\\, \\simplify{{soly2dp}})$, $\\frac{d^2y}{dx^2} = \\simplify{{check2}}$, so it is a maximum.

Use this link to find some resources which will help you revise this topic.

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There is a minimum point at ([[0]], [[1]]) and a maximum point at ([[2]] , [[3]]).

(Give the coordinates of the stationary points to 2 decimal places where necessary.)

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Calculating the derivative of a function of the form $\\sin(ax^m+bx^n)$ using the chain rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the derivative of $y=\\simplify[all]{sin({a}*x^{n}+{b}*x^{m})}$.

", "advice": "

If we have a function of the form $y=f(g(x))$, sometimes described as a function of a function, to calculate its derivative we need to use the chain rule:

\\[ \\frac{dy}{dx} = \\frac{du}{dx} \\times \\frac{dy}{du}.\\]

This can be split up into steps:

Let $u=g(x)$;
Rewrite $y$ in terms of $u$, such that $y=f(u)$;
Calculate $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Write $\\frac{dy}{dx}$ as a product of $\\frac{du}{dx}$ and $\\frac{dy}{du}$;
Make sure $\\frac{dy}{dx}$ is only in terms of $x$. Ensure any $u$ terms have been replaced using the initial substitution.

Following this process, we must first identify $g(x)$. Since the function is of the form $y=f(g(x))$, we are looking for the 'inner' function.

So, for $y=\\simplify[all,fractionNumbers]{sin({a}*x^{n}+{b}*x^{m})}$, \\[g(x)=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}.\\]

If we now set $u=g(x)$, we can rewrite $y$ in terms of $u$ such that $y=f(u)$:

\\[y=\\simplify[all, fractionNumbers,unitFactor]{sin(u)}.\\]

Next, we calculate the two derivatives $\\frac{du}{dx}$ and $\\frac{dy}{du}$:

\\[\\frac{du}{dx}=\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}, \\quad \\frac{dy}{du}=\\simplify[all, fractionNumbers, unitFactor]{cos(u)}.\\]

Plugging these into the chain rule:

\\[ \\begin{split} \\frac{dy}{dx} &= \\frac{du}{dx} \\times \\frac{dy}{du}, \\\\&=(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}}) \\times\\simplify[all, fractionNumbers, unitFactor]{cos(u)}. \\end{split} \\]

Finally, we need to express $\\frac{dy}{dx}$ only in terms of $x$, so we must replace the $u$ term using the initial substitution $u=\\simplify[all, fractionNumbers, unitFactor]{{a}*x^{n}+{b}*x^{m}}$:

\\[ \\frac{dy}{dx} =(\\simplify[all,fractionNumbers]{{a*n}x^{n-1}+{b*m}x^{m-1}})\\simplify[all, fractionNumbers, unitFactor]{cos({a}*x^{n}+{b}*x^{m})}.\\]

Use this link to find some resources which will help you revise this topic.

$\\frac{dy}{dx}=$[[0]]

Calculating the derivative a function of the form $ax^n \\sin(bx)$ using the product rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of \\[ \\simplify{y={a}x^{n} sin({b}x)}. \\]

", "advice": "

If we have a function of the form $y=u(x)v(x)$, to calculate its derivative we need to use the product rule:

\\[ \\dfrac{dy}{dx} = u(x) \\times \\dfrac{dv}{dx} + v(x) \\times\\dfrac{du}{dx}.\\]

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$;
Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
Substitute these into the formula for the product rule to obtain an expression for $\\tfrac{dy}{dx}$;
Simplify $\\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$.

As \\[ \\simplify{y={a}x^{n} sin({b}x)}, \\]

let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{sin({b}x)}.\\]

Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

\\[ \\dfrac{du}{dx} = \\simplify{{a*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}cos({b}x)}.\\]

Substituting these results into the product rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{du}{dx}\\times v(x) + u(x) \\times\\dfrac{dv}{dx} \\\\ &\\,=\\simplify{{a*n}x^{n-1}} \\times\\simplify{sin({b}x)} +\\simplify{{a}x^{n}} \\times \\simplify{{b}cos({b}x)}. \\end{split}\\]

Simplifying,

\\[\\dfrac{dy}{dx} = \\simplify{{n*a}x^{n-1}sin({b}x) + {a*b}x^{n}cos({b}x)}. \\]

Use this link to find some resources which will help you revise this topic

$\\dfrac{dy}{dx}=$[[0]]

Calculating the derivative of a function of the form $\\frac{ax^n}{bx+c}$ using the quotient rule.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the derivative of \\[ \\simplify{y={a}x^{n}/({b}x+{c})}. \\]

", "advice": "

If we have a function of the form $y=\\tfrac{u(x)}{v(x)}$, to calculate its derivative we need to use the quotient rule:

\\[ \\dfrac{dy}{dx} = \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2}\\,.\\]

This can be split up into steps:

Identify the functions $u(x)$ and $v(x)$;
Calculate their derivatives $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$;
Substitute these into the formula for the quotient rule to obtain an expression for $\\tfrac{dy}{dx}$;
Simplify $\\tfrac{dy}{dx}$ where possible.

Following this process, we must first identify $u(x)$ and $v(x)$.

As \\[ \\simplify{y={a}x^{n}/({b}x+{c})}, \\]

let \\[ u(x) = \\simplify{{a}x^{n}} \\quad \\text{and} \\quad v(x)=\\simplify{{b}x+{c}}.\\]

Next, we need to find the derivatives, $\\tfrac{du}{dx}$ and $\\tfrac{dv}{dx}$:

\\[ \\dfrac{du}{dx} = \\simplify{{a*n}x^{n-1}}\\quad \\text{and} \\quad\\dfrac{dv}{dx}=\\simplify{{b}}.\\]

Substituting these results into the quotient rule formula we can obtain an expression for $\\tfrac{dy}{dx}$:

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,= \\dfrac{v(x) \\times \\frac{du}{dx} - u(x) \\times\\frac{dv}{dx}}{[v(x)]^2} \\\\ \\\\&\\,=\\dfrac{(\\simplify{{b}x+{c}}) \\times\\simplify{{a*n}x^{n-1}} - \\simplify{{a}x^{n}} \\times \\simplify{{b}}}{\\simplify{({b}x+{c})^2}}. \\end{split}\\]

Simplifying,

\\[ \\begin{split} \\dfrac{dy}{dx} &\\,=\\dfrac{(\\simplify{{b}x+{c}})\\simplify{{a*n}x^{n-1}} - \\simplify{{b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\&\\,=\\dfrac{\\simplify[all,!cancelTerms]{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}}\\\\ \\\\ &\\,=\\dfrac{\\simplify{{b*a*n}x^{n}+{c*a*n}x^{n-1} - {b*a}x^{n}}}{\\simplify{({b}x+{c})^2}} \\\\ \\\\ &\\,=\\dfrac{\\simplify{{simp}x^{n-1}({(b*a*n-b*a)/simp}x+{c*a*n/simp})}}{\\simplify{({b}x+{c})^2}} \\end{split} \\]

Use this link to find some resources which will help you revise this topic.

$\\dfrac{dy}{dx}=$[[0]]

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Calculating the integral of a function of the form $a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3}$ using a table of integrals.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the integral of $f(x)=\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4}}$.

", "advice": "

From the Table of Integrals we see that a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int x^n dx = \\frac{x^{n+1}}{n+1}+ c,\\]

and \\[\\int kf(x) dx = k \\int f(x) dx.\\]

Additionally, the integral of the sum or difference of two or more functions is equal to the sum or difference of the integrals of each function: \\[ \\int(f(x)\\pm g(x))\\, dx = \\int f(x)\\, dx \\pm \\int g(x) \\, dx.\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,unitPower]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4}},\\]

the integral is

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{int({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}+{a_4},x)} &\\,= \\simplify{{a_1}int(x^{b_1},x)+{a_2}int(x^{b_2},x)+{a_3}int(x^{b_3},x)+int({a_4},x)} \\\\&\\,= \\simplify[all,fractionNumbers]{({a_1}*x^{b_1+1})/{b_1+1}+({a_2}*x^{b_2+1})/{b_2+1}+({a_3}*x^{b_3+1})/{b_3+1}+{a_4}x}+c.\\end{split} \\]

Note: You only need to put one $c$ term here, you do not need to put a separate constant term for each calculation.

Use this link to find some resources which will help you revise this topic.

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionb": {"name": "solutionb", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} +(\\\\var[fractionNumbers]{a_3}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutionc": {"name": "solutionc", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} +(\\\\var[fractionNumbers]{a_2}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

\"", "description": "", "templateType": "long string", "can_override": false}, "solutiond": {"name": "solutiond", "group": "Ungrouped variables", "definition": "\"

So, for the function \\\\[y=\\\\simplify[all, fractionNumbers]{{a_1}x^{b_1}+{a_2}x^{b_2}+{a_3}x^{b_3}} \\\\] the derivative is \\\\begin{split}\\\\frac{dy}{dx} &= (\\\\var[fractionNumbers]{a_1}\\\\times\\\\var[fractionNumbers]{b_1})x^{\\\\var[fractionNumbers]{b_1}-1} -(\\\\var[fractionNumbers]{abs(a_2)}\\\\times\\\\var[fractionNumbers]{b_2})x^{\\\\var[fractionNumbers]{b_2}-1} -(\\\\var[fractionNumbers]{abs(a_3)}\\\\times\\\\var[fractionNumbers]{b_3})x^{\\\\var[fractionNumbers]{b_3}-1},\\\\\\\\ \\\\\\\\&= \\\\simplify[all, fractionNumbers]{{a_1*b_1}x^{b_1-1} +{a_2*b_2}x^{b_2-1} +{a_3*b_3}x^{b_3-1}}.\\\\end{split}

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[[0]]

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{a_1}x^{{b_1}+1}/{b_1+1}+{a_2}x^{{b_2}+1}/{b_2+1}+{a_3}x^{{b_3}+1}/{b_3+1}+{a_4}x", "answerSimplification": "all, fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{a_1}x^{{b_1}+1}/{b_1+1}+{a_2}x^{{b_2}+1}/{b_2+1}+{a_3}x^{{b_3}+1}/{b_3+1}+{a_4}x+c", "answerSimplification": "all, fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI02 Definite integration", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}], "tags": [], "metadata": {"description": "

Calculating the definite integral $\\int_{n_1}^{n_2}a_1x^{b_1}+a_2x^{b_2}+a_3x^{b_3} dx$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Evaluate \\[ \\int_{\\var{n_1}}^{\\var{n_2}}\\simplify[unitFactor, unitPower, fractionNumbers]{{a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3}} \\,dx.\\]

", "advice": "

Integrating a function of the form \\[ f(x)=x^n \\] has the integral \\[ \\int_a^b x^n dx = \\left[\\frac{x^{n+1}}{n+1}\\right]_a^b,\\]

and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

Therefore,

\\[ \\begin{split}\\simplify[unitFactor,unitPower]{defint({a_1}*x^{b_1}+{a_2}*x^{b_2}+{a_3}*x^{b_3},x,{n_1},{n_2})} &\\,= \\simplify{{a_1}defint(x^{b_1},x,{n_1},{n_2})+{a_2}defint(x^{b_2},x,{n_1},{n_2})+{a_3}defint(x^{b_3},x,{n_1},{n_2})} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers]{{a_1}x^{b_1+1}/{b_1+1}+{a_2}x^{b_2+1}/{b_2+1}+{a_3}x^{b_3+1}/{b_3+1}}\\right]_\\var{n_1}^\\var{n_2} \\\\ &\\,= \\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_2^(b_1+1)}/{b_1+1}+{a_2*n_2^(b_2+1)}/{b_2+1}+{a_3*n_2^(b_3+1)}/{b_3+1}}\\right] -\\left[\\simplify[all,fractionNumbers,!collectNumbers]{{a_1*n_1^(b_1+1)}/{b_1+1}+{a_2*n_1^(b_2+1)}/{b_2+1}+{a_3*n_1^(b_3+1)}/{b_3+1}}\\right] \\\\ &\\,= \\simplify[!collectNumbers]{{eval2a}-{eval1a}} \\\\ &\\,=\\var{sol1} \\end{split} \\]

Use this link to find some resources on areas under curves which will help you revise this topic.

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[[0]] (Give answers to 2 decimal places where necessary)

Calculating the integral of a function of the form $\\frac{c}{(x+a)(x+b)}$ using partial fractions.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral

\\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)} .\\]

", "advice": "

In order to integrate the function \\[ \\simplify{int({c}/((x^2+{aPlusb}x+{ab})),x)}, \\] we want to rewrite it in terms of its partial fractions.

First we need to factorise the denominator so we have

\\[ \\simplify{{c}/((x+{a})(x+{b}))}. \\]

Now to write this as a partial fraction, we want to set the function equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:

\\[ \\simplify{{c}/((x+{a})(x+{b}))} = \\simplify{A/(x+{a}) + B/(x+{b})}.\\]

To find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives

\\[ \\simplify{{c}=A(x+{b})+B(x+{a})}.\\]

To find $A$, we can eliminate $B$ by setting $\\simplify{x={-a}}$:

\\[ \\simplify{{c}=A{b-a}} \\implies \\simplify[fractionNumbers]{A={c/(b-a)}}.\\]

Similarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:

\\[ \\simplify{{c}=B{a-b}} \\implies \\simplify[fractionNumbers]{B={c/(a-b)}}.\\]

Therefore,

{check1}

and

{check2}

Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{Asol}/(x+{a})+{Bsol}/(x+{b})},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Sol2": {"name": "Sol2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b}))},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "int1": {"name": "int1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({Asol}/(x+{a})+{Bsol}/(x+{b}),x)}\\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "int2": {"name": "int2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b})),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)} \\\\\\\\ \\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

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[[0]]

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": true, "answer": "{Asol} ln (x+{a})+{Bsol} ln (x+{b}) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b}))", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI04 Integration - trig identities", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Using the various versions of $\\cos{2x}$ identity to integrate $\\sin^2{x}$ and $\\cos^2{x}$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Integrate $f(x)=\\var{Func}$.

", "advice": "

We can't integrate $\\var{Coeff}\\sin^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=1-2\\sin^2(x)$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\sin^2(x)=\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x).\\]

From the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{-Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{-Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2-\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x-{Coeff/4} sin(2x)+c}\\]

We can't integrate $\\var{Coeff}\\cos^2(x)$ directly so first we have to use the double angle formula $\\cos(2x)=2\\cos^2(x)-1$. We re-arrange using the double angle formula to give us,

\\[\\var{Coeff}\\cos^2(x)=\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x).\\]

From the Table of Integrals we see that a function of the form \\[ f(x)= \\cos(nx) \\] has the integral \\[ \\int \\cos(nx) dx = \\frac{1}{n}\\sin(nx)+c\\]

So, for the function

\\[f(x)=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)},\\]

the integral is

\\[ \\begin{split} \\int\\simplify[unitFactor,fractionNumbers]{{Coeff/2}cos(2x)} dx \\,= \\simplify[unitFactor,fractionNumbers]{{Coeff/2}int(cos(2x),x)} &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}(1/2 sin({2}x))} +c, \\\\ &\\,=\\simplify[unitFactor,fractionNumbers]{{Coeff/4} sin(2x)+c}. \\end{split} \\]

The integral of $\\frac{\\var{Coeff}}2$ is

\\[\\int\\frac{\\var{Coeff}}2dx=\\frac{\\var{Coeff}}2x+c,\\]

so combining these our final answer is

\\[\\int\\frac{\\var{Coeff}}2+\\frac{\\var{Coeff}}2\\cos(2x)dx=\\simplify[unitFactor,fractionNumbers]{{Coeff/2}x+{Coeff/4} sin(2x)+c}\\]

Use this link to find some resources which will help you revise this topic.

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It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "1/2{Coeff}*(x+{OneIfCosMinusOneIfSine}/2*sin(2x))", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "1/2{Coeff}*(x+{OneIfCosMinusOneIfSine}/2*sin(2x))+c", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI05 Integration - Substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculating the integral of a function of the form $\\frac{nx^{n-1}}{x^n+a}$ using integration by substitution.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate \\[ \\simplify[all]{int(({n}x^{n-1})/(x^{n}+{a}),x)}\\]

by using the substitution \\[ \\simplify[all]{u=x^{n}+{a}}.\\]

", "advice": "

Since this integral is of the form \\[ \\int g'(x)f(g(x))\\,dx,\\] we can use the method of substitution to calculate the solution.

Firstly, we must make a change of variables from $x$ to $u$, where $u$ is equal to the 'inner' function $g(x)$.

So, for \\[\\simplify[fractionNumbers]{int(({n}x^{n-1})/((x^{n}+{a})),x)}\\]

let $\\color{red}{u=\\simplify[fractionNumbers]{x^{n}+{a}}}.$

Now, we need to calculate the differential, $du$, where \\[ du = \\left(\\frac{du}{dx}\\right)dx. \\]

Differentiating $u$ with respect to $x$:

\\[ \\frac{du}{dx}= \\simplify[fractionNumbers]{{n}x^{n-1}}.\\]

Therefore, \\[ \\color{blue}{du = \\simplify[fractionNumbers]{{n}x^{n-1}}\\, dx}.\\]

We can now rewrite the original integral in terms of $u$:

\\[ \\int \\frac{\\color{blue}{\\simplify{{n}x^{n-1}}}}{\\color{red}{\\simplify{x^{n}+{a}}}}\\color{blue}{\\text{d}x} = \\int \\frac{1}{\\color{red}{u}}\\color{blue}{\\text{d}u}.\\]

(Note: It is important to see that both the function we are integrating, and the variable we are integrating with respect to, has changed.)

\\[ \\simplify[fractionNumbers]{int(1/u,u) = ln(abs(u)) + c}.\\]

Finally, we must rewrite our solution back in terms of the original variable $x$:

\\[ \\simplify[fractionNumbers]{ln(abs(u)) + c = ln(abs(x^{n}+{a})) + c}.\\]

Use this link to find some resources which will help you revise this topic.

[[0]]

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Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": false, "answer": "ln(x^{n}+{a})+c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alternative using \"+k\"", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "ln(abs(x^{n}+{a})) + k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Alternative using brackets and \"+k\"", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

Technically we should use the absolute value symbols for the logs. This can be done in NUMBAS by using \"abs(*function*)\".

", "useAlternativeFeedback": false, "answer": "ln(x^{n}+{a})+k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "ln(abs(x^{n}+{a}))", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "ln(abs(x^{n}+{a}))+c", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": "0.01", "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "CI06 Integration - Parts", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculating the integral of a function of the form $ax^2 \\cos(bx)$ using integration by parts.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the integral \\[ \\simplify{int({a}x^2 cos({b}x),x)}\\]

", "advice": "

If we have a function of $x$ which is the product of two functions of $x$, to integrate such a function it is often necessary to use Integration by Parts. The formula for Integration by Parts is:

\\[ \\int u(x) \\frac{dv}{dx} dx = u(x)v(x) - \\int v(x) \\frac{du}{dx} dx.\\]

Using this method can be broken down into steps:

Identify $u(x)$ and $\\tfrac{dv}{dx}$ (The function you pick for each is important, in general you want $u(x)$ to become simpler when differentiating it, and you must be able to integrate $\\tfrac{dv}{dx}$ to find $v(x)$);
Calculate $\\tfrac{du}{dx}$ and $v(x)$;
Put the functions $u(x)$, $v(x)$, and their derivatives into the Integration by Parts formula;
Calculate the integral $\\int v(x) \\tfrac{du}{dx} dx$ (This may require you to use Integration by Parts again, this is OK!);
Simplify your answer where possible and don't forget to add the constant of integration.

For the integral

\\[ \\simplify{int({a}x^2 cos({b}x),x)},\\]

we must first identify $u(x)$ and $\\tfrac{dv}{dx}$. In this case, let \\[ u(x)=\\simplify{{a}x^2},\\quad \\frac{dv}{dx}= \\simplify{cos({b}x)}. \\]

Next, we need to calculate $\\tfrac{du}{dx}$ and $v(x)$:

\\[ \\begin{split} u(x) = \\var{a}x^2 \\quad &\\implies \\frac{du}{dx} = \\simplify{{2a}x}; \\\\ \\frac{dv}{dx} = \\cos(\\var{b}x) &\\implies v(x) = \\simplify[fractionNumbers]{1/{b} sin({b}x)}. \\end{split} \\]

Plugging these 4 terms into the integration by parts formula:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) - int({2a/b}x sin({b}x),x)}, \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x sin({b}x),x)}.\\end{split} \\]

Since the integral on the right-hand side is still the product of two functions of $x$, we need to use integration by parts again.

So, for

\\[ \\simplify{int(x sin({b}x),x)}, \\]

Let $u=x$ and $\\tfrac{dv}{dx} = \\sin(\\var{b}x)$. Therefore, $\\tfrac{du}{dx}=1$ and $v(x)=\\simplify{-1/{b} cos({b}x)}$.

Hence,

\\[ \\begin{split} \\simplify{int(x sin({b}x),x)} &\\,= \\simplify{-1/{b}x cos({b}x)- int(-1/{b} cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify{-1/{b}x cos({b}x)+1/{b^2}sin({b}x)}. \\end{split}\\]

Plugging this back into the original calculation:

\\[ \\begin{split} \\simplify{int({a}x^2 cos({b}x),x)} &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}int(x cos({b}x),x)} \\\\ \\\\ &\\,= \\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) -{2a/b}[-1/{b}x cos({b}x)+1/{b^2}sin({b}x)]} \\\\ \\\\ &\\,=\\simplify[fractionNumbers]{{a/b}x^2 sin({b}x) +{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)} + c.\\end{split} \\]

Use this link to find some resources which will help you revise this topic.

[[0]]

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Correct Answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": true, "customName": "Alt constant +k", "marks": "1", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)+k", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "k", "value": ""}, {"name": "x", "value": ""}]}, {"type": "jme", "useCustomName": true, "customName": "Forgotten constant", "marks": "0.5", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "

It looks like you forgot to include the integration constant. You should always remember the \"+C\" when doing an indefinite integral.

", "useAlternativeFeedback": false, "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)", "answerSimplification": "all,!collectLikeFractions,fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "{a/b}x^2 sin({b}x)+{2a/b^2}x cos({b}x)-{2a/b^3}sin({b}x)+c", "answerSimplification": "fractionNumbers, basic", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": [{"name": "c", "value": ""}, {"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Geometry - Angles and Trigonometry", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], []], "questions": [{"name": "GA01 Angles in triangles", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Find the missing angle in a triangle using the fact that the angles add to 180 degrees.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the size of the missing angle $C$? All angles are measured in degrees.

{geogebra_applet('https://www.geogebra.org/m/akwnfkfr',[a: a, b: b])}

", "advice": "

Recall that the angles in a triangle add up to $180^{\\circ}$.

We can add together two angles we know and subtract the result from $180$ to find the size of our missing angle,

\\[ \\begin{split} 180 - (\\var{a} + \\var{b}) &\\, = 180 - (\\var{a+b}) \\\\ &\\, = \\var{180-(a+b)}^{\\circ}. \\end{split} \\]

Use this link to find resources to help you revise properties of triangles.

The size of the missing angle is [[0]]$^{\\circ}$.

Draws a triangle based on 3 side lengths and randomises asking for hypotenuse or not.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{statement}

Find $x$.

", "advice": "

Only round your final answer to 1 decimal place.

{advice}

Use this link to find some resources to help you revise how to use pythagoras' theorem.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"setup": {"name": "setup", "group": "Varying q and advice", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "answerside": {"name": "answerside", "group": "Varying q and advice", "definition": "sh1", "description": "", "templateType": "anything", "can_override": false}, "answerhyp": {"name": "answerhyp", "group": "Varying q and advice", "definition": "hyp", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Varying q and advice", "definition": "if(setup=1,answerside,answerhyp)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "Varying q and advice", "definition": "if(setup=1,advice1,advice2)", "description": "", "templateType": "anything", "can_override": false}, "advice2": {"name": "advice2", "group": "Varying q and advice", "definition": "\"

Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = \\\\var{sh1}$, $b =\\\\var{sh2}$ and $c = x$

\\n

So

\\\\[\\\\var{sh1}^2 + \\\\var{sh2}^2 = x^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{sh1^2} + \\\\var{sh2^2}\\\\]

\\\\[x = \\\\sqrt{\\\\var{sh1^2} + \\\\var{sh2^2}}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{sh1^2+sh2^2}}\\\\]

$x = \\\\var{hyp}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "advice1": {"name": "advice1", "group": "Varying q and advice", "definition": "\"

Avoid using rounded values in calculations and just round for the final answer.

Pythagoras Theorem states that, in a right angled triangle, with hypotenuse $c$:

\\n

\\\\[a^2 + b^2 = c^2\\\\]

\\n

Let\\'s call the unknown value $x$, therefore we can write:

\\n

$a = x$, $b =\\\\var{sh2}$ and $c = \\\\var{hyp}$

\\n

\\\\[x^2 + \\\\var{sh2}^2 = \\\\var{hyp}^2\\\\]

\\n

and therefore

\\n

\\\\[x^2 = \\\\var{hyp^2} - \\\\var{sh2^2}\\\\]

\\n

\\\\[x = \\\\sqrt{\\\\var{hyp^2-sh2^2}}\\\\]

$x = \\\\var{sh1}$ to 1 d.p.

\"", "description": "", "templateType": "long string", "can_override": false}, "statement1": {"name": "statement1", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/zhu32wjq',[sh1: sh1, sh2: sh2])}", "description": "

", "templateType": "anything", "can_override": false}, "statement2": {"name": "statement2", "group": "Varying q and advice", "definition": "{geogebra_applet('https://www.geogebra.org/m/b8wz2vn9',[sh1: sh1, sh2: sh2])}", "description": "", "templateType": "anything", "can_override": true}, "statement": {"name": "statement", "group": "Varying q and advice", "definition": "if(setup=1,statement1,statement2)", "description": "", "templateType": "anything", "can_override": false}, "hyp": {"name": "hyp", "group": "Unnamed group", "definition": "precround(sqrt(sh1^2+sh2^2),1)", "description": "", "templateType": "anything", "can_override": false}, "sh1_gen": {"name": "sh1_gen", "group": "Unnamed group", "definition": "random(5 .. 10#0.1)", "description": "

one of two shortest sides for calculations.

", "templateType": "randrange", "can_override": false}, "sh1": {"name": "sh1", "group": "Unnamed group", "definition": "precround(sh1_gen,1)", "description": "", "templateType": "anything", "can_override": false}, "sh2_gen": {"name": "sh2_gen", "group": "Unnamed group", "definition": "random(4 .. 9#0.1)", "description": "", "templateType": "randrange", "can_override": true}, "sh2": {"name": "sh2", "group": "Unnamed group", "definition": "precround(sh2_gen,1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "10"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["hyp", "sh1_gen", "sh1", "sh2_gen", "sh2"]}, {"name": "Varying q and advice", "variables": ["setup", "answerside", "answerhyp", "ans", "advice", "advice2", "advice1", "statement1", "statement2", "statement"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$x=$[[0]] to 1 d.p.

Find the diagonal or one side of a rectangle using Pythagoras' theorem.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the height of the rectangle below (all measurements given in $cm$)? Please give your answer to one decimal place.

{geogebra_applet('https://www.geogebra.org/m/jk3n6sxh',[base: base, hyp: hyp])}

", "advice": "

You can see that the rectangle contains a right-angled triangle. We also have the lengths of the base and the hypoteneuse of the triangle. This means we can use Pythagoras' theorem to calculate the last remaining side of the triangle which is also the height of the rectangle.

\\[ \\begin{split} Height &\\, = \\sqrt{hypoteneuse^2 - base^2} \\\\ &\\, = \\sqrt{\\var{hyp}^2-\\var{base}^2} \\\\ &\\, = \\sqrt{\\var{{hyp}^2}-\\var{{base}^2}} \\\\ &\\, = \\sqrt{\\var{{{hyp}^2}-{{base}^2}}}\\\\ &\\, = \\var{ans}\\\\ &\\, = \\var{ansr} \\text{ to 1 d.p.} \\end{split} \\]

Use this link to find resources to help you revise Pythagoras' theorem.

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[[0]]$cm$ (give your answer to 1 d.p.)

Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{max_height(25,diagram)}

", "advice": "

Avoid using rounded values in calculations and just round for the final answer.

{advice}

Use this link to find some resources to help you revise how to answer trigonometry questions that ask you to find a missing side.

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"can_override": false}, "d_c_s_1": {"name": "d_c_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , a..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "diagram": {"name": "diagram", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n random(d_s_a_1,d_s_a_2),\n random(d_s_s_1,d_s_s_2)),\n if(SCT='t',\n if(AngORside='ang',\n random(d_t_a_1,d_t_a_2),\n random(d_t_s_1,d_t_s_2)),\n if(SCT='c',\n if(AngORside='ang',\n random(d_c_a_1,d_c_a_2),\n random(d_c_s_1,d_c_s_2)),'X')))\n ", "description": "", "templateType": "anything", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "random(32..72)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_1": {"name": "d_s_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , b..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_1": {"name": "d_c_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , a..c label(ac + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_1": {"name": "d_s_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..c label(ac + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle+'\u00b0')\n , b..c label(bc + 'cm')\n , a..c label('x cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x\u00b0')\n , a..c label(ac + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'side'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle+'\u00b0')\n , a..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_2": {"name": "d_s_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_2": {"name": "d_s_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle+'\u00b0')\n , b..c label('x cm')\n , a..b label(ab + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_2": {"name": "d_t_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0), \n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x\u00b0')\n , b..c label(bc + 'cm')\n , a..c label(ac + 'cm')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n angle,\n bc),\n if(SCT='t',\n if(AngORside='ang',\n angle,\n ac),\n if(AngORside='ang',\n angle,ac)))", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "advice", "definition": "if(SCT='s',\n if(AngORside='ang',\n {sin_a},\n {sin_bc}),\n if(SCT='c',\n if(AngORside='ang',\n {cos_a},\n {cos_ac}),\n if(AngORside='ang',\n {tan_a},{tan_ac})))", "description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse $\\\\sin$\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\sin^{-1}(\\\\var{bc}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse $\\\\cos$\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\cos^{-1}(\\\\var{ac}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]

We need to use the \\'inverse $\\\\tan$\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\tan^{-1}(\\\\var{bc}/\\\\var{ac})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}}, "variablesTest": {"condition": "precround(180*(arcsin(bc/(ab)))/pi,1) = precround(angle,1)", "maxRuns": "6"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "angle", "SCT", "AngORside", "answer"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Given a right angled triangle as shown calculate the value of x.

Angles are given in degrees (make sure you calculator is in the right mode)

Give your answer correct to 2 decimal place.

", "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": "100", "precisionMessage": "", "strictPrecision": true, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA05 Trigonometry - missing angle", "extensions": ["eukleides"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{max_height(25,diagram)}

", "advice": "

Avoid using rounded values in calculations and just round for the final answer.

{advice}

Use this link to find resources to help you revise how to answer trigonometry questions that ask you to find the missing angle.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"ab": {"name": "ab", "group": "Unnamed group", "definition": "precround(sqrt(ac^2+bc^2),1)", "description": "", "templateType": "anything", "can_override": false}, "ac": {"name": "ac", "group": "Unnamed group", "definition": "precround(gen_ac,1)", "description": "", "templateType": "anything", "can_override": false}, "bc": {"name": "bc", "group": "Unnamed group", "definition": "precround(gen_bc,1)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_1": {"name": "d_t_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_1": {"name": "d_c_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "diagram": {"name": "diagram", "group": "Unnamed group", "definition": "if(SCT='s',\n if(AngORside='ang',\n random(d_s_a_1,d_s_a_2),\n random(d_s_s_1,d_s_s_2)),\n if(SCT='t',\n if(AngORside='ang',\n random(d_t_a_1,d_t_a_2),\n random(d_t_s_1,d_t_s_2)),\n if(SCT='c',\n if(AngORside='ang',\n random(d_c_a_1,d_c_a_2),\n random(d_c_s_1,d_c_s_2)),'X')))\n ", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_1": {"name": "d_s_s_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_1": {"name": "d_c_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_1": {"name": "d_s_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_1": {"name": "d_t_a_1", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_s_2": {"name": "d_t_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , b..c label(bc)\n , a..c label('x')\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_c_a_2": {"name": "d_c_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , a..c label(ac)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "SCT": {"name": "SCT", "group": "Unnamed group", "definition": "random('s','c','t')", "description": "", "templateType": "anything", "can_override": false}, "AngORside": {"name": "AngORside", "group": "Unnamed group", "definition": "'ang'", "description": "", "templateType": "anything", "can_override": false}, "d_c_s_2": {"name": "d_c_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label(angle)\n , a..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_a_2": {"name": "d_s_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label('x')\n , b..c label(bc)\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_s_s_2": {"name": "d_s_s_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,-ac),\n b, point(-bc,0),\n c, point(0,0),\n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(c,a,b) label(angle)\n , b..c label('x')\n , a..b label(ab)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "d_t_a_2": {"name": "d_t_a_2", "group": "triangle types", "definition": "eukleides(\"A right-angled triangle\",\n let(\n a, point(0,ac),\n b, point(-bc,0),\n c, point(0,0), \n\n [\n \n a..b..c\n , angle(a,c,b) right\n , angle(b,a,c) label('x')\n , b..c label(bc)\n , a..c label(ac)\n \n ]\n ),\n[\"angle\":{angle}]\n)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Unnamed group", "definition": "precround(angle*180/pi,2)", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "advice", "definition": "if(SCT='s',\n if(AngORside='ang',\n {sin_a},\n {sin_bc}),\n if(SCT='c',\n if(AngORside='ang',\n {cos_a},\n {cos_ac}),\n if(AngORside='ang',\n {tan_a},{tan_ac})))", "description": "", "templateType": "anything", "can_override": false}, "sin_a": {"name": "sin_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(x) = \\\\frac{\\\\var{bc}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arcsin$ or notated $\\\\sin^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arcsin(\\\\var{bc}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arcsin(bc/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_a": {"name": "cos_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Adjacent} = \\\\var{ac}$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(x) = \\\\frac{\\\\var{ac}}{\\\\var{ab}}\\\\]

We need to use the \\'inverse cos\\' button on the calculator (also called $\\\\arccos$ or notated $\\\\cos^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arccos(\\\\var{ac}/\\\\var{ab})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arccos(ac/(ab)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_a": {"name": "tan_a", "group": "advice", "definition": "\"

In this situation $x$ is an angle. We label the known sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we are interested in:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = \\\\var{ac}$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(x) = \\\\frac{\\\\var{bc}}{\\\\var{ac}}\\\\]

We need to use the \\'inverse sin\\' button on the calculator (also called $\\\\arctan$ or notated $\\\\tan^{-1}$) in order to isolate $x$:

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\arctan(\\\\var{bc}/\\\\var{ac})\\\\]

\\n

\\\\[ x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,4)}\\\\]

\\n

Round as required:

\\n

\\\\[x = \\\\var{precround(180*(arctan(bc/(ac)))/pi,2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sin_bc": {"name": "sin_bc", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = x$
$\\\\text{Hypotenuse} = \\\\var{ab}$

We have \\'O\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\sin$ formula:

\\n

\\\\[ \\\\sin(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\sin(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\sin(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*sin(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "cos_ac": {"name": "cos_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Hypotenuse} = \\\\var{ab}$
$\\\\text{Adjacent} = x$

We have \\'A\\' and \\'H\\' in SOHCAHTOA, so we know we need to use the $\\\\cos$ formula:

\\n

\\\\[ \\\\cos(\\\\text{Angle}) = \\\\frac{\\\\text{Adjacent}}{\\\\text{Hypotenuse}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\cos(\\\\var{angle}) = \\\\frac{x}{\\\\var{ab}}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\var{ab} \\\\times \\\\cos(\\\\var{angle}) \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(ab*cos(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "tan_ac": {"name": "tan_ac", "group": "advice", "definition": "\"

In this situation $x$ is a side. We label the relevant sides of the triangle \\'opposite\\', \\'adjacent\\' and \\'hypotenuse\\' in relation to the angle we know:

\\n

$\\\\text{Opposite} = \\\\var{bc}$
$\\\\text{Adjacent} = x$

We have \\'O\\' and \\'A\\' in SOHCAHTOA, so we know we need to use the $\\\\tan$ formula:

\\n

\\\\[ \\\\tan(\\\\text{Angle}) = \\\\frac{\\\\text{Opposite}}{\\\\text{Adjacent}}\\\\]

\\n

Now we subsitute the values we have in this particular question

\\n

\\\\[ \\\\tan(\\\\var{angle}) = \\\\frac{\\\\var{bc}}{x}\\\\]

and rearrange to give:

\\n

\\\\[ x = \\\\frac{\\\\var{bc}}{\\\\tan(\\\\var{angle})} \\\\]

Make sure your calculator is set to \\'degree\\' mode, if you get an odd answer you are likely in the wrong mode!

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),4)}\\\\]

\\n

Round as required:

\\n

\\\\[ x = \\\\var{precround(bc/tan(pi*angle/180),2)}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "angle": {"name": "angle", "group": "Unnamed group", "definition": "If(SCT='c',arccos(ac/ab),if(SCT = 's',arcsin(bc/ab),arctan(bc/ac)))", "description": "", "templateType": "anything", "can_override": false}, "gen_ac": {"name": "gen_ac", "group": "Unnamed group", "definition": "random(3 .. 12#0.1)", "description": "", "templateType": "randrange", "can_override": false}, "gen_bc": {"name": "gen_bc", "group": "Unnamed group", "definition": "random(5 .. 15#0.1)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "300"}, "ungrouped_variables": [], "variable_groups": [{"name": "Unnamed group", "variables": ["ab", "ac", "bc", "diagram", "SCT", "AngORside", "answer", "angle", "gen_ac", "gen_bc"]}, {"name": "triangle types", "variables": ["d_t_a_2", "d_t_s_1", "d_s_a_1", "d_c_a_1", "d_c_s_1", "d_s_s_1", "d_c_s_2", "d_t_a_1", "d_t_s_2", "d_s_a_2", "d_s_s_2", "d_c_a_2"]}, {"name": "advice", "variables": ["advice", "tan_a", "sin_a", "cos_a", "sin_bc", "cos_ac", "tan_ac"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Given a right angled triangle as shown calculate the value of x.

Give your answer in degrees (make sure you calculator is in the right mode), correct to 2 decimal place.

", "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "2", "precisionPartialCredit": "100", "precisionMessage": "", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GA06 Trigonometry - non-right angled trig", "extensions": ["geogebra"], "custom_part_types": [], "resources": ["question-resources/Picture1_caMIdF1.png", "question-resources/Picture2_6KE4ZpW.png"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Wishart", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1461/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": [], "metadata": {"description": "

Draws a triangle based on 3 side lengths. Randomises asking angle or side.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{diagram}

Find x.

", "advice": "

{Advice}

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"Ruleuse": {"name": "Ruleuse", "group": "Question structure", "definition": "random('s','c','s','c')", "description": "", "templateType": "anything", "can_override": false}, "ANGorSIDE": {"name": "ANGorSIDE", "group": "Question structure", "definition": "random('ang','side')", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEadvice": {"name": "cosSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of two sides given and the angle specifically between those two sides. Further to this, the instruction is to find the other missing side. These are the conditions for when to use the $\\\\textit{cosine rule}$.

\\n

The formula for a missing side using the cosine rule is:

\\n

\\\\[ a^2 = b^2 + c^2 - 2bc \\\\cos(A)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the angle given and it\\'s opposite side.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A=\\\\var{Cang},\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x^2 = \\\\var{a}^2 +\\\\var{b}^2-2\\\\times\\\\var{a}\\\\times\\\\var{b}\\\\times\\\\cos{(\\\\var{Cang})},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\sqrt{\\\\var{a^2 +b^2-2*a*b*(cos(Cang))}}\\\\]

\\n

\\\\[x=\\\\var{c}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 1: missing side in the cosine rule.

", "templateType": "long string", "can_override": false}, "cosANGadvice": {"name": "cosANGadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that there are the lengths of all three sides given. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{cosine rule}$ but in its rearranged form to find an angle. You need to identify which side is \\\"$a$\\\" as being the one opposite the angle you are asked to find.

\\n

The formula for a missing angle using the cosine rule is:

\\n

\\\\[ A = \\\\arccos\\\\left(\\\\frac{b^2+c^2-a^2}{2bc}\\\\right)\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the $a$ (side) and $A$ (angle) labels are applied to the side opposite the angle that is asked for and the angle that is asked for.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad c=\\\\var{b}, \\\\text{and} \\\\quad A= x,\\\\]

\\n

where the choice of which way round $b$ and $c$ are assigned doesn\\'t matter.

\\n

So, we now have:

\\n

\\\\[x = \\\\arccos\\\\left(\\\\frac{\\\\var{a}^2+\\\\var{b}^2-\\\\var{c_round}^2}{2\\\\times\\\\var{a}\\\\times\\\\var{b}}\\\\right),\\\\]

\\n

hence,

\\n

\\\\[x=\\\\var{(180/pi)*arccos((a^2 +b^2-c_round^2)/(2*a*b))}\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "sinSIDEadvice": {"name": "sinSIDEadvice", "group": "Question structure", "definition": "\"

First recognise that the diagram is a non-right angled triangle and that a single length is provided, along with two angles, crucially including the angle opposite the given side. Further to this, the instruction is to find the a missing angle. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding a side using the sine rule can be written as:

\\n

\\\\[ \\\\frac{a}{\\\\sin(A)}=\\\\frac{b}{\\\\sin(B)}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the side being asked for is in the above notation $a$.

\\n

In this case:

\\n

\\\\[ a=x, \\\\quad b=\\\\var{a}, \\\\quad A=\\\\var{Cang}, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{x}{\\\\sin{(\\\\var{Cang})}}=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\frac{\\\\var{a}}{\\\\sin{(\\\\var{Aang_round})}}\\\\times\\\\sin{(\\\\var{Cang})},\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\"", "description": "

case 3

", "templateType": "long string", "can_override": false}, "sinANGadvice": {"name": "sinANGadvice", "group": "Question structure", "definition": "safe(\"

First recognise that the diagram is a non-right angled triangle and that two lengths are provided, along with an angle, crucially including an angle opposite a given side. Further to this, the instruction is to find the a missing side. These are the conditions for when to use the $\\\\textit{sine rule}$. The sine rule uses the sides and angles in pairs and uses two pairs for any given calculation

\\n

The formula for finding an angle using the sine rule can be written as:

\\n

\\\\[ \\\\frac{\\\\sin(A)}{a}=\\\\frac{\\\\sin(B)}{b}\\\\]

\\n

The labels of $a$, $b$ and $c$ can be misleading. The critical thing is that regardless of the letters used in the diagram, the angle being asked for is in the above notation $A$.

\\n

In this case:

\\n

\\\\[ a=\\\\var{c_round}, \\\\quad b=\\\\var{a}, \\\\quad A= x, \\\\text{and} \\\\quad B= \\\\var{Aang_round}.\\\\]

\\n

So, we now have:

\\n

\\\\[\\\\frac{\\\\sin{(x)}}{\\\\var{c_round}}=\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}},\\\\]

\\n

hence,

\\n

\\\\[x=\\\\arcsin\\\\left(\\\\var{c_round}\\\\times\\\\frac{\\\\sin{(\\\\var{Aang_round})}}{\\\\var{a}}\\\\right),\\\\]

\\n

\\\\[x=\\\\var{ans}\\\\text{ to 1 decimal place.}\\\\]

\")", "description": "

case 4

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side length a

", "templateType": "randrange", "can_override": false}, "b": {"name": "b", "group": "Quantities", "definition": "random(5 .. 10#0.1)", "description": "

side length b

", "templateType": "randrange", "can_override": false}, "Cang": {"name": "Cang", "group": "Quantities", "definition": "random(40..140 except 85..95)", "description": "

C angle in degrees

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angle A in degrees

", "templateType": "anything", "can_override": false}, "Bang": {"name": "Bang", "group": "Quantities", "definition": "180-(Aang+Cang)", "description": "", "templateType": "anything", "can_override": false}, "cosSIDEans": {"name": "cosSIDEans", "group": "Quantities", "definition": "c", "description": "", "templateType": "anything", "can_override": false}, "cosANGans": {"name": "cosANGans", "group": "Quantities", "definition": "arccos((a^2+b^2-c_round^2)/(2*a*b))*180/pi", "description": "

Calculated answer for c from rounded values - as these will be seen information by student.

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$x =$[[0]]

Rewriting a trigonometric expression of the form $A\\sin(\\theta)-B\\cos(\\theta)$ to $R\\sin(\\theta-\\alpha)$ by calculating $R$ and $\\alpha$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If \\[ \\simplify[unitFactor]{{A}sin(theta)-{B}cos(theta)} = R \\sin (\\theta - \\alpha),\\]

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

To find $R$ and $\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\sin(a-b)=\\sin(a)\\cos(b)-\\sin(b)\\cos(a)$:

\\[ \\begin{split}\\simplify[unitFactor]{{A}sin(theta)-{B}cos(theta)} &\\,= R \\sin(\\theta-\\alpha) \\\\ &\\,= R(\\sin(\\theta)\\cos(\\alpha) - \\sin(\\alpha)\\cos(\\theta)) \\\\ &\\,= R\\sin(\\theta)\\cos(\\alpha) - R\\sin(\\alpha)\\cos(\\theta). \\end{split} \\]

By comparing the coefficients of $\\sin(\\theta)$ and $\\cos(\\theta)$, we find that

\\[ R\\cos(\\alpha) = \\var{A},\\quad \\text{and} \\quad R\\sin(\\alpha) = \\var{B}. \\]

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\sin^2(\\alpha)+\\cos^2(\\alpha) = 1$:

{Rsol}

Similarly, to find $\\alpha$ we can divide $R\\sin(\\alpha) = \\var{B}$ by $R\\cos(\\alpha) = \\var{A}$, and use the identity $\\tan(\\alpha) = \\frac{\\sin(\\alpha)}{\\cos(\\alpha)}$:

\\[ \\frac{R\\sin(\\alpha)}{R\\cos(\\alpha)} = \\frac{\\var{B}}{\\var{A}} \\implies \\tan(\\alpha) = \\simplify[fractionNumbers]{{B/A}}.\\]

Therefore, \\[ \\begin{split} \\alpha &\\,= \\tan^{-1}\\left(\\simplify[fractionNumbers]{{B/A}}\\right) \\\\ &\\,= \\var{alpharound} \\text{ (2 d.p.)}. \\end{split} \\]

Use this link to find some resources which will help you revise this topic.

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\\\\[ \\\\begin{split} R^2\\\\cos^2(\\\\alpha) + R^2 \\\\sin^2(\\\\alpha) &\\\\,= \\\\var{A}^2+\\\\var{B}^2 \\\\\\\\ R^2 (\\\\cos^2(\\\\alpha) +\\\\sin^2(\\\\alpha)) &\\\\,= \\\\var{A^2+B^2} \\\\\\\\ R &\\\\,= \\\\var{R}. \\\\end{split} \\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["A", "B", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

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Rewriting a trigonometric expression of the form $A\\cos(\\theta)\\pm B\\sin(\\theta)$ to either $R\\sin(\\theta+\\alpha)$ or $R\\cos(\\theta+\\alpha)$ by calculating $R$ and $\\alpha$.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{question}

find the values for $R$ and $\\alpha$, given $R>0$ and $0<\\alpha<\\frac{\\pi}{2}$.

", "advice": "

{answer}

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"A": {"name": "A", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything", "can_override": false}, "B": {"name": "B", "group": "Ungrouped variables", "definition": "random(1..5 except A)", "description": "", "templateType": "anything", "can_override": false}, "R": {"name": "R", "group": "Ungrouped variables", "definition": "sqrt(A^2+B^2)", "description": "", "templateType": "anything", "can_override": false}, "Rround": {"name": "Rround", "group": "Ungrouped variables", "definition": "precround(R,2)", "description": "", "templateType": "anything", "can_override": false}, "alpha": {"name": "alpha", "group": "Ungrouped variables", "definition": "arctan(B/A)", "description": "", "templateType": "anything", "can_override": false}, "Rsol": {"name": "Rsol", "group": "Ungrouped variables", "definition": "if(R=round(R),'{Rsol1}','{Rsol2}')", "description": "", "templateType": "anything", "can_override": false}, "Rsol1": {"name": "Rsol1", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "Rsol2": {"name": "Rsol2", "group": "Ungrouped variables", "definition": "\"

\"", "description": "", "templateType": "long string", "can_override": false}, "alpharound": {"name": "alpharound", "group": "Ungrouped variables", "definition": "precround(alpha,2)", "description": "", "templateType": "anything", "can_override": false}, "question": {"name": "question", "group": "Ungrouped variables", "definition": "if(Q=1,'{q1}','{q2}')", "description": "", "templateType": "anything", "can_override": false}, "Q": {"name": "Q", "group": "Ungrouped variables", "definition": "random(1,2)", "description": "", "templateType": "anything", "can_override": false}, "sign": {"name": "sign", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything", "can_override": false}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} = \\\\simplify[unitFactor]{R sin (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "\"

\\\\[ \\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} = \\\\simplify[unitFactor]{R cos (theta+{sign}*alpha)},\\\\]

\"", "description": "", "templateType": "long string", "can_override": false}, "a1": {"name": "a1", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify[unitFactor]{sin(a+{sign}*b)=sin(a)cos(b)+{sign}*sin(b)cos(a)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}sin(theta)+{sign*B}cos(theta)} &\\\\,= \\\\simplify{R sin(theta+{sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(sin(theta)cos(alpha) + {sign}*sin(alpha)cos(theta))} \\\\\\\\ &\\\\,= \\\\simplify{Rsin(theta)cos(alpha) + {sign}*R sin(alpha)cos(theta)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\sin(\\\\theta)$ and $\\\\cos(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

Similarly, to find $\\\\alpha$ we can divide $R\\\\sin(\\\\alpha) = \\\\var{B}$ by $R\\\\cos(\\\\alpha) = \\\\var{A}$, and use the identity $\\\\tan(\\\\alpha) = \\\\frac{\\\\sin(\\\\alpha)}{\\\\cos(\\\\alpha)}$:

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "a2": {"name": "a2", "group": "Ungrouped variables", "definition": "\"

To find $R$ and $\\\\alpha$ we want to first rewrite our equation using the double-angle formula, $\\\\simplify{cos(a+{sign}*b)=cos(a)cos(b)-{sign}*sin(a)sin(b)}$:

\\n

\\\\[ \\\\begin{split}\\\\simplify[unitFactor]{{A}cos(theta)-{sign*B}sin(theta)} &\\\\,= \\\\simplify[unitFactor]{R cos (theta + {sign}*alpha)} \\\\\\\\ &\\\\,= \\\\simplify{R(cos(theta)cos(alpha) - {sign}*sin(theta)sin(alpha))} \\\\\\\\ &\\\\,= \\\\simplify{Rcos(theta)cos(alpha) - {sign}*R sin(theta)sin(alpha)}. \\\\end{split} \\\\]

\\n

By comparing the coefficients of $\\\\cos(\\\\theta)$ and $\\\\sin(\\\\theta)$, we find that

\\n

\\\\[ R\\\\cos(\\\\alpha) = \\\\var{A},\\\\quad \\\\text{and} \\\\quad R\\\\sin(\\\\alpha) = \\\\var{B}. \\\\]

\\n

To calculate $R$, we want to square these results and add them together, allowing us to make use of $\\\\sin^2(\\\\alpha)+\\\\cos^2(\\\\alpha) = 1$:

\\n

{Rsol}

\\n

\\\\[ \\\\frac{R\\\\sin(\\\\alpha)}{R\\\\cos(\\\\alpha)} = \\\\frac{\\\\var{B}}{\\\\var{A}} \\\\implies \\\\tan(\\\\alpha) = \\\\simplify[fractionNumbers]{{B/A}}.\\\\]

\\n

Therefore, \\\\[ \\\\begin{split} \\\\alpha &\\\\,= \\\\tan^{-1}\\\\left(\\\\simplify[fractionNumbers]{{B/A}}\\\\right) \\\\\\\\ &\\\\,= \\\\var{alpharound} \\\\text{ (2 d.p.)}. \\\\end{split} \\\\]

\\n

\"", "description": "", "templateType": "long string", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "if(Q=1,'{a1}','{a2}')", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["Q", "A", "B", "sign", "R", "Rround", "alpha", "alpharound", "Rsol", "Rsol1", "Rsol2", "question", "q1", "q2", "answer", "a1", "a2"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$R=$[[0]]

$\\alpha=$[[1]]

(Give your answers to 2 decimal places where necessary.)

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Match the graphs to the functions. No randomisation. Multiple choice.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

This is about knowledge of graphs. Generally with trigonometric graphs it is best to start with making sure you know and understand the graphs of the functionts $\\sin(x)$, $\\cos(x)$ and $\\tan(x)$. From there you can use knowledge of where they are zero to work out the position of the asymptotes in the graphs of $\\sec(x)$, $\\text{cosec}(x)$ and $\\cot(x)$. However, you still need really to be able to recall the shape of each graph for some purposes and be confident about where the zeros and turning points are.

Use this link to find some resources to help you familiarise yourself with these graphs.

Match the graph to its function.

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Finding the area of a circle when given the diameter of the circle.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the area of a circle with diameter $\\var{d}$ cm giving your answer to 1 decimal place.

{geogebra_applet('https://www.geogebra.org/m/ngcchpcj',[d: d])}

", "advice": "

To calculate the area of a circle we want to use the formula \\[ A = \\pi r^2, \\]

where $r$ is the radius of the circle.

So, if the diameter, d, is $\\var{d}$ cm, then the radius is, $r=\\frac{d}{2}=\\var{{d}/2}$ cm, then

\\[ \\begin{split} Area &\\,=\\var{{d}/2}^2 \\times \\pi \\text{ cm}^2 \\\\ &\\,= \\simplify[all, fractionNumbers]{{{{d}^2/4}}pi} \\text{ cm}^2 \\\\ &\\,= \\var{precround({d}^2/4*pi,1)} \\text{ cm}^2. \\end{split} \\]

Use this link to find some resources to help you revise how to calculate the area of a circle.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"d": {"name": "d", "group": "Ungrouped variables", "definition": "random(6,8,10,12,14,16,18,20)", "description": "", "templateType": "anything", "can_override": true}, "t": {"name": "t", "group": "Ungrouped variables", "definition": "random(2..6)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["d", "t"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$Area=$ [[0]] $\\text{ cm}^2$

", "gaps": [{"type": "jme", "useCustomName": true, "customName": "Gap 0", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": false, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "precround({{d/2}}^2*pi,1)", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "answer": "precround({{d/2}}^2*pi,1)", "answerSimplification": "fractionNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "GM02 Volume of a triangular prism", "extensions": [], "custom_part_types": [], "resources": ["question-resources/sqbasedpyramid_sEpkGzO.svg", "question-resources/triangularprism.svg", "question-resources/cylinder.svg", "question-resources/cuboid.svg"], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Aiden McCall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1592/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": ["3D shapes", "cuboid", "Cylinder", "cylinder", "pyramid", "taxonomy", "triangular prism", "volume", "Volume", "volume of a cuboid", "volume of a cylinder", "volume of a pyramid", "volume of a triangular prism"], "metadata": {"description": "

Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a triangular prism, we first need to find the area of one of the faces then multiply this area by the depth of the prism.
In this example the easiest way to calculate the volume is to take the area of the triangular face first with $\\mathrm{base} = \\var{w6}m$ and $\\mathrm{height} = \\var{h6}m\\thinspace$.

\\begin{align}
\\mathrm{Area\\thinspace_\\triangle} &= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\\\
&= \\frac{\\var{w6} \\times \\var{h6}}{2} \\\\
&= \\var{0.5*w6*h6}\\, \\mathrm{m}^2\\,.
\\end{align}

Now that we have the area of the triangular face ($\\mathrm{Area\\thinspace_\\triangle}$) we can multiply this by the $\\mathrm{depth} = \\var{d6}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\var{0.5*w6*h6} \\times \\var{d6} \\\\
&= \\var{0.5*w6*h6*d6}\\, \\mathrm{m}^2\\,.
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a triangular prism.

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Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

", "templateType": "anything", "can_override": false}, "d8": {"name": "d8", "group": "Square based pyramid", "definition": "random(3..6#0.1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h8": {"name": "h8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

Height of pyramid.

", "templateType": "anything", "can_override": false}, "w7": {"name": "w7", "group": "Cylinder", "definition": "random(7..15#0.1)", "description": "

Depth of cylinder.

", "templateType": "anything", "can_override": false}, "d6": {"name": "d6", "group": "Triangular prism", "definition": "random(9..15#0.1)", "description": "

Depth of triangular prism.

", "templateType": "anything", "can_override": false}, "r7": {"name": "r7", "group": "Cylinder", "definition": "random(2..6#1)", "description": "

Radius of the cylinder.

", "templateType": "anything", "can_override": false}, "h4": {"name": "h4", "group": "Cuboid ", "definition": "random(2..5#1 except d4)", "description": "

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w4": {"name": "w4", "group": "Cuboid ", "definition": "random(5.5..8#0.1)", "description": "

Width of cuboid.

", "templateType": "anything", "can_override": false}, "w8": {"name": "w8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h6": {"name": "h6", "group": "Triangular prism", "definition": "random(2..5#1)", "description": "

Height of traingle.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": [], "variable_groups": [{"name": "Cuboid ", "variables": ["w4", "d4", "h4"]}, {"name": "Triangular prism", "variables": ["w6", "h6", "d6"]}, {"name": "Cylinder", "variables": ["r7", "w7"]}, {"name": "Square based pyramid", "variables": ["h8", "w8", "d8"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the $\\mathrm{Volume}$ of the following triangular prism.

$\\mathrm{Volume} =$[[0]]$\\mathrm{m}^3$.

", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Volume of a triangular prism:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\triangle} \\times \\mathrm{depth} \\\\
&= \\frac{\\mathrm{base} \\times \\mathrm{height}}{2} \\times \\mathrm{depth}
\\end{align}

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Calculate the volume of different 3D shapes, given the units and measurements required. The formulae for the volume of each shape are available as steps if required.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

For a cylinder, we first need to find the area of the circular face then multiply this area by the depth of the cylinder.
In this example the radius of the circular face is $\\mathrm{radius} = \\var{r7}m$ which can be used to calculate the area of the circular face.

\\begin{align}
\\mathrm{Area\\thinspace_\\bigcirc} &= \\pi \\times \\mathrm{radius}^2 \\\\
&= \\pi \\times \\var{r7}^2 \\\\
&= \\var{pi * (r7)^2}\\, \\mathrm{m}^2 \\,.
\\end{align}

Now that we have the area of the circular face ($\\mathrm{Area\\thinspace_\\bigcirc}$) we can multiply this by the $\\mathrm{depth} =\\var{w7}m\\thinspace$.

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\var{pi*(r7)^2} \\times \\var{w7} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 5)} \\\\
&= \\var{dpformat(pi*w7*(r7)^2, 1)}\\, \\mathrm{m}^2\\,. \\quad \\text{1 d.p.}
\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a cylinder.

Side of square in cuboid.

", "templateType": "anything", "can_override": false}, "w6": {"name": "w6", "group": "Triangular prism", "definition": "random(5..9#1)", "description": "

Creates base of triangle.

", "templateType": "anything", "can_override": false}, "d8": {"name": "d8", "group": "Square based pyramid", "definition": "random(3..6#0.1)", "description": "

One side of square base.

", "templateType": "anything", "can_override": false}, "h8": {"name": "h8", "group": "Square based pyramid", "definition": "random(3..7#1)", "description": "

Height of pyramid.

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Depth of cylinder.

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Depth of triangular prism.

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Radius of the cylinder.

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Side of square in cuboid.

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Width of cuboid.

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One side of square base.

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Height of traingle.

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Calculate the $\\mathrm{Volume}$ of the following cylinder.

$\\mathrm{Volume} =$[[0]] $\\mathrm{m}^3$. Round your answer to 1 decimal place.

Volume of a cylinder:

\\begin{align}
\\mathrm{Volume} &= \\mathrm{Area\\thinspace_\\bigcirc} \\times \\mathrm{depth} \\\\
&= \\pi \\times \\mathrm{r}^2 \\times \\mathrm{depth}
\\end{align}

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Find the volume of a semicylinder from a diagram.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate the volume of this (all lengths are in $cm$):

{geogebra_applet('https://www.geogebra.org/m/vdbvgwkf',[height: height,radius: radius])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a semi-circle. Find the area of a circle and then half it.

The area of a semi-circle is given by:

\\begin{align} \\frac{\\pi\\times r^2}{2} \\end{align}

where $r$ is the radius of the circle.

\\begin{align} \\frac{\\pi\\times\\var{radius}^2}{2} = \\var{precround(semiarea,2)}... \\quad cm^2 \\end{align}

Then to calculate the volume you multiply the cross-sectional area by the length,

\\begin{align} \\frac{\\pi\\times r^2}{2} \\times l \\end{align}

\\begin{align} \\var{precround(semiarea,2)}... \\times \\var{height} = \\var{precround(answer,2)}cm^3.\\end{align}

Use this link to find resources to help you revise how to calculate the volume of a prism.

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[[0]]$cm^3$

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Find the volume of a prism with a trapezium as a cross section from a diagram.

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Calculate the volume of this (all lengths are in $cm$):

{geogebra_applet('https://www.geogebra.org/m/qvcktek2',[basew: basew, topw: topw, h: h, l: l])}

", "advice": "

In order to work out the volume of a prism you need to work out the cross sectional area first. In this question the cross section is a trapezium. Find the area of a trapezium,

\\begin{align} \\frac{\\var{basew}+\\var{topw}}{2}\\times \\var{h} = \\var{traparea} cm^2 \\end{align}

Then to calculate the volume you times the cross-sectional area by the length,

\\begin{align} \\var{traparea} \\times \\var{l} = \\var{answer}cm^3\\end{align}.

Use this link to find resources to help you revise how to calculate the volume of a prism.

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[[0]]$cm^3$

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Commutative

The definition of commutativity can be written in the following way:

$$
a \\times b = b \\times a.
$$

There are varying degress of technical detail that can be included in this definition depending on what area you are studying. The key idea is that an operation is said to be commutative if the order in which you write the two elements being operated on does not matter. Multiplication of real numbers is commutative because as we know $2 \\times 3 = 3 \\times 2 = 6$ for example. The most common example of something being non-commutative is multiplication for matrices. In general for two matrices $A$ and $B$, $AB \\neq BA$ (in fact sometimes one of these things can be calculated and the other does not even exist).

Assosciative

The definition of associativity can be written in the following way:

$$
(ab)c = a(bc).
$$

In other words it doesn't matter if you first work out $a$ times $b$ and then take the result and times it by $c$, or if you first work out $b$ times $c$ and then pre-multiply the result by $a$.

Distributive

The definition of distributive can be written in the following way:

$$
a \\times (b + c) = a \\times b + a \\times c.
$$

As with the others there are increasing levels of detail that can be put into this definition (such as including ideas such as right-distributive and left-distributive) but the key idea is that you can \"expand brackets\" as you can in elementary algebra, if an operator is distributive.

For more reading on this try (for example) this link.

Match the word to its correct definition

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The following questions are designed to explore the dimensions of matrices and what you can and can't do with matrices of differing dimensions.

", "advice": "

Rows and Columns

The convention in Matrix notation is to give the dimensions of a matrix in the order \"rows\" by \"columns\".

For $\\var{Dimensions}$ there are $\\var{rows[0]}$ rows and $\\var{columns[0]}$ columns. We write this as \"this is a $\\var{rows[0]}$X$\\var{columns[0]}$ matrix\".

When can you add and subtract matrices?

Two Matrices can be added or subtracted if they have the exact same dimensions as each other. For example $\\var{canadd1}$ and $\\var{canadd2}$ are both $\\var{rows[1]}$X$\\var{columns[1]}$ matrices and therefore can be added (or subtracted). However, $\\var{cantaddsub1}$ is a $\\var{rows[3]}$X$\\var{columns[3]}$ matrix and $\\var{cantaddsub2}$ is a $\\var{rows[3]}$X$\\var{columns[3]+1}$ matrix. Since these dimensions are different these matrices cannot be added or subtracted.

Multiplying Dimensions

When you multiply two matrices together the number of columns in the first matrix must match the number of rows in the second matrix. For example in the calculation $\\var{Mult3}$X$\\var{Mult4}$ the first matrix has $3$ columns and the second matrix has $3$ rows so they can be multiplied. In addition to this when multiplying two matrices (that can be multiplied) the result will be a single matrix that has the number of rows of the first matrix and the number of columns of the second matrix. In this example the first matrix has $\\var{rows[0]}$ rows and the second matrix has $\\var{columns[1]}$ columns, so the result of multiplying the two matrices will be a $\\var{rows[0]}$X$\\var{columns[1]}$ matrix.

Use this link to find some resources which will help you revise this topic.

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What are the dimensions of the following matrix?

$\\var{dimensions}$

[[0]]X[[1]]

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Which of the following calculations are defined?

(Indicate ALL possible answers by ticking the corresponding box(es))

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Is this calculation defined?

$\\var{Mult1}$X$\\var{Mult2}$

[[0]]

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What will be the dimensions of the matrix you get when you multiply these two matrices?

$\\var{Mult3}$X$\\var{Mult4}$.

[[0]]X[[1]]

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Rows", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "rows[0]", "maxValue": "rows[0]", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "Cols", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "columns[1]", "maxValue": "columns[1]", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LM02 Adding matrices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Generates two sets of matrices that can be added and asks the student to add them. Will ensure that the dimensions in part a) are different to the dimensions in part b) to offer some variety.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Adding Matrices

Calculate the following:

", "advice": "

Matrix addition is fairly straightforward in that each entry is calculated by adding the corresponding entries. For example the top left entry of the answer to part a) can be calculated as $\\var{A[0][0]} + \\var{Apair[0][0]} = \\var{A[0][0] + Apair[0][0]}$. So for this question:

a) $\\var{A}+\\var{Apair} = \\var{A+Apair}$

b) $\\var{B}+\\var{Bpair} = \\var{B+Bpair}$

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"A": {"name": "A", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[0]),row[0]))", "description": "", "templateType": "anything", "can_override": false}, "col": {"name": "col", "group": "Ungrouped variables", "definition": "repeat(random(2..3),3)", "description": "", "templateType": "anything", "can_override": false}, "row": {"name": "row", "group": "Ungrouped variables", "definition": "repeat(random(2..3),3)", "description": "", "templateType": "anything", "can_override": false}, "B": {"name": "B", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[1]),row[1]))", "description": "", "templateType": "anything", "can_override": false}, "C": {"name": "C", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[2]),row[2]))", "description": "", "templateType": "anything", "can_override": false}, "Apair": {"name": "Apair", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[0]),row[0]))", "description": "", "templateType": "anything", "can_override": false}, "Bpair": {"name": "Bpair", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[1]),row[1]))", "description": "", "templateType": "anything", "can_override": false}, "Cpair": {"name": "Cpair", "group": "Ungrouped variables", "definition": "matrix(repeat(repeat(random(-9..9),col[2]),row[2]))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "col[0]<>row[0] AND col[0]<>col[1]", "maxRuns": 100}, "ungrouped_variables": ["col", "row", "A", "Apair", "B", "Bpair", "C", "Cpair"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "matrix", "useCustomName": true, "customName": "a)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{A} + \\var{Apair}=$

", "correctAnswer": "{A}+{Apair}", "correctAnswerFractions": false, "numRows": "{row[0]}", "numColumns": "{col[0]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "b)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{B} +\\var{Bpair}$

", "correctAnswer": "{B}+{Bpair}", "correctAnswerFractions": false, "numRows": "{row[1]}", "numColumns": "{col[1]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LM03 Subtracting matrices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Generates two sets of matrices that can be added and asks the student to add them. Will ensure that the dimensions in part a) are different to the dimensions in part b) to offer some variety.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Subtracting Matrices

Calculate the following:

", "advice": "

Matrix subtraction is fairly straightforward in that each entry is calculated by subtracting the corresponding entries. For example the top left entry of the answer to part a) can be calculated as $\\var{A[0][0]} - \\var{Apair[0][0]} = \\var{A[0][0] - Apair[0][0]}$. So for this question:

a) $\\var{A}-\\var{Apair} = \\var{A-Apair}$

b) $\\var{B}-\\var{Bpair} = \\var{B-Bpair}$

Use this link to find some resources which will help you revise this topic.

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$\\var{A} - \\var{Apair}=$

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$\\var{B} -\\var{Bpair}$

", "correctAnswer": "{B}-{Bpair}", "correctAnswerFractions": false, "numRows": "{row[1]}", "numColumns": "{col[1]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LM04 Multiplying matrices", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Multiplying Matrices

Calculate the following:

", "advice": "

When you are multiplying two matrices together you need to know what elements of each matrix multiply with each other to produce the answer. In order to look at this let's define the two matrices:

$A=\\var{MatrixA1}$

and

$B=\\var{MatrixA2}.$

These two matrices match the dimensions of the two given in is part a) of this question. When they are multiplied together as we have seen the answer will be a $\\var{row[0]}$ by $\\var{col[0]}$ matrix. So we can denote this matrix in the following way:

$\\var{MatrixA1}$X$\\var{MatrixA2}=\\var{Answermatrixa}$

The way in which this is calculated is as follows in terms of a formula:

$\\var{Answermatrixa}=\\var{matrixmultcalc}$

If you are trying to calculate $c_{11}$ for example one way to think of this is to imagine picking up the first column of the second matrix and tipping it on its side and laying it on top of the top row of the first matrix so, for example, $a_{11}$ gets paired with $b_{11}$ and $a_{12}$ gets paired with $b_{21}$ etc. As we can see in the above formula.

a) $\\var{A}$X$\\var{Apair} = \\var{A*Apair}$

b) $\\var{B}$X$\\var{Bpair} = \\var{B*Bpair}$

b) $\\var{C}$X$\\var{Cpair} = \\var{C*Cpair}$

Use this link to find some resources which will help you revise this topic.

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\"Latexcodebits\" is a JME function written in the Extensions and scripts section. It is a useful function for creating a list of symbols such as \"a_11\" to populate a matrix for the explanation section.

", "templateType": "anything", "can_override": false}, "row": {"name": "row", "group": "Ungrouped variables", "definition": "repeat(random(2..4),3)", "description": "", "templateType": "anything", "can_override": false}, "answermatrixa": {"name": "answermatrixa", "group": "Advice", "definition": "latex('\\\\begin{pmatrix}'+ stringify(rawanswermatrixa) + '\\\\end{'+'pmatrix}')", "description": "

\"stringify\" is a javascript written in the Extensions and scripts section - this converts a list of things into a string suitable for latex to read as a matrix.

", "templateType": "anything", "can_override": false}, "RawmatrixA1": {"name": "RawmatrixA1", "group": "Advice", "definition": "map('a_{'+'{j}'+'{l}}'+latexcodebits(row[0],match[0])[j-1][l-1],[j,l],product(1..row[0],1..match[0]))", "description": "", "templateType": "anything", "can_override": false}, "RawmatrixA2": {"name": "RawmatrixA2", "group": "Advice", "definition": "map('b_{'+'{j}'+'{l}}'+Latexcodebits(match[0],col[0])[j-1][l-1],[j,l],product(1..match[0],1..col[0]))", "description": "", "templateType": "anything", "can_override": false}, "matrixA1": {"name": "matrixA1", "group": "Advice", "definition": "latex('\\\\begin{pmatrix}'+ stringify(RawmatrixA1) + '\\\\end{'+'pmatrix}')", "description": "", "templateType": "anything", "can_override": false}, "matrixA2": {"name": "matrixA2", "group": "Advice", "definition": "latex('\\\\begin{pmatrix}'+ stringify(RawmatrixA2) + '\\\\end{'+'pmatrix}')", "description": "", "templateType": "anything", "can_override": false}, "Rawmatrixmultcalc": {"name": "Rawmatrixmultcalc", "group": "Advice", "definition": "map('a_\\{{r+1}{t+1}\\}'+'\\\\cdot'+' b_\\{{t+1}{s+1}\\}'+latexcodebitsproductcalc(row[0],match[0],col[0])[r][s][t],[r,s,t],product(0..(row[0]-1),0..(col[0]-1),0..(match[0]-1)))", "description": "", "templateType": "anything", "can_override": false}, "matrixmultcalc": {"name": "matrixmultcalc", "group": "Advice", "definition": "latex('\\\\begin{pmatrix}'+ stringify(rawmatrixmultcalc) +'\\\\end{'+'pmatrix}')", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["col", "row", "match", "A", "Apair", "B", "Bpair", "C", "Cpair"], "variable_groups": [{"name": "matrix notation", "variables": []}, {"name": "Advice", "variables": ["RawmatrixA1", "matrixA1", "RawmatrixA2", "matrixA2", "Rawanswermatrixa", "answermatrixa", "Rawmatrixmultcalc", "matrixmultcalc"]}], "functions": {"Latexcodebits": {"parameters": [["m", "number"], ["n", "number"]], "type": "list", "language": "jme", "definition": "repeat(repeat('&',n-1)+['\\\\\\\\'],m-1)+[repeat('&',n-1)+['']]"}, "stringify": {"parameters": [["input", "list"]], "type": "string", "language": "javascript", "definition": "var output = '';\nvar i;\nfor (i = 0; i < input.length; i++) {\n output += input[i];\n} \nreturn output;"}, "Latexcodebitsproductcalc": {"parameters": [["m", "number"], ["k", "number"], ["n", "number"]], "type": "anything", "language": "jme", "definition": "repeat(repeat(repeat('+',k-1)+['&'],n-1)+[repeat('+',k-1)+['\\\\\\\\']],m-1)+[repeat(repeat('+',k-1)+['&'],n-1)+[repeat('+',k-1)+['']]]"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "matrix", "useCustomName": true, "customName": "a)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{A}$X$\\var{Apair}=$

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$\\var{B}$X$\\var{Bpair}$

", "correctAnswer": "{B}*{Bpair}", "correctAnswerFractions": false, "numRows": "{row[1]}", "numColumns": "{col[1]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "c)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{C}$X$\\var{Cpair}$

", "correctAnswer": "{C}*{Cpair}", "correctAnswerFractions": false, "numRows": "{row[2]}", "numColumns": "{col[2]}", "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LM05 Determinant of a 2x2 matrix", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The determinant of a 2x2 matrix:

Calculate the determinant of the following 2X2 matrices.

", "advice": "

Determinant

The detrminant of a 2x2 matrix can be calculated using the following process:

$$
\\begin{vmatrix}
a & b \\\\
c & d \\\\
\\end{vmatrix} = ad-bc.
$$

Example

For $A = \\var{nonzerodet2},$

we have that,

$$
\\begin{vmatrix}
\\var{nonzerodet2[0][0]} & \\var{nonzerodet2[0][1]}\\\\
\\var{nonzerodet2[1][0]} & \\var{nonzerodet2[1][1]}\\\\
\\end{vmatrix}=(\\var{nonzerodet2[0][0]})(\\var{nonzerodet2[1][1]}) - (\\var{nonzerodet2[0][1]})(\\var{nonzerodet2[1][0]}) =\\var{detnonzerodet2}
$$

Use this link to find some resources which will help you revise this topic.

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$$
A=\\var{nonzerodet2}
$$

$
\\text{What is the value of}\\begin{vmatrix}
A
\\end{vmatrix}?
$

answer:[[0]]

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$$
m=\\var{zerodet1}
$$

$
\\text{What is the value of}\\begin{vmatrix}
M
\\end{vmatrix}?
$

answer:[[0]]

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The inverse of a 2x2 matrix

", "advice": "

Can the inverse of a 2X2 Matrix be found?

The inverse of a matrix can be found if the matrix is a square matrix and has a non-zero determinant. In the first part of this question all the matruces are 2X2 square matrices. The detrminant can be calculated by doing:

$$
\\begin{vmatrix}
a & b \\\\
c & d \\\\
\\end{vmatrix} = ad-bc
$$

If this comes out to be non-zero then the determinant can be found.

Finding the Inverse of a 2x2 Matrix

If there are two 2x2 matrices $A$ and $B$ such that:

$$
AB=BA=
\\left( \\begin{matrix}
1 & 0 \\\\
0 & 1 \\\\
\\end{matrix}\\right)
$$

then we can say that $A$ and $B$ are inverses of each other. The notation for this is that the inverse of a matrix $C$ is written as $C^{-1}$.

If,

$$
C=
\\left(\\begin{matrix}
a & b \\\\
c & d \\\\
\\end{matrix}\\right),
$$

then, the inverse of $C$ is given by the formula:

$$
C^{-1}=\\frac{1}{ad-bc}
\\left(\\begin{matrix}
d & -b \\\\
-c & a \\\\
\\end{matrix}\\right),
$$

So for part 2) of this question let's call the given matrix $D$:

$D=\\var{nonzerodet1}.$

The inverse of $D$ is calculated as follows:

$$
D^{-1} = \\frac{1}{(\\var{nonzerodet1[0][0]})(\\var{nonzerodet1[1][1]})-(\\var{nonzerodet1[1][0]})(\\var{nonzerodet1[0][1]})}\\left(\\begin{matrix}
\\var{nonzerodet1[1][1]} & \\var{-nonzerodet1[0][1]} \\\\
\\var{-nonzerodet1[1][0]} & \\var{nonzerodet1[0][0]} \\\\
\\end{matrix}\\right)
= \\var{inversenonzerodet1}
$$

Use this link to find some resources which will help you revise this topic.

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For which of the following matrices can the inverse be calculated?

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What is the inverses of the following matrix?

(Enter your answers as fractions or decimals to 3 significant figures).

$\\var{nonzerodet1}$

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Find the transpose of the following matrices:

", "advice": "

Transposing a matrix

To transpose a matrix you have to swap over the rows and columns. So for example a 3 by 2 matrix will have a transpose that is a 2 by 3 matrix. The elements of the transpose of a matrix also swap places following the same rule.

Example:

From part b) of the question you have the matrix $M$ given by:

$M=\\var{Transpose1}.$

Swapping the rows with the columns (for example meaning the first row becomes the first column in the transpose), gives the answer:

$M^{T}=\\var{answer}.$

Use this link to find some resources which will help you revise this topic.

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$\\var{squaretranspose}$

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$\\var{Transpose1}$

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Cofactors Determinant and inverse of a 3x3 matrix.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

The Determinant of a 3x3 matrix

", "advice": "

Determinant of a 3X3 matrix

Minors

In order to understand the process of finding a determinant for a 3x3 (or larger square) matrix we introduce the idea of a minor.

For example we will look at the matrix, $M$, defined as

$$
M = \\var{Example}.
$$

Each element of $M$ has an associated minor. The minor is formed from finding the detrminant of the remaining matrix after you have removed the row and column containg that element. For example, consider the minor for element $m_{12} = 2$. We remove the row and column containg $m_{12}$ (the top row and the second column) leaving:

$$
\\begin{vmatrix}
4 & 6\\\\
7 & 9
\\end{vmatrix} = 4*9-6*7 = -6
$$

Cofactors

The next important concept is a cofactor (you don't need to calculate ALL of the cofactors for finding a determinant but you will need them to go on and find the inverse of a 3x3 matrix). A cofactor is the a minor with a sign attached. The appropriate sign comes from the pattern of alternating signs:

$$
\\begin{array}{ccc}
+ & - & +\\\\
- & + & - \\\\
+ & - & +\\\\
\\end{array}
$$

So to continue the example above we would say the cofactor for entry $m_{12} = -(-6) = 6$.

The determinant is then calculated by choosing a row or column and taking the sum of the entries multiplied by their cofactors.

Putting it all together

For simplicity we will choose the top row for this example.

$$
\\begin{aligned}
\\det{M} &= 1*\\begin{vmatrix}
5 & 6 \\\\
8 & 9 \\\\
\\end{vmatrix} - 2* \\begin{vmatrix}
4 & 6 \\\\
7 & 9 \\\\
\\end{vmatrix} + 3* \\begin{vmatrix}
4 & 5 \\\\
7 & 8 \\\\
\\end{vmatrix} \\\\
&= 1 \\times -3 - \\left(2 \\times -6 \\right) + 3 \\times -3 \\\\
&= 0
\\end{aligned}
$$

Worked solution

For the question given the same calculation can be carried out as follows:

$$
\\begin{aligned}
\\det{A}
&= \\var{matrixA[0][0]}
\\begin{vmatrix}
\\var{matrixA[1][1]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][1]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} - \\var{matrixA[0][1]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} + \\var{matrixA[0][2]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][1]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{matrixA[0][0]} \\times \\var{min11} - \\left(\\var{matrixA[0][1]} \\times \\var{min12}\\right) + \\var{matrixA[0][2]} \\times \\var{min13} \\\\
&= \\var{answer}
\\end{aligned}
$$

Use this link to find some resources which will help you revise this topic.

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Let:

$$
A=\\var{matrixA}
$$

Find the determinant of $A$

$\\det A =$ [[0]]

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Cofactors Determinant and inverse of a 3x3 matrix.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Finding the Inverse of a 3x3 Matrix

Follow the steps in the questions to find the inverse of a $3 \\times 3$ matrix

", "advice": "

For simplicity, we will use the expansion of the first row to find the determinant

$$
\\begin{aligned}
\\det{A} &= a_1A_1 + b_1B_1+ c_1C_1 \\\\
&= \\var{matrixA[0][0]}
\\begin{vmatrix}
\\var{matrixA[1][1]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][1]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} - \\var{matrixA[0][1]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][2]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][2]} \\\\
\\end{vmatrix} + \\var{matrixA[0][2]} \\begin{vmatrix}
\\var{matrixA[1][0]} & \\var{matrixA[1][1]} \\\\
\\var{matrixA[2][0]} & \\var{matrixA[2][1]} \\\\
\\end{vmatrix} \\\\
&= \\var{matrixA[0][0]} \\times \\var{cof11} - \\left(\\var{matrixA[0][1]} \\times \\var{cof12}\\right) + \\var{matrixA[0][2]} \\times \\var{cof13} \\\\
&= \\var{deta}
\\end{aligned}
$$

Given arbitrary matrix

$$
A = \\begin{pmatrix}
a & b & c \\\\
d & e & f \\\\
g & h & j \\\\
\\end{pmatrix}
$$

It's cofactors are given by

$$
\\begin{aligned}
A_{11} &= +\\begin{vmatrix}
e & f \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a22} & \\var{a23} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof11}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{12} &= -\\begin{vmatrix}
d & f \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a21} & \\var{a23} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof12}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{13} &= +\\begin{vmatrix}
d & e \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a21} & \\var{a22} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof13}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{21} &= -\\begin{vmatrix}
b & c \\\\
h & j \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a32} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof21}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{22} &= +\\begin{vmatrix}
a & c \\\\
g & j \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a31} & \\var{a33} \\\\
\\end{vmatrix} \\\\
&= \\var{cof22}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{23} &= -\\begin{vmatrix}
a & b \\\\
g & h \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a31} & \\var{a32} \\\\
\\end{vmatrix} \\\\
&= \\var{cof23}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{31} &= +\\begin{vmatrix}
b & c \\\\
e & f \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a12} & \\var{a13} \\\\
\\var{a22} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof31}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{32} &= -\\begin{vmatrix}
a & c \\\\
d & f \\\\
\\end{vmatrix} \\\\
&= -\\begin{vmatrix}
\\var{a11} & \\var{a13} \\\\
\\var{a21} & \\var{a23} \\\\
\\end{vmatrix} \\\\
&= \\var{cof32}
\\end{aligned}
$$

$$
\\begin{aligned}
A_{33} &= +\\begin{vmatrix}
a & b \\\\
d & e \\\\
\\end{vmatrix} \\\\
&= +\\begin{vmatrix}
\\var{a11} & \\var{a12} \\\\
\\var{a21} & \\var{a22} \\\\
\\end{vmatrix} \\\\
&= \\var{cof33}
\\end{aligned}
$$

Using our answer from the previous question, we simply write the cofactors in the form

$$
\\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\\\
A_{21} & A_{22} & A_{23} \\\\
A_{31} & A_{32} & A_{33} \\\\
\\end{pmatrix}
$$

Giving us our matrix of cofactors

$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof12} & \\var{cof13} \\\\
\\var{cof21} & \\var{cof22} & \\var{cof23} \\\\
\\var{cof31} & \\var{cof32} & \\var{cof33} \\\\
\\end{pmatrix}
$$

The transposition process turns rows into columns and columns into rows

Carrying out this process on our matrix of cofactors gives us the adjugate

$$
\\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix}
$$

We can find the inverse of $A$ using our determinant and adjugate, using the formula

$$
A^{-1} = \\frac{1}{\\det A}(adj \\; A)
$$

Therefore, we can calculate $A^{-1}$ by

$$
\\begin{aligned}
A^{-1} &= \\frac{1}{\\var{deta}} \\begin{pmatrix}
\\var{cof11} & \\var{cof21} & \\var{cof31} \\\\
\\var{cof12} & \\var{cof22} & \\var{cof32} \\\\
\\var{cof13} & \\var{cof23} & \\var{cof33} \\\\
\\end{pmatrix} \\\\
&= \\var{inverseA}
\\end{aligned}
$$

Use this link to find some resources which will help you revise this topic.

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cof23

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Let:

$$
A=\\var{matrixA}
$$

Find the determinant of $A$

$\\det A =$ [[0]]

Calculate the nine cofactors of $A$

The cofactor $A_{ij}$ denotes the cofactor in row $i$ and column $j$

$A _{11}=$ [[0]]

$A_{12}=$ [[1]]

$A_{13}=$ [[2]]

$A_{21}=$ [[3]]

$A_{22}=$ [[4]]

$A_{23}=$ [[5]]

$A_{31}=$ [[6]]

$A_{32}=$ [[7]]

$A_{33}=$ [[8]]

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false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Using your answer to part $b)$, state the matrix of cofactors

[[0]]

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Find the transpose of your matrix from part $c)$, giving us the adjugate

[[0]]

Using your answers from all the previous parts, find the inverse of $A$

Elements will be accepted as fractions or correct to 2 decimal places

$A^{-1}=$ [[0]]

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What is the $L^{0}$ norm of the following vectors

", "advice": "

The L-zero \"norm\"

The $L^{0}$ norm of a vector is not actually a norm (by the formal definition). However, it is a useful tool with practical usage in computing. It simply counts the number of non-zero elements of a vector. So for example you can set up code that returns \"true\" (=0) and \"false\" (=1) to check if a username and password combination are correct. If they are both correct then the vector would be $(0,0)$ in which case the $L^{0}$ norm would return the value $0$ and the log in would be successful. In all other cases (such as correct username but incorrect password) the $L^{0}$ norm would return a non-zero value and you could use this to ensure the login is unsuccessful.

For more information on norms read (for example) this.

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$\\var{vector1}$

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$\\var{vector2}$

", "minValue": "answerb", "maxValue": "answerb", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{number} \\times \\var{vector1}$

", "minValue": "answerc", "maxValue": "answerc", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LN02 The L1 Norm", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the $L^{1}$ norm of the following vectors

", "advice": "

The L-one norm

The $L^{1}$ norm of a vector is found by adding up the absolute value of all of the elements of the vector.

So, introducing the $\\left| \\bf{v} \\right|_1$ notation for the $L^{1}$ norm, gives the definition:

$$
\\begin{vmatrix}
\\left(
\\begin{array}{l}
x\\\\
y\\\\
z\\\\
\\end{array}
\\right)
\\end{vmatrix}_1 =\\left|x\\right| + \\left| y \\right| + \\left| z \\right|
$$

Therefore, for example, a) has the answer:

$$
\\begin{vmatrix}
\\var{vector1}
\\end{vmatrix}_1 = \\var{advice} = \\var{answera}
$$

For more information on norms read (for example) this.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"vector1": {"name": "vector1", "group": "Ungrouped variables", "definition": "vector(repeat(random(-4..4),rowsa))", "description": "", "templateType": "anything", "can_override": false}, "vector2": {"name": "vector2", "group": "Ungrouped variables", "definition": "vector(repeat(random(-6..2),rowsb))", "description": "", "templateType": "anything", "can_override": false}, "answera": {"name": "answera", "group": "Ungrouped variables", "definition": "sumabsval(vector1)", "description": "", "templateType": "anything", "can_override": false}, "answerb": {"name": "answerb", "group": "Ungrouped variables", "definition": "sumabsval(vector2)", "description": "", "templateType": "anything", "can_override": false}, "answerc": {"name": "answerc", "group": "Ungrouped variables", "definition": "abs(number)*answera", "description": "", "templateType": "anything", "can_override": false}, "rowsa": {"name": "rowsa", "group": "Ungrouped variables", "definition": "random(2..5)", "description": "", "templateType": "anything", "can_override": false}, "rowsb": {"name": "rowsb", "group": "Ungrouped variables", "definition": "random(2,3,4)", "description": "", "templateType": "anything", "can_override": false}, "number": {"name": "number", "group": "Ungrouped variables", "definition": "random(-4,5,-6,3,-2,7)", "description": "", "templateType": "anything", "can_override": false}, "rawadviceabs": {"name": "rawadviceabs", "group": "Ungrouped variables", "definition": "map('|\\\\' + 'var{'+'vector1[{i-1}]'+'}|'+ latexcodebits(rowsa)[i-1],[dummy,i],product(1..1,1..rowsa))", "description": "", "templateType": "anything", "can_override": false}, "advice": {"name": "advice", "group": "Ungrouped variables", "definition": "latex(stringify(rawadviceabs))", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["vector1", "vector2", "answera", "answerb", "answerc", "rowsa", "rowsb", "number", "rawadviceabs", "advice"], "variable_groups": [], "functions": {"sumabsval": {"parameters": [["vector", "vector"]], "type": "number", "language": "javascript", "definition": "var output = 0;\nvar i;\n let count = 0;\n for (let i = 0; i < vector.length; i++) {\n if (vector[i] !== 0) {\n output = output + Math.abs(vector[i]);\n }\n }\n return output;"}, "Latexcodebits": {"parameters": [["n", "number"]], "type": "anything", "language": "jme", "definition": "repeat('+',n-1)+['']"}, "Stringify": {"parameters": [["input", "list"]], "type": "string", "language": "javascript", "definition": "var output = '';\nvar i;\nfor (i = 0; i < input.length; i++) {\n output += input[i];\n} \nreturn output;"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{vector1}$

$\\var{vector2}$

$\\var{number} \\times \\var{vector1}$

", "minValue": "answerc", "maxValue": "answerc", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LN03 The L2 Norm", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the $L^{2}$ norm of the following vectors

", "advice": "

The L-two norm

The $L^{2}$ norm of a vector is found by doing pythagoras' theorem and is our usual definition of \"length\".

So, introducing the $\\left| \\bf{v} \\right|_2$ notation for the $L^{2}$ norm, gives the definition (for a vector with 3 elements):

$$
\\begin{vmatrix}
\\left(
\\begin{array}{l}
x\\\\
y\\\\
z\\\\
\\end{array}
\\right)
\\end{vmatrix}_2 =\\sqrt{x^2 + y^2 + z ^2}
$$

Therefore, for example, a) has the answer:

$$
\\begin{vmatrix}
\\var{vector1}
\\end{vmatrix}_2 = \\sqrt{\\var{advice}} = \\var{answera} = \\var{roundeda}
$$

For more information on norms read (for example) this.

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$\\var{vector1}$

", "minValue": "answera", "maxValue": "answera", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "sigfig", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{vector2}$

", "minValue": "answerb", "maxValue": "answerb", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "sigfig", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\var{number} \\times \\var{vector1}$

", "minValue": "answerc", "maxValue": "answerc", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "sigfig", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LN04 The LInfinity Norm", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the $L^{\\infty}$ norm of the following vectors

", "advice": "

The L-two norm

The $L^{\\infty}$ norm of a vector is found by picking the maximum of the absolute value of the elements of the vector.

So, introducing the $\\left| \\bf{v} \\right|_\\infty$ notation for the $L^{\\infty}$ norm, gives the definition (for a vector with 3 elements):

$$
\\begin{vmatrix}
\\left(
\\begin{array}{l}
x\\\\
y\\\\
z\\\\
\\end{array}
\\right)
\\end{vmatrix}_\\infty =\\max(\\left|x\\right|,\\left|y\\right|,\\left|z\\right|)
$$

Therefore, for example, a) has the answer:

$$
\\begin{vmatrix}
\\var{vector1}
\\end{vmatrix}_\\infty = \\max(\\var{advice})=\\var{answera}
$$

For more information on norms read (for example) this.

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$\\var{vector1}$

$\\var{vector2}$

$\\var{number} \\times \\var{vector1}$

", "minValue": "answerc", "maxValue": "answerc", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Linear Algebra - Vectors", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], []], "questions": [{"name": "LV01 Adding and subtracting vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Vector Arithmetic

Work through the following questions to ensure you know how to add and subtract vectors in 2D.

For the whole of this question:

$\\bf{a} = \\var{a}$ and $\\bf{b}=\\var{b}$.

", "advice": "

The vectors in this question have two dimensions but the idea of addition and subtraction of vectors works in any number of dimensions (as long as all the vectors being added or subtracted have the same dimensions as each other).

To add two vectors you simply add their corresponding elements. In general:

$$
\\left(\\begin{array}{c}
a \\\\
b \\\\
\\end{array}\\right) +
\\left(\\begin{array}{c}
c \\\\
d \\\\
\\end{array}\\right) =
\\left(\\begin{array}{c}
a+c \\\\
b+d \\\\
\\end{array}\\right).
$$

Subtraction works in the same way so we have:

$$
\\var{a} + \\var{b} = \\var{a+b}.
$$

$$
\\var{a} - \\var{b} = \\var{a-b}.
$$

In order to undertstand the third part of the question you need to know what a \"position vector\" and \"direction vector\" are.

A position vector is defined as a vector that symbolises the location of any given point with respect to the origin. It can be thought of as a coordinate point, but written as a column vector - top entry is the \"x-coordinate\" and the bottome entry is the \"y-coordinate\".

A direction vector is defined as a vector that symbolises a direction and a distance in that direction but with no specified \"starting point\". In 2D it can be summarized as an instruction to go the top element number of units left or right based on the sign of the element and the bottom element number of units up or down based on the sign of the element.

So the direction vector from $A$ to $B$ can be worked out by looking at a route from $A$ to $B$ that travels along the position vectors given. Starting at $A$ we have to go backwards down $\\bf{a}$ to the origin and then forwards along $\\bf{b}$. This corresponds to doing \"minus\" $\\bf{a}$ and \"positive\" $\\bf{b}$:

$$
\\vec{AB} = (-)\\bf{a} + \\bf{b} = \\bf{b}-\\bf{a} = \\var{b}-\\var{a} = \\var{b-a}.
$$

Use this link to find some resources which will help you revise this topic.

Calculate $\\bf{a}+\\bf{b}$.

", "correctAnswer": "answeradd", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "matrix", "useCustomName": true, "customName": "2)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate $\\bf{a} - \\bf{b}$.

Let $\\bf{a}$ be the position vector of point $A$ and $\\bf{b}$ be the position vector of point $B$. Find the direction vector $\\vec{AB}$.

", "correctAnswer": "answerab", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV02 Scalar multiplication of vectors", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Vector Arithmetic

Work through the following questions exploring how to multiply a vector by a scalar.

For the whole of this question:

$\\bf{a} = \\var{a}$ and $\\bf{b}=\\var{b}$.

", "advice": "

The vectors in this question have two dimensions but the ideas herein work in any number of dimensions.

To multiply a vector by a scalar (number) you just multiply each element by that scalar:

$$
k\\left(\\begin{array}{c}
a \\\\
b \\\\
\\end{array}\\right) =
\\left(\\begin{array}{c}
ak \\\\
bk \\\\
\\end{array}\\right).
$$

So we have:

$$
\\var{m}\\var{a} = \\var{m*a}.
$$

The second and third part of this question just combine this idea of multiplying a vector by a scalar and the idea that addition and subtraction work by just calculating element by element (as long as all the vectors involved have the same dimensions).

$$
\\var{p}\\var{a} + \\var{q}\\var{b} =
\\left(\\begin{array}{c}
\\var{p} \\times \\var{a[0]} + \\var{q} \\times \\var{b[0]} \\\\
\\var{p}\\times \\var{a[1]} + \\var{q} \\times \\var{b[1]}\\\\
\\end{array}\\right) = \\var{p*a+q*b}.
$$

$$
\\var{r}\\var{a} - \\var{s}\\var{b} =
\\left(\\begin{array}{c}
\\var{r} \\times \\var{a[0]} - \\var{s} \\times \\var{b[0]} \\\\
\\var{s}\\times \\var{a[1]} - \\var{s} \\times \\var{b[1]}\\\\
\\end{array}\\right) = \\var{r*a-s*b}.
$$

Use this link to find some resources which will help you revise this topic.

Calculate $\\var{m}\\bf{a}$.

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Calculate $\\var{p}\\bf{a}+\\var{q}\\bf{b}$.

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Calculate $\\var{r}\\bf{a} - \\var{s}\\bf{b}$.

", "correctAnswer": "answersub", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "LV03 Vectors and geometry", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

This question is about understanding the use of vectors to prove that a shape is a trapezium.

The quadrilateral ABCD is shown below.

$$
A = (\\var{a[0]},\\var{a[1]})\\\\
B = (\\var{b[0]},\\var{b[1]})\\\\
C = (\\var{c[0]},\\var{c[1]})\\\\
D = (\\var{d[0]},\\var{d[1]})
$$

{geogebra_applet('https://www.geogebra.org/m/xb9bvcaa',defs)}

", "advice": "

Vectors between points

In order to undertstand how to find the vector between two points it is helpful to know what a \"position vector\" and \"direction vector\" are. This advice should cover parts a), b), e) and f) of this question.

The point $B$ has coordinates $(\\var{B[0]},\\var{B[1]})$ and it has position vector, denoted $\\bf{b}$, given as $\\bf{b} = \\var{b}$.

So the direction vector from $B$ to $C$ can be worked out by looking at a route from $B$ to $C$ that travels along the position vectors given. Starting at $B$ we have to go backwards down $\\bf{b}$ to the origin and then forwards along $\\bf{c}$. This corresponds to doing \"minus\" $\\bf{b}$ and \"positive\" $\\bf{c}$:

$$
\\vec{BC} = (-)\\bf{b} + \\bf{c} = \\bf{c}-\\bf{b} = \\var{c}-\\var{b} = \\var{c-b}.
$$

Parallel vectors

If one vector is a multiple of another then they are vectors that point in the same direction. This means they are parallel. You just need to check what the multplier is between corresponding elements in each vector. If it is the same for both pairs of elements then the vectors are parallel (and if not then they are not).

For example, $\\vec{BC} = \\var{bc}$ and $\\vec{AD} = \\var{k*BC}.$ Since $\\frac{\\var{k*BC[0]}}{\\var{BC[0]}} = \\var{k}$ which gives the same multiplier as $\\frac{\\var{k*BC[1]}}{\\var{BC[1]}} = \\var{k}$ then $\\vec{BC}$ and $\\vec{AD}$ are parallel.

Conclusions about shapes

This question is looking at a trapezium specifically. The key properties of a trapezium are that it is a quadrilateral and there is one pair of parallel sides. This question goes through establishing that one pair of sides are parallel and then does the calculations to show that the other pair is not parallel. At this point we can conclude that $ABCD$ is a trapezium.

Use this link to find some resources which will help you revise this topic.

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Find the vector $\\vec{BC}$.

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Find the vector $\\vec{AD}$.

", "correctAnswer": "-a+d", "correctAnswerFractions": false, "numRows": "2", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}, {"type": "numberentry", "useCustomName": true, "customName": "c)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Hence, it is possible to write $\\vec{AD} = k \\times \\vec{BC}$. Find the value of $k$.

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The fact that $\\vec{AD}$ can be written as a multiplier times by $\\vec{BC}$ means:

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Find the vector $\\vec{AB}$.

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Find the vector $\\vec{CD}$.

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Up to this point in the question we have now shown which of the following:

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the scalar (or dot) product of $\\bf{a}$ and $\\bf{b}$.

", "advice": "

It is important to note that for vectors there is more than one type of multiplication. This question is specifically about the scalar (or dot) product.

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2},$ the scalar (or dot) product is defined as

$$
\\mathbf{v \\cdot w} = v_1 \\times w_1 + v_2 \\times w_2.
$$

So for this question:

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b}\\\\
\\bf{a} \\cdot \\bf{b} = \\var{a[0]}\\times\\var{b[0]} + \\var{a[1]}\\times\\var{b[1]} = \\var{adotb}.
$$

Use this link to find some resources which will help you revise this topic.

$\\bf{a} \\cdot \\bf{b} =$ [[0]]

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the scalar (or dot) product of $\\bf{a}$ and $\\bf{b}$.

", "advice": "

It is important to note that for vectors there is more than one type of multiplication. This question is specifically about the scalar (or dot) product.

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2 \\\\ v_3},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2 \\\\ w_3},$ the scalar (or dot) product is defined as

$$
\\mathbf{v \\cdot w} = v_1 \\times w_1 + v_2 \\times w_2 + v_3 \\times w_3.
$$

So for this question:

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b}\\\\
\\bf{a} \\cdot \\bf{b} = \\var{a[0]}\\times\\var{b[0]} + \\var{a[1]}\\times\\var{b[1]} + \\var{a[2]}\\times\\var{b[2]} = \\var{adotb}.
$$

Use this link to find some resources which will help you revise this topic.

$\\bf{a} \\cdot \\bf{b} =$ [[0]]

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the angle between $\\bf{a}$ and $\\bf{b}$.

", "advice": "

To answer these questions, we want to use the equations for the scalar product. Recall:

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2},$

\\[ \\begin{split} \\mathbf{v \\cdot w} &\\,= v_1 \\times w_1 + v_2 \\times w_2 \\\\\\\\ \\mathbf{v \\cdot w} &\\,= |\\mathbf v| |\\mathbf w | \\cos(\\theta), \\end{split} \\]

where $|\\mathbf v|$ and $|\\mathbf w|$ are the magnitudes of the vectors, and $\\theta$ is the angle between the vectors.

For

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b},
$$

we have

$$
\\bf{a} \\cdot \\bf{b} = \\var{adotb},
$$

$$
|\\bf{a}| = \\sqrt{\\var{a[0]^2 +a[1]^2}} \\approx \\var{precround(asize,2)},\\\\
|\\bf{b}| = \\sqrt{\\var{b[0]^2 +b[1]^2}} \\approx \\var{precround(bsize,2)}.
$$

So we have:
$$
\\theta = \\cos^{-1}\\left(\\frac{\\var{adotb}}{\\sqrt{\\var{a[0]^2 +a[1]^2}}\\times\\sqrt{\\var{b[0]^2 +b[1]^2}}}\\right) = \\var{precround(angle,1)}.
$$

Use this link to find some resources which will help you revise this topic.

The angle between $\\bf{a}$ and $\\bf{b}$ is [[0]] degrees.

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\var{b}$.

Find the angle between $\\bf{a}$ and $\\bf{b}$.

", "advice": "

To answer these questions, we want to use the equations for the scalar product. Recall:

For the vectors $ \\mathbf v = \\pmatrix{v_1 \\\\ v_2 \\\\ v_3},\\, \\mathbf w = \\pmatrix{w_1 \\\\ w_2 \\\\ w_3},$

\\[ \\begin{split} \\mathbf{v \\cdot w} &\\,= v_1 \\times w_1 + v_2 \\times w_2 + v_3 \\times w_3 \\\\\\\\ \\mathbf{v \\cdot w} &\\,= |\\mathbf v| |\\mathbf w | \\cos(\\theta), \\end{split} \\]

where $|\\mathbf v|$ and $|\\mathbf w|$ are the magnitudes of the vectors, and $\\theta$ is the angle between the vectors.

For

$$
\\bf{a} = \\var{a} \\qquad \\text{and} \\qquad \\bf{b} = \\var{b},
$$

we have

$$
\\bf{a} \\cdot \\bf{b} = \\var{adotb},
$$

and

$$
|\\bf{a}| = \\sqrt{\\var{a[0]^2 +a[1]^2 +a[2]^2}} \\approx \\var{precround(asize,2)},\\\\
|\\bf{b}| = \\sqrt{\\var{b[0]^2 +b[1]^2 +b[2]^2}}\\approx \\var{precround(bsize,2)}.
$$

So we have:
$$
\\theta = \\cos^{-1}\\left(\\frac{\\var{adotb}}{\\sqrt{\\var{a[0]^2 +a[1]^2 +a[2]^2}}\\times\\sqrt{\\var{b[0]^2 +b[1]^2 +b[2]^2}}}\\right) = \\var{precround(angle,1)}.
$$

Use this link to find some resources which will help you revise this topic.

The angle between $\\bf{a}$ and $\\bf{b}$ is [[0]] degrees.

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\pmatrix{\\var{b1} \\\\ k}$ and that the two vectors are perpendicular.

Find the value of $k$.

", "advice": "

The key thing to understand for this question is that for perpendicular vectors the scalar (or dot) product will give a result of zero.

In this question we have,

\\begin{alignat}{2}
&\\quad
&\\var{a}\\cdot\\pmatrix{\\var{b1} \\\\ k}
& = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]}\\times\\var{b1} + \\var{a[1]} \\times k & = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]*b1} + \\var{a[1]}k & = 0.
\\end{alignat}

Solving this then gives,

$$
k = \\var{k}.
$$

Use this link to find some resources which will help you revise this topic.

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It is given $\\bf{a} = \\var{a}$ and $\\bf{b} = \\pmatrix{\\var{b1} \\\\ k \\\\ \\var{b3}}$ and that the two vectors are perpendicular.

Find the value of $k$.

", "advice": "

The key thing to understand for this question is that for perpendicular vectors the scalar (or dot) product will give a result of zero.

In this question we have,

\\begin{alignat}{2}
&\\quad
&\\var{a}\\cdot\\pmatrix{\\var{b1} \\\\ k \\\\ \\var{b3}}
& = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]}\\times\\var{b1} + \\var{a[1]} \\times k + \\var{a[2]} \\times \\var{b3} & = 0 \\\\
&\\Rightarrow\\quad
&\\var{a[0]*b1 + a[2]*b3} + \\var{a[1]}k & = 0.
\\end{alignat}

Solving this then gives,

$$
k = \\var{k}.
$$

Use this link to find some resources which will help you revise this topic.

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The vectors $\\bf a$ and $\\bf b$ are defined as follows,
$$
{\\bf a} = \\var{a} \\qquad \\text{and} \\qquad \\bf b = \\var{b}.
$$

", "advice": "

The vector product (also referred to as the cross product) of two vectors can be calculated as follows:

$$
\\pmatrix{a_1 \\\\ a_2 \\\\ a_3} \\times \\pmatrix{b_1 \\\\ b_2 \\\\ b_3} = \\pmatrix{a_2b_3-a_3b_2 \\\\ -(a_1b_3-a_3b_1) \\\\ a_1b_2-a_2b_1}.
$$

A way to think of this is as the following determinant:

$$
\\begin{vmatrix}
\\bf i & \\bf j & \\bf k\\\\
a_1 & a_2 & a_3 \\\\
b_1 & b_2 & b_3
\\end{vmatrix},
$$

where $\\bf i$, $\\bf j $, and $\\bf k$ are the standard unit basis vectors.

In this question we therefore have:

$$
\\begin{align*}
\\var{a} \\times \\var{b} & = \\pmatrix{\\var{a[1]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[1]} \\\\ -(\\var{a[0]} \\times \\var{b[2]} - \\var{a[2]} \\times \\var{b[0]}) \\\\ \\var{a[0]} \\times \\var{b[1]} - \\var{a[1]} \\times \\var{b[0]}}\\\\
& = \\var{answer}.
\\end{align*}
$$

Use this link to find some resources which will help you revise this topic.

Find $\\bf a \\times \\bf b$.

", "correctAnswer": "answer", "correctAnswerFractions": false, "numRows": "3", "numColumns": 1, "allowResize": false, "tolerance": 0, "markPerCell": false, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0, "prefilledCells": ""}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Number - Arithmetic", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "NA01 Money Calculation - Make Change", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculate the change from a note given two costs.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

You buy a sandwich for {PriceS} and a drink for {PriceD}.

You pay with a £5 bank note, what is the correct change in pounds?

", "advice": "\n\n\n\n\n\n

To find the total cost of the sandwich and drink, add them together:

{PriceS} $+$ {PriceD} $=$ {Sumtext}

If you pay with a £5 banknote, your change will be £5 minus the sum of the sandwich and drink:

£5.00 $-$ {Sumtext} = {answertext}

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"PriceD": {"name": "PriceD", "group": "Prices", "definition": "currency(PD,\"\u00a3\",\"p\")", "description": "", "templateType": "anything", "can_override": false}, "PriceS": {"name": "PriceS", "group": "Prices", "definition": "currency(PS,\"\u00a3\",\"p\")", "description": "", "templateType": "anything", "can_override": false}, "PS": {"name": "PS", "group": "Prices", "definition": "random(0.5 .. 2.5#0.01)", "description": "", "templateType": "randrange", "can_override": false}, "PD": {"name": "PD", "group": "Prices", "definition": "random(0.5 .. 2.5#0.01)", "description": "", "templateType": "randrange", "can_override": false}, "SUM": {"name": "SUM", "group": "For explanation", "definition": "PS+PD", "description": "", "templateType": "anything", "can_override": false}, "Sumtext": {"name": "Sumtext", "group": "For explanation", "definition": "currency(SUM,\"\u00a3\",\"p\")", "description": "", "templateType": "anything", "can_override": false}, "ANSWER": {"name": "ANSWER", "group": "For explanation", "definition": "5 - sum", "description": "", "templateType": "anything", "can_override": false}, "Answertext": {"name": "Answertext", "group": "For explanation", "definition": "currency(answer,\"\u00a3\",\"p\")", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "PS>PD", "maxRuns": "20"}, "ungrouped_variables": [], "variable_groups": [{"name": "Prices", "variables": ["PD", "PriceD", "PriceS", "PS"]}, {"name": "For explanation", "variables": ["Sumtext", "SUM", "ANSWER", "Answertext"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

£[[0]]

Whole number division in a context of number of tablets per day based on a tablet size and a daily prescribed amount.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

You prescribe a patient {p} milligrams (mg) per day of a particular drug. The drug is available in tablets of {t} milligrams (mg). How many tablets should you instruct the patient to take each day?

", "advice": "\n\n\n\n\n\n

Since each tablet is {t} mg of the necessary drug, we must prescribe enough tablets to add up to {p} milligrams. By dividing {p} by {t} we can find out how many {t} mg tablet make up {p} mg, which gives {answer}.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"p": {"name": "p", "group": "Ungrouped variables", "definition": "random(300 .. 600#50)", "description": "", "templateType": "randrange", "can_override": false}, "t": {"name": "t", "group": "Ungrouped variables", "definition": "random(100 .. 300#50)", "description": "", "templateType": "randrange", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "p/t", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "mod(p,t)=0", "maxRuns": 100}, "ungrouped_variables": ["p", "t", "answer"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The daily number of tablets prescribed should be: [[0]]

Division resulting in decimals in the context of a dose per hour.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

You are required to give an infusion of a drug of {d}ml over {t}hours, what is the rate per hour?

(answer to 1 decimal place)

", "advice": "\n\n\n\n\n\n

If there is {d}ml in {t} hours, then {d} should be divided by {t} to get {answert}ml per hour.

Make use of the following resource(s) to revise if necessary:

Use this link to find some resources which will help you revise this topic

If you were unsure how to round your answer then look at the following resource(s):

Use this link to find some resources which will help you revise this topic

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"d": {"name": "d", "group": "Ungrouped variables", "definition": "random(250 .. 1000#50)", "description": "", "templateType": "randrange", "can_override": false}, "t": {"name": "t", "group": "Ungrouped variables", "definition": "random(4 .. 10#1)", "description": "", "templateType": "randrange", "can_override": false}, "answert": {"name": "answert", "group": "Ungrouped variables", "definition": "dpformat(answer,1)", "description": "", "templateType": "anything", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "d/t", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["d", "t", "answert", "answer"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]] ml per hour

The answer is either 'True' or 'False'

", "help_url": "", "input_widget": "radios", "input_options": {"correctAnswer": "if(eval(settings[\"correct_answer_expr\"]), 0, 1)", "hint": {"static": true, "value": ""}, "choices": {"static": true, "value": ["True", "False"]}}, "can_be_gap": true, "can_be_step": true, "marking_script": "mark:\nif(studentanswer=correct_answer,\n correct(),\n incorrect()\n)\n\ninterpreted_answer:\nstudentAnswer=0\n\ncorrect_answer:\nif(eval(settings[\"correct_answer_expr\"]),0,1)", "marking_notes": [{"name": "mark", "description": "This is the main marking note. It should award credit and provide feedback based on the student's answer.", "definition": "if(studentanswer=correct_answer,\n correct(),\n incorrect()\n)"}, {"name": "interpreted_answer", "description": "A value representing the student's answer to this part.", "definition": "studentAnswer=0"}, {"name": "correct_answer", "description": "", "definition": "if(eval(settings[\"correct_answer_expr\"]),0,1)"}], "settings": [{"name": "correct_answer_expr", "label": "Is the answer \"True\"", "help_url": "", "hint": "", "input_type": "mathematical_expression", "default_value": "true", "subvars": false}], "public_availability": "always", "published": true, "extensions": []}], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}, {"name": "Merryn Horrocks", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4052/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Identify the truth value of an inequality (T/F) between two numbers

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

The symbol is a comparitor that $<$ says that the object to its left is less than the object to its right. Similarly, $>$ says the object to its left is greater than the object to its right.

In each case the sentence can be read in two ways e.g. $a<b$ can be read as \"$a$ is less than $b$\" OR \"$b$ is greater than $a$\".

The symbol $\\leq$ just changes the sentence to include \"...or equal to...\". This can be of particular relevance when dealing with integers (whole numbers) e.g. $x \\geq 4$ and $x$ is a whole number means that $x$ could be $4, 5, 6,$ or $7$ and so on. Whereas, $x>4$ and $x$ is a whole number means that $x$ could be $5,6,$... etc. Notably in the second case $x$ cannot be $4$.

Use this link to find some resources which will help you revise this topic.

Is the following statement true or false?

\$\\var{disp_values[0]}\\var{disp_op}\\var{disp_values[1]}\$

", "alternatives": [{"type": "true-false", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "settings": {"correct_answer_expr": "eval(expression(disp_values[0]+if(disp_op=\"\\\\gt\",\">\",\">=\")+disp_values[1]))"}}], "settings": {"correct_answer_expr": "eval(expression(disp_values[0]+if(disp_op=\"\\\\gt\",\">\",\">=\")+disp_values[1]))"}}, {"type": "true-false", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Is the following statement true or false?

\$\\var{disp_values2[0]}\\var{disp_op2}\\var{disp_values2[1]}\$

", "settings": {"correct_answer_expr": "eval(expression(disp_values2[0]+if(disp_op2=\"\\\\lt\",\"<\",\"<=\")+disp_values2[1]))"}}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA05 Number sequences - find a missing number", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Marie Nicholson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1799/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Number sequences - find a missing number

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Find the missing number from the following sequence

\\[\\var{a}, \\var{a+d}, \\var{a+2*d}, \\var{a+3d}, ?, \\var{a+5d}, \\var{a+6d}, \\var{a+7d}, \\ldots\\]

", "advice": "

In this sequence it is important to spot that the terms are changing by the same amount each time ($\\var{d}$ in this case). This type of sequence is called an Arithmetic Sequence.

Use this link to find some resources which will help you revise this topic.

[[0]]

Multiples, factors, lowest common multiples and highest common factors.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

Factors multiply together in pairs to give the original number (e.g., the factors of 10 are 1, 2, 5 and 10, because 1x10=10 and 2x5=10.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"ca2": {"name": "ca2", "group": "Ungrouped variables", "definition": "random(answers except ca1)", "description": "", "templateType": "anything", "can_override": false}, "ca1": {"name": "ca1", "group": "Ungrouped variables", "definition": "random(answers)", "description": "", "templateType": "anything", "can_override": false}, "hcf": {"name": "hcf", "group": "Ungrouped variables", "definition": "gcd(k,l)", "description": "", "templateType": "anything", "can_override": false}, "lcm4": {"name": "lcm4", "group": "Ungrouped variables", "definition": "lcm(p,q)", "description": "", "templateType": "anything", "can_override": false}, "lcm2": {"name": "lcm2", "group": "Ungrouped variables", "definition": "lcm(n,m)", "description": "", "templateType": "anything", "can_override": false}, "product": {"name": "product", "group": "Ungrouped variables", "definition": "a*b*c", "description": "", "templateType": "anything", "can_override": false}, "answers": {"name": "answers", "group": "Ungrouped variables", "definition": "[a,b,c,a*b,a*c,b*c,1,product]", "description": "", "templateType": "anything", "can_override": false}, "hcf3": {"name": "hcf3", "group": "Ungrouped variables", "definition": "gcd(p,q)", "description": "", "templateType": "anything", "can_override": false}, "hcf2": {"name": "hcf2", "group": "Ungrouped variables", "definition": "gcd(n,m)+1", "description": "", "templateType": "anything", "can_override": false}, "lcm": {"name": "lcm", "group": "Ungrouped variables", "definition": "lcm(k,l)", "description": "", "templateType": "anything", "can_override": false}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(1,2,3)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(5,7)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(2,3)", "description": "", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(1..12)", "description": "", "templateType": "anything", "can_override": false}, "g": {"name": "g", "group": "Ungrouped variables", "definition": "random(2..10 except b)", "description": "", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(1..100 except [d*1,d*2,d*3,d*4,d*5,d*6,d*7,d*8,d*9,d*10,f*1,f*2,f*3,f*4,f*5,f*6,f*7,f*8,f*9,f*10,g*1,g*2,g*3,g*4,g*5,g*6,g*7,g*8,g*9,g*10])", "description": "", "templateType": "anything", "can_override": false}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random(4..12)", "description": "", "templateType": "anything", "can_override": false}, "j": {"name": "j", "group": "Ungrouped variables", "definition": "random(1..50 except [d*1,d*2,d*3,d*4,d*5,d*6,d*7,d*8,d*9,d*10,f*1,f*2,f*3,f*4,f*5,f*6,f*7,f*8,f*9,f*10,g*1,g*2,g*3,g*4,g*5,g*6,g*7,g*8,g*9,g*10,h])", "description": "", "templateType": "anything", "can_override": false}, "m": {"name": "m", "group": "Ungrouped variables", "definition": "random(2..15 except [k,l,lcm])", "description": "", "templateType": "anything", "can_override": false}, "l": {"name": "l", "group": "Ungrouped variables", "definition": "random(4..15 except k)", "description": "", "templateType": "anything", "can_override": false}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "random(6..15 except [k,l,lcm,m])", "description": "", "templateType": "anything", "can_override": false}, "q": {"name": "q", "group": "Ungrouped variables", "definition": "random(4..36 except p)", "description": "", "templateType": "anything", "can_override": false}, "p": {"name": "p", "group": "Ungrouped variables", "definition": "random(6..24)", "description": "", "templateType": "anything", "can_override": false}, "wa2": {"name": "wa2", "group": "Ungrouped variables", "definition": "random(1..product except [a,b,c,a*b,a*c,b*c,1,product,wa1])", "description": "", "templateType": "anything", "can_override": false}, "wa1": {"name": "wa1", "group": "Ungrouped variables", "definition": "random(1..product except answers)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["hcf2", "lcm2", "a", "b", "c", "product", "answers", "wa1", "wa2", "ca1", "ca2", "d", "f", "g", "h", "j", "k", "l", "lcm", "m", "n", "hcf", "p", "q", "hcf3", "lcm4"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Choose all of the factors of $\\var{product}$.

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "checkbox", "displayColumns": 0, "minAnswers": 0, "maxAnswers": "0", "warningType": "warn", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["

$\\var{ca1}$

", "

$\\var{ca2}$

", "

$\\var{wa1}$

", "

$\\var{wa2}$

"], "matrix": ["1", "1", 0, 0], "distractors": ["", "", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA07 HCF", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "Calculating the LCM and HCF of numbers by using prime factorisation.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

By considering the prime factorisation of $\\var{x}$ and $\\var{y}$, or otherwise, find the highest common factor (HCF) of $\\var{x}$ and $\\var{y}$.

", "advice": "

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$

For HCF of $\\var{x}$ and $\\var{y}$ we need to multiply each prime factor the least number of times it occurs in either $\\var{x}$ or $\\var{y}$

i.e. HCF$(x,y) = \\var{show_factors(hcf_xy)}=\\var{hcf_xy}$

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x_powers": {"name": "x_powers", "group": "Ungrouped variables", "definition": "[random(1..4),random(0..4),random(0..3),random(0..3)]", "description": "", "templateType": "anything", "can_override": false}, "y_powers": {"name": "y_powers", "group": "Ungrouped variables", "definition": "[random(0..4),random(1..4),random(0..3),random(0..2)]", "description": "", "templateType": "anything", "can_override": false}, "x": {"name": "x", "group": "Ungrouped variables", "definition": "2^x_powers[0]*3^x_powers[1]*5^x_powers[2]*7^x_powers[3]", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "2^y_powers[0]*3^y_powers[1]*5^y_powers[2]*7^y_powers[3]", "description": "", "templateType": "anything", "can_override": false}, "hcf_xy": {"name": "hcf_xy", "group": "Ungrouped variables", "definition": "2^min(x_powers[0],y_powers[0])*3^min(x_powers[1],y_powers[1])*5^min(x_powers[2],y_powers[2])*7^min(x_powers[3],y_powers[3])", "description": "", "templateType": "anything", "can_override": false}, "lcm_xy": {"name": "lcm_xy", "group": "Ungrouped variables", "definition": "2^max(x_powers[0],y_powers[0])*3^max(x_powers[1],y_powers[1])*5^max(x_powers[2],y_powers[2])*7^max(x_powers[3],y_powers[3])", "description": "", "templateType": "anything", "can_override": false}, "primes": {"name": "primes", "group": "Ungrouped variables", "definition": "[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "x_powers[0]+x_powers[1]+x_powers[2]+x_powers[3]<5\nand\nx_powers[0]+x_powers[1]+x_powers[2]+x_powers[3]>2\nand\ny_powers[0]+y_powers[1]+y_powers[2]+y_powers[3]<5\nand\ny_powers[0]+y_powers[1]+y_powers[2]+y_powers[3]>2\nand (x-y)<>0\nand hcf_xy>6", "maxRuns": "500"}, "ungrouped_variables": ["x_powers", "y_powers", "x", "y", "hcf_xy", "lcm_xy", "primes"], "variable_groups": [], "functions": {"show_factors": {"parameters": [["n", "number"]], "type": "string", "language": "jme", "definition": "latex( // mark the output as a string of raw LaTeX\n join(\n map(\n if(a=1,p,p+'^{'+a+'}'), // when the exponent is 1, return p, otherwise return p^{exponent}\n [p,a],\n filter(x[1]>0,x,zip(primes,factorise(n))) // for all the primes p which are factors of n, return p and its exponent\n ),\n ' \\\\times ' // join all the prime powers up with \\times symbols\n )\n)"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "hcf_xy", "maxValue": "hcf_xy", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA08 LCM", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "Calculating the LCM and HCF of numbers by using prime factorisation.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

By considering the prime factorisation of $\\var{x}$ and $\\var{y}$, or otherwise, find the lowest common multiple (LCM) of $\\var{x}$ and $\\var{y}$.

", "advice": "

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$.

For LCM of $\\var{x}$ and $\\var{y}$ we need to multiply each factor the greatest number of times it occurs in either $\\var{x}$ or $\\var{y}$.

i.e. LCM$(x,y) = \\var{show_factors(lcm_xy)}=\\var{lcm_xy}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x_powers": {"name": "x_powers", "group": "Ungrouped variables", "definition": "[random(1..4),random(0..4),random(0..3),random(0..3)]", "description": "", "templateType": "anything", "can_override": false}, "y_powers": {"name": "y_powers", "group": "Ungrouped variables", "definition": "[random(0..4),random(1..4),random(0..3),random(0..2)]", "description": "", "templateType": "anything", "can_override": false}, "x": {"name": "x", "group": "Ungrouped variables", "definition": "2^x_powers[0]*3^x_powers[1]*5^x_powers[2]*7^x_powers[3]", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "2^y_powers[0]*3^y_powers[1]*5^y_powers[2]*7^y_powers[3]", "description": "", "templateType": "anything", "can_override": false}, "hcf_xy": {"name": "hcf_xy", "group": "Ungrouped variables", "definition": "2^min(x_powers[0],y_powers[0])*3^min(x_powers[1],y_powers[1])*5^min(x_powers[2],y_powers[2])*7^min(x_powers[3],y_powers[3])", "description": "", "templateType": "anything", "can_override": false}, "lcm_xy": {"name": "lcm_xy", "group": "Ungrouped variables", "definition": "2^max(x_powers[0],y_powers[0])*3^max(x_powers[1],y_powers[1])*5^max(x_powers[2],y_powers[2])*7^max(x_powers[3],y_powers[3])", "description": "", "templateType": "anything", "can_override": false}, "primes": {"name": "primes", "group": "Ungrouped variables", "definition": "[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "x_powers[0]+x_powers[1]+x_powers[2]+x_powers[3]<5\nand\nx_powers[0]+x_powers[1]+x_powers[2]+x_powers[3]>2\nand\ny_powers[0]+y_powers[1]+y_powers[2]+y_powers[3]<5\nand\ny_powers[0]+y_powers[1]+y_powers[2]+y_powers[3]>2\nand (x-y)<>0\nand hcf_xy>6", "maxRuns": "500"}, "ungrouped_variables": ["x_powers", "y_powers", "x", "y", "hcf_xy", "lcm_xy", "primes"], "variable_groups": [], "functions": {"show_factors": {"parameters": [["n", "number"]], "type": "string", "language": "jme", "definition": "latex( // mark the output as a string of raw LaTeX\n join(\n map(\n if(a=1,p,p+'^{'+a+'}'), // when the exponent is 1, return p, otherwise return p^{exponent}\n [p,a],\n filter(x[1]>0,x,zip(primes,factorise(n))) // for all the primes p which are factors of n, return p and its exponent\n ),\n ' \\\\times ' // join all the prime powers up with \\times symbols\n )\n)"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "lcm_xy", "maxValue": "lcm_xy", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA09 BIDMAS without a division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

And is a way for us to remember guidance about the order in which calculations are carried out to ensure that everyone doing teh same sum gets the same answer. In this case the first thing that is in the question is Multiplication.

First work through the expression from left to right, evaluating any multiplication as you come to them. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{a}-\\var{b} \\times \\var{c}\\]

\\[=\\var{a}-\\var{b*c}\\]

\\[=\\var{a-b*c}\\]

Use this link to find some resources which will help you revise this topic.

Calculate

$\\var{a}-\\var{b} \\times\\var{c}$

", "minValue": "{a-b*c}", "maxValue": "{a-b*c}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA10 BIDMAS with a division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Questions testing understanding of the precedence of operators using BIDMAS, applied to integers. These questions only test DMAS. That is, only Division/Multiplcation and Addition/Subtraction.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following expression:

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

First work through the expression from left to right, evaluating any division as you come to it. You should be left with an expression involving only pluses and minuses. Evaluate this expression, again working from left to right. Thus:

\\[\\var{h}-\\var{a2*b2} \\div \\var{b2}\\]

\\[=\\var{h}-\\var{a2}\\]

\\[=\\var{h-a2}\\]

Use this link to find some resources which will help you revise this topic.

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$\\var{h}-\\var{a2*b2} \\div \\var{b2}$

", "minValue": "{h-a2}", "maxValue": "{h-a2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA11 BIDMAS with a division 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Applying the order of operators.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

To calculate the following expression you press a sequence of buttons on your calculator.

\\begin{align}\\frac{\\var{num}}{\\var{a}\\times\\var{b}}\\end{align}

Which of the following would give the WRONG answer?

", "advice": "

BIDMAS stands for:

Brackets

Indices

Division

Multiplication

Addition

Subtraction

This is the standardized order of operations that we carry out and is part of how the calculator is designed to work. The most effective way to use most modern calculators is to use either the fraction button (on scientific calculators) or as is hinted at in this question, use brackets.

Use this link to find some resources which will help you revise this topic.

Calculations with negative numbers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate $\\var{x}+(\\var{y})$.

", "advice": "

When you add a negative number that is the same as subtracting the number so

\\[\\var{x}+(\\var{y})=\\var{x}-\\var{-y}=\\var{x+y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(1..50)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-50..-1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x}+{y}", "maxValue": "{x}+{y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA13 Negatives (add/subtract) 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculations with negative numbers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate $\\var{x}-(\\var{y})$.

", "advice": "

When you subtract a negative number that is the same as adding the number so

\\[\\var{x}-(\\var{y})=\\var{x}+\\var{-y}=\\var{x-y}\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(1..50)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-50..-1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x}-{y}", "maxValue": "{x}-{y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA14 Multiplying Negatives 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculations with negative numbers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate $\\var{x}\\times(\\var{y})$.

", "advice": "

When you multiply by a negative number that is the same as doing the multiplication as if the numbers were positive and then making the result negative. This means we have

\\[\\var{x}\\times(\\var{y})=-(\\var{x}\\times\\var{-y})=\\var{x*y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(1..10)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-10..-1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x*y}", "maxValue": "{x*y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA15 Multiplying Negatives 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculations with negative numbers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate $(\\var{x})\\times(\\var{y})$.

", "advice": "

Multiplying two negative numbers gives a positive so we just calculate the multiplication as if both numbers were positive. This means we have

\\[(\\var{x})\\times(\\var{y})=\\var{-x}\\times\\var{-y}=\\var{x*y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(-10..-1)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-10..-1)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x*y}", "maxValue": "{x*y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA16 Dividing Negatives", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculations with negative numbers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Calculate $(\\var{x})\\div(\\var{y})$.

", "advice": "

When we divide two numbers the rule is,

plus $\\div$ plus = plus
minus $\\div$ plus = minus
plus $\\div$ minus = minus
minus $\\div$ minus = plus

In this calculation we have

\\[(\\var{x})\\div(\\var{y})=\\var{x/y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(-10..10)*y", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(-10..10 except 0)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{x/y}", "maxValue": "{x/y}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NA17 Worded proportion", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Proportional calculation in context.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

If {n1} adult cinema tickets cost £{n1price}, how much would it cost for {n2} adults to buy tickets to the cinema?

", "advice": "

If {n1} tickets cost £{n1price} then we can calculate that one ticket costs

\\[£\\var{n1price}\\div\\var{n1}=£\\var{price}.\\]

This means that {n2} tickets cost $\\var{n2}\\times£\\var{price}=£\\var{n2price}$

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"n1": {"name": "n1", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "templateType": "anything", "can_override": false}, "price": {"name": "price", "group": "Ungrouped variables", "definition": "precround(random(2..9 # 0.01),2)", "description": "", "templateType": "anything", "can_override": false}, "n2": {"name": "n2", "group": "Ungrouped variables", "definition": "random(2..9 except n1)", "description": "", "templateType": "anything", "can_override": false}, "n1price": {"name": "n1price", "group": "Ungrouped variables", "definition": "precround(n1*price,2)", "description": "", "templateType": "anything", "can_override": false}, "n2price": {"name": "n2price", "group": "Ungrouped variables", "definition": "precround(n2*price,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["n1", "price", "n2", "n1price", "n2price"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

£[[0]]

Round numbers to a given number of decimal places.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

We can approximate numbers by rounding.

Round $\\var{c1}$ to a given number of decimal places.

", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

When you're asked to round your answer to a number of decimal places, you need to decide whether to keep the last digit the same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

It is important to keep in mind that if the digit we are increasing is 9, it becomes zero and we increase the previous digit instead. If this digit is 9 as well, we move along to the left side until we find a digit less than 9.

To round a number to a given number $n$ of decimal places, we look at the $n$th digit after the decimal point.

We have $\\var{c1}$.

We look at the first digit after the decimal point. This is $\\var{cdig[4]}$ and the following digit is $\\var{cdig[3]}$ so we round updown to get $\\var{precround(c1, 1)}$.

ii)

The second digit after the decimal point is $\\var{cdig[3]}$. It is followed by $\\var{cdig[2]}$ so we round updown to get $\\var{precround(c1, 2)}$.

iii)

The 3rd decimal place is $\\var{cdig[2]}$, followed by $\\var{cdig[1]}$. We get $\\var{precround(c1, 3)}$. The 4th decimal place is $\\var{cdig[1]}$, followed by $\\var{cdig[0]}$. We get $\\var{precround(c1, 4)}$.

Use this link to find some resources which will help you revise this topic

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Random number with 5 decimal places.

", "templateType": "anything", "can_override": false}, "cdig": {"name": "cdig", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 5)", "description": "", "templateType": "anything", "can_override": false}, "dp": {"name": "dp", "group": "Ungrouped variables", "definition": "random(3..4)", "description": "

Number of decimal places to round.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "100"}, "ungrouped_variables": ["dp", "cdig", "c1"], "variable_groups": [], "functions": {"n_from_digits": {"parameters": [["digits", "list"]], "type": "number", "language": "jme", "definition": "if(\n len(digits)=0,\n 0,\n digits[0]+10*n_from_digits(digits[1..len(digits)])\n)"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

i) $\\var{c1}$ rounded to 1 decimal place is: [[0]]

ii) $\\var{c1}$ rounded to 2 decimal places is: [[1]]

iii) $\\var{c1}$ rounded to {dp} decimal places is: [[2]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "precround(c1, 1)", "maxValue": "precround(c1, 1)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "precround(c1, 2)", "maxValue": "precround(c1, 2)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "precround(c1, dp)", "maxValue": "precround(c1, dp)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF02 Rounding SF (integer)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["rounding"], "metadata": {"description": "

Round numbers to a given number of significant figures.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

When you're asked to round your answer to a number of significant figures, you need to decide whether to keep the last digit same (rounding down) or increase it by 1 (rounding up). If the following digit is less than 5 we round down and we round up when the next digit is 5 or more.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

The last digit we need to keep will depend on how many zeros there are. We don't consider leading zeros to be significant,
i.e. 0.03 and 0.3 both have 1 significant figure (but 0.30 has two significant figures, since the second zero isn't a 'leading' zero).

We round $\\var{d1}$ to 1 significant figure. The first non-zero digit is $\\var{ddig[5]}$. The following digit is $\\var{ddig[4]}$ so we round updown to get $\\var{dpformat(siground(d1, 1), 0)}$.

ii)

We round $\\var{d1}$ to {sf} significant figures. The first non-zero digit is $\\var{ddig[5]}$. The second following digit is $\\var{ddig[4]}$, the third following digit is $\\var{ddig[3]}$ and the fourth following digit is $\\var{ddig[2]}$. The digit following the last digit we are keeping is $\\var{ddig[3]}$$\\var{ddig[2]}$$\\var{ddig[1]}$, so we round to get $\\var{sigformat(d1, sf)}$. These are our {sf} significant figures.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"edig": {"name": "edig", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 5)", "description": "", "templateType": "anything", "can_override": false}, "d1": {"name": "d1", "group": "Ungrouped variables", "definition": "n_from_digits(ddig)", "description": "

Random integer.

", "templateType": "anything", "can_override": false}, "e1": {"name": "e1", "group": "Ungrouped variables", "definition": "n_from_digits(edig)*10^(random(-6,-7,-8))", "description": "

Random number with 7 decimal places.

", "templateType": "anything", "can_override": false}, "ddig": {"name": "ddig", "group": "Ungrouped variables", "definition": "repeat(random(1..9), 6)", "description": "", "templateType": "anything", "can_override": false}, "sf": {"name": "sf", "group": "Ungrouped variables", "definition": "3", "description": "

Number of significant figures to round.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "100"}, "ungrouped_variables": ["sf", "ddig", "edig", "d1", "e1"], "variable_groups": [], "functions": {"n_from_digits": {"parameters": [["digits", "list"]], "type": "number", "language": "jme", "definition": "if(\n len(digits)=0,\n 0,\n digits[0]+10*n_from_digits(digits[1..len(digits)])\n)"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Round $\\var{d1}$

i) $\\var{d1}$ rounded to 1 significant figure is: [[0]]

ii) $\\var{d1}$ rounded to {sf} significant figures is: [[1]]

Round numbers to a given number of significant figures.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

The first thing to do when we are rounding numbers is to identify the last digit we are keeping.

To write it down in steps:

Identify the last digit we need to keep.
Look at the following digit.
If it's 5 or more, increase the previous digit by one.
If it's 4 or less, keep the previous digit the same.
Fill any spaces to the right of the digit with zeros if needed.

We round $\\var{e1}$ to 1 significant figure. The first non-zero digit is $\\var{edig[4]}$, followed by $\\var{edig[3]}$. This is lower than 5 so we round downmore than 5 so we round up to get $\\var{sigformat(e1,1)}$.

ii)

We round $\\var{e1}$ to {sf} significant figures. The first non-zero digit is $\\var{edig[4]}$. The second following digit is $\\var{edig[3]}$, the third following digit is $\\var{edig[2]}$ and the fourth following digit is $\\var{edig[1]}$. The digit following the last digit we are keeping is $\\var{edig[2]}$$\\var{edig[1]}$$\\var{edig[0]}$, so we round to get $\\var{sigformat(e1, sf)}$. These are our {sf} significant figures.

Use this link to find some resources which will help you revise this topic.

Random integer.

", "templateType": "anything", "can_override": false}, "e1": {"name": "e1", "group": "Ungrouped variables", "definition": "n_from_digits(edig)*10^(random(-6,-7,-8))", "description": "

Random number with 7 decimal places.

Number of significant figures to round.

Round $\\var{e1}$

iii) $\\var{e1}$ rounded to 1 significant figure is: [[0]]

iv) $\\var{e1}$ rounded to {sf} significant figures is: [[1]]

State the Upper and lower bound of a distance that has been rounded to either the nearest 10 or 100 miles.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The distance between two towns had been rounded to the nearest {x} miles in an aticle in the newspaper. If they reported that the distance was {y} miles, what are the upper and lower bound for the reported number?

", "advice": "

If a number like {y} has been rounded to the nearest {x} then {y} would have been rounded down if it was less than {y+x/2} because {y} is the nearest multiple of {x}.

Similarly {y} would have been rounded up if it was larger than or equal to {y-x/2}. This means the lower bound is {y-x/2} and the upper bound is {y+x/2}.

Use this link to find some resources which will help you revise this topic.

Upper bound:

[[0]]

Lower bound:

[[1]]

Given a student discount, calculate a discounted price.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{pname} is buying a new {item}. The price of the model he picked is £{price}. On a website with discounts for students, he found a voucher for a discount of {percentage}%.

", "advice": "

There are multiple methods to approach this problem. The first method involves working out the discounted price as a percentage of the original, while the second method calculates the value of the discount and subtracts that from the listed price.

Method 1

There is a {percentage}% decrease in price. This means that the new price will be {100-percentage}% of the old price.

\\[\\begin{align} \\frac{\\var{100-percentage}}{100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)} \\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

Or, using the multiplier method,

\\[\\begin{align} \\var{(100-percentage)/100} \\times \\var{price} &= \\var{dpformat((100-percentage)/100*price,4)}\\\\&= \\var{dpformat((100-percentage)/100*price, 2)}\\text{.} \\end{align}\\]

When we are talking about money, it is usually assumed that we will round the answer to 2 decimal places.

Method 2

We find the discount first. This is

\\[\\frac{\\var{percentage}}{100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Or using a decimal multiplier,

\\[\\var{(percentage)/100} \\times \\var{price} = \\var{dpformat((percentage)/100*price,4)}\\text{.}\\]

Then we subtract the discount from the original price to get the new price:

\\[ \\var{price} - \\var{dpformat(discount,2)} = \\var{dpformat(price - discount, 2)}\\text{.} \\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"discount": {"name": "discount", "group": "Ungrouped variables", "definition": "percentage*price/100", "description": "", "templateType": "anything", "can_override": false}, "pname": {"name": "pname", "group": "Ungrouped variables", "definition": "random(\"Adair\",\"Aya\",\"Bergen\",\"Dua\",\"Fadhili\",\"Harper\",\"Kaden\",\"Ola\",\"Pat\",\"Skylar\",\"Wren\",\"Zendaya\")", "description": "

Names.

", "templateType": "anything", "can_override": false}, "discountrounded": {"name": "discountrounded", "group": "Ungrouped variables", "definition": "precround(discount,2)", "description": "", "templateType": "anything", "can_override": false}, "price": {"name": "price", "group": "Ungrouped variables", "definition": "switch(\n item = \"TV\", random(170.99..1199.99), \n item = \"laptop\", random(200.99..799.99),\n item = \"smartphone\", random(100.99..799.99),\n item = \"PC\", random(200.99..969.99),\n item = \"gaming console\", random(80.99..349.99),\n random(110.99..649.99))\n", "description": "

Price of an item.

", "templateType": "anything", "can_override": false}, "item": {"name": "item", "group": "Ungrouped variables", "definition": "random(\"TV\", \"laptop\", \"smartphone\", \"PC\", \"gaming console\", \"fridge\")", "description": "

The bought item.

", "templateType": "anything", "can_override": false}, "percentage": {"name": "percentage", "group": "Ungrouped variables", "definition": "random(5..40 #5)", "description": "

Discount percentage.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": "1000"}, "ungrouped_variables": ["item", "pname", "price", "percentage", "discount", "discountrounded"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

What will the discounted price of the {item} be?

Round your answer to the nearest penny.

£ [[0]]

Your answer does not make sense in real life, we cannot divide a penny any further. Shops always round their prices for items. That is why you should have rounded your answer to $\\var{precround((100-percentage)/100*price, 2)}$.

", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF06 Percentage increase", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Calculate the percentage increase (as a percentage) given a number and the size of the increase.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is the percentage increase in a class of {total} if {additional} more are added to it?

Give your answer to 2 decimal places.

", "advice": "

To calculate a percentage increase you need to find how much the increase is as a percentage of the original number. In this question the increase is {additional} and the original number is {total} so the percentage is

\\[ \\frac{\\var{additional}}{\\var{total}}\\times100\\%=\\var{dpformat(additional/total,4)}\\times 100\\%=\\var{dpformat(percentage,2)}\\%\\]

Use this link to find some resources which will help you revise this topic.

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[[0]]%

Compound percentage change.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The value of a car is initially {StartingPrice}. If the value decreases by {dec}%, and then increases by {inc}%, what is the final value?

Give your answer correct to two decimal places.

", "advice": "

There is a {dec}% decrease in price. This means that price after the decrease will be {100-dec}% of the old price.

\\[\\frac{\\var{100-dec}}{100} \\times \\var{StartingPrice} = \\var{(100-dec)/100*StartingPrice}\\]

Then there is a {inc}% increase in price. This means the final price will be {100+inc}% of the price after the decrease.

\\[\\frac{\\var{100+inc}}{100} \\times \\var{(100-dec)/100*StartingPrice} = £\\var{dpformat((100+inc)/100*(100-dec)/100*StartingPrice,2)}\\]

Use this link to find some resources which will help you revise this topic.

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£[[0]]

Find the original price before a discount by dividing the new price by the percentage discount.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{name1} and {name2} are friends. {name1} noticed {name2}'s new {item} when he came over to visit her house. He immediately knew he wanted to buy the same model. When he got home, he bought the {item} online for £{newprice}.

", "advice": "

We need to find the original price paid by {name2}. This value represents 100%.

By the time {name1} bought the {item}, the price had decreased by {percentage}%.

{name1} therefore paid {100-percentage}% of the price {name2} paid.

We use the unitary method to find the original price. We know the price paid by {name1}.

\\[\\var{100-percentage}\\text{%} = \\var{newprice} \\text{.}\\]

Divide both sides by {100-percentage} to get

\\[\\begin{align} 1\\text{%} &= \\var{newprice} \\div \\var{100-percentage} \\\\&= \\var{newprice/(100-percentage)} \\text{.} \\end{align}\\]

Multiply both sides by 100 to get

\\[\\begin{align} 100\\text{%} &= \\var{newprice/(100-percentage)} \\times 100 \\\\&= \\var{newprice/(100-percentage)*100} \\\\&= \\var{oldprice}\\text{.} \\end{align}\\]

This is the original price paid by {name2} before the {percentage}% decrease.

We can check our answer with a different method.

\\[\\begin{align} \\var{100-percentage}\\text{% of } \\var{oldprice} &= \\var{(100-percentage)/100} \\times \\var{oldprice} \\\\&= \\var{(100-percentage)/100*oldprice} \\\\&= \\var{precround((100-percentage)/100*oldprice, 2)} \\text{.} \\end{align}\\]

Use this link to find some resources which will help you revise this topic.

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Discount percentage.

The bought item.

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "precround(precround(oldprice*(100-percentage)/100,2)*100/(100-percentage),2) = oldprice", "maxRuns": "1000"}, "ungrouped_variables": ["item", "name1", "percentage", "name2", "oldprice", "newprice"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

When {name1} told {name2} how much he had paid for the {item}, {name2} said the price had decreased by {percentage}% since she bought it.

How much did {name2} pay for the {item}?

£ [[0]]

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Find a percentage of an amount.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

What is {x}% of {y}?

", "advice": "

Taking {x}% of {y} is calculated by multiplying,

\\[\\frac{\\var{x}}{100}\\times\\var{y}.\\]

For this question we can calculate this by noticing that 10% of {y} is {y*0.1} and then since $\\var{x}\\%=\\var{x/10}\\times10\\%$ we can calculate {x}% of {y} as,
\\[\\var{x/10}\\times10\\%\\times \\var{y}=\\var{x/10}\\times\\var{y*0.1}=\\var{x/100*y}.\\]

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(10..90 #10)", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(10.. 100 # 10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "y*(x/100)", "maxValue": "y*(x/100)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF10 Percentage of amount 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Calculate one number as percentage of another.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To find the percentage of a number we can use the formula:

\\[ \\text{New value } = \\text{Original value } \\times \\text{Percentage in decimal form} \\]

Firstly, to convert a percentage into decimal form we need to divide by $100$:

\\[ \\var{p} \\% = \\var{p/100} \\]

Therefore,

\\[ \\begin{split} \\var{p} \\% \\,\\text{ of } \\var{og} &\\,= \\var{og} \\times \\var{p/100} \\\\ &\\,= \\var{ans} \\end{split} \\]

Use this link to find resources to help you revise how to work out percentages.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"p": {"name": "p", "group": "Ungrouped variables", "definition": "random(1..99)", "description": "", "templateType": "anything", "can_override": false}, "og": {"name": "og", "group": "Ungrouped variables", "definition": "random(101..999)", "description": "", "templateType": "anything", "can_override": false}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "og*p*0.01", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["p", "og", "ans"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find {p}% of {og}

", "minValue": "ans", "maxValue": "ans", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF11 One number as a percentage of another", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Don Shearman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/680/"}, {"name": "Adelle Colbourn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2083/"}, {"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "

Given the number of international students enrolled on a course of $n$ students, calculate the percentage of 'home' students.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{num_students} of the {class_size} students enrolled on a course are international students. What percentage are 'home' students?

", "advice": "

First work out the number of students who are not international. In this case it is {class_size} - {num_students} = {class_size-num_students} students.

Then write this as a fraction out of {class_size}. $ \\frac{\\var{class_size-num_students}} {\\var{class_size}} $

Then convert this to a percentage. You should put this fraction into your calculator and then multiply by 100:

$ \\frac{\\var{class_size-num_students}} {\\var{class_size}} \\times 100 = \\var{(class_size-num_students)/class_size*100}\\%$

Use this link to find resources to help you revise how to calculate percentages.

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The number of students in the class who do speak a language other than English.

", "templateType": "anything", "can_override": false}, "class_size": {"name": "class_size", "group": "Ungrouped variables", "definition": "random(80..300)", "description": "", "templateType": "anything", "can_override": false}, "per": {"name": "per", "group": "Ungrouped variables", "definition": "random(5..90 except 50)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "num_students = precround(num_students,0) AND (num_students<>class_size/2 AND class_size<>100)", "maxRuns": 100}, "ungrouped_variables": ["num_students", "class_size", "per"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["$\\var{class_size-num_students}\\%$", "$\\var{num_students*100/class_size}\\%$", "$\\var{num_students}\\%$", "$\\var{(class_size-num_students)/class_size*100}\\%$"], "matrix": [0, 0, 0, "1"], "distractors": ["Have you converted this to a percentage? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Click on Reveal Answer and scroll down for Advice regarding this question.", "How many students do NOT speak a language other than English at home? Then convert this to a percentage. Click on Reveal Answer and scroll down for Advice regarding this question.", "Well done!"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF12 Simplify (cancel down) Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "Calculating the LCM and HCF of numbers by using prime factorisation.", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Express the fraction below in its simplest form:

\\[\\frac{\\var{x}}{\\var{y}}\\]

", "advice": "

To simplify a fraction we need to divide both numbers by their common factors.

We can write $\\var{x}$ and $\\var{y}$ as a product of prime factors as follows:

$\\var{x}=\\var{show_factors(x)}$

$\\var{y}=\\var{show_factors(y)}$.

So to fully simplify the fraction we need to divide both $\\var{x}$ and $\\var{y}$ by

\\[\\var{show_factors(hcf_xy)}.\\]

This gives us the fraction

\\[\\frac{\\var{x/hcf_xy}}{\\var{y/hcf_xy}}\\]

Use this link to find some resources which will help you revise this topic.

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This question tests the student's ability to identify equivalent fractions through spotting a fraction which is not equivalent amongst a list of otherwise equivalent fractions. It also tests the students ability to convert mixed numbers into their equivalent improper fractions. It then does the reverse and tests their ability to convert an improper fraction into an equivalent mixed number.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

A mixed number is a number consisting of an integer and a proper fraction, i.e. a number in the form $ a \\displaystyle \\frac{b}{c}$ where $a$ is an integer and $\\displaystyle\\frac{b}{c}$ is a proper fraction: $b$ is smaller than $c$.

An improper fraction is a fraction where the numerator is larger than the denominator, i.e. a number of the form $\\displaystyle\\frac{d}{e}$ where the numerator, $d$, is greater than the denominator, $e$.

To convert a mixed number into an improper fraction, multiply the integer part of the mixed number, $a$, by the denominator, $c$.

The numerator of the improper fraction will be equal to this added to what was already on the numerator of the proper fraction.

The denominator of the proper fraction will stay the same when it converts to an improper fraction to give a final answer of

$\\displaystyle\\frac{({a}\\times{c})+b}{c}$.

\\[
{\\var{f}\\frac{\\var{g_coprime}}{\\var{h_coprime}}} = \\frac{({\\var{f}}\\times{\\var{h_coprime}})+{\\var{g_coprime}}}{{\\var{h_coprime}}}=\\simplify{{num}/{h_coprime}}\\text{.}
\\]

Use this link to find some resources which will help you revise this topic.

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PART C

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PART C

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numerator for the improper fraction c(i)

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gcd of num and h

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Random number between 1 and 5

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Write the mixed number as an improper fraction and reduce it down to its simplest form.

$\\displaystyle{\\var{f}\\frac{\\var{g_coprime}}{\\var{h_coprime}}} =$ [[0]] [[1]] .

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following addition, giving the fraction in its simplest form.

", "advice": "

$\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}+\\frac{\\var{c_coprime}}{\\var{d_coprime}}$

To add or subtract fractions, we need to have a common denominator on both fractions.

To get a common denominator, we need to find the lowest common multiple of the two denominators.

The lowest common multiple of $\\var{b_coprime}$ and $\\var{d_coprime}$ is $\\var{lcm}.$

This will be the new denominator, and we need to multiply each fraction individually to ensure we get this denominator.

For $\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}$, we need to multiply the fraction by $\\displaystyle\\frac{\\var{lcm_b}}{\\var{lcm_b}}$ to give $\\displaystyle\\frac{\\var{alcm_b}}{\\var{lcm}}.$

For $\\displaystyle\\frac{\\var{c_coprime}}{\\var{d_coprime}}$, we need to multiply the fraction by $\\displaystyle\\frac{\\var{lcm_d}}{\\var{lcm_d}}$ to give $\\displaystyle\\frac{\\var{clcm_d}}{\\var{lcm}}.$

Now that we have each fraction in terms of a common denominator, we can now add the fractions together.

$\\displaystyle\\frac{\\var{alcm_b}}{\\var{lcm}}+\\frac{\\var{clcm_d}}{\\var{lcm}}=\\frac{(\\var{alcm_b}+\\var{clcm_d})}{\\var{lcm}}=\\frac{\\var{alcmclcm}}{\\var{lcm}}.$

From this, we can try to simplify the result down by finding the greatest common divisor of the numerator and denominator and dividing the whole fraction by this amount.

The greatest common divisor of $\\var{alcmclcm}$ and $\\var{lcm}$ is $\\var{gcd}.$

Simplifying using this value gives a final answer of $\\displaystyle\\frac{\\var{num}}{\\var{denom}}.$

Therefore, the expression cannot be simplified further, and $\\displaystyle\\frac{\\var{num}}{\\var{denom}}$ is the final answer.

Find out more about this topic using our resource

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PART A variable c times variable b

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PART A simplification of fractions in the question.

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PART A variable a times variable d

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PART A variable c times the lcm of b and d, divided by d

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PART A greatest common divisor of the variables alcmclcm and lcm

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PART A

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PART A lowest common multiple of variable b_coprime and variable d_coprime.

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$\\displaystyle\\frac{\\var{a_coprime}}{\\var{b_coprime}}+\\frac{\\var{c_coprime}}{\\var{d_coprime}}=$ [[0]] [[1]]

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Evaluate the following additions and subtractions, giving each fraction in its simplest form. Write the numerator (the top number) as negative if the fraction is negative.

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$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}-\\frac{\\var{h_coprime}}{\\var{j_coprime}}+2.$

The two fractions can be individually multiplied to achieve a common denominator of the lowest common multiple, $\\var{lcm2}.$

$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}$ becomes $\\displaystyle\\frac{\\var{flcm2_g}}{\\var{lcm2}}$ and $\\displaystyle\\frac{\\var{h_coprime}}{\\var{j_coprime}}$ becomes $\\displaystyle\\frac{\\var{hlcm2_j}}{\\var{lcm2}}.$

We can now subtract the second fraction from the first.

$\\displaystyle\\frac{\\var{flcm2_g}}{\\var{lcm2}}-\\frac{\\var{hlcm2_j}}{\\var{lcm2}}=\\frac{\\var{flcmhlcm}}{\\var{lcm2}}.$

Find out more about this topic using our resource.

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PART B

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$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}-\\frac{\\var{h_coprime}}{\\var{j_coprime}}=$ [[0]] [[1]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "flcmhlcm", "maxValue": "flcmhlcm", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "lcm2", "maxValue": "lcm2", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF16 Multiply Fractions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Lauren Richards", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1589/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Oliver Spenceley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23557/"}], "tags": ["improper fractions", "mixed numbers", "multiplication of fractions", "multiplying fractions", "squared fraction", "taxonomy"], "metadata": {"description": "

Several problems involving the multiplication of fractions, with increasingly difficult examples, including a mixed fraction and a squared fraction. The final part is a word problem.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following multiplication, giving the answer in its simplest form.

", "advice": "

To multiply $\\displaystyle\\frac{\\var{a_coprime}}{\\var{c_coprime}}\\times\\frac{\\var{b_coprime}}{\\var{d_coprime}}$, address the numerators and denominators separately.

Multiply the numerators across both fractions.

$\\var{a_coprime}\\times\\var{b_coprime}=\\var{ab}$,

and then multiply the denominators across both fractions.

$\\var{c_coprime}\\times\\var{d_coprime}=\\var{cd}$.

The values of the multiplied numerators and denominators will be the numerator and denominator of the new fraction: $\\displaystyle\\frac{\\var{ab}}{\\var{cd}}$.

This answer may need simplifying down, and to do this, find the greatest common divisor in both the numerator and denominator and divide by this number.

The greatest common divisor of $\\var{ab}$ and $\\var{cd}$ is $\\var{gcd}$.

By using $\\var{gcd}$ to cancel down the fraction, the final answer is $\\displaystyle\\simplify{{ab}/{cd}}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"k": {"name": "k", "group": "Part b", "definition": "random(1..7 except j)", "description": "

Random number between 1 and 20

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Random number between 1 and 20.

", "templateType": "randrange", "can_override": false}, "cd": {"name": "cd", "group": "Part a", "definition": "c_coprime*d_coprime", "description": "

Variable c times variable d.

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Random number from 1 to 12.

", "templateType": "anything", "can_override": false}, "d": {"name": "d", "group": "Part a", "definition": "random(2 .. 12#1)", "description": "

Random number from 1 to 12.

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Numerator of the improper fraction converted from a mixed number.

", "templateType": "anything", "can_override": false}, "gcd_gh": {"name": "gcd_gh", "group": "Part b", "definition": "gcd(g,h)", "description": "", "templateType": "anything", "can_override": false}, "fh": {"name": "fh", "group": "Part b", "definition": "f*h_coprime", "description": "

Variable f times variable h

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Random number between 1 and 4 - integer part of the mixed number.

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Random number from 1 to 12.

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PART A

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Denominator of new fraction.

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Random number between 1 and 20.

", "templateType": "randrange", "can_override": false}, "num": {"name": "num", "group": "Part b", "definition": "k_coprime*{numif/gcda}", "description": "

Numerator of gap 0

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gcd of the numerator of the improper fraction

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Random number from 1 to 12.

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Random number between 1 and 20

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Variable a times variable b

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$\\displaystyle\\frac{\\var{a_coprime}}{\\var{c_coprime}}\\times\\frac{\\var{b_coprime}}{\\var{d_coprime}}$ = [[0]] [[1]]

Several problems involving dividing fractions, with increasingly difficult examples, including mixed numbers and complex fractions.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Evaluate the following sums involving division of fractions. Simplify your answers where possible.

", "advice": "

When faced with dividing fractions, it much easier to switch one of the fractions around and multiply them together instead of divide them.

\\[ \\left( \\frac{\\var{f_coprime}}{\\var{g_coprime}}\\div\\frac{\\var{h_coprime}}{\\var{j_coprime}} \\right) \\equiv \\left( \\frac{\\var{f_coprime}}{\\var{g_coprime}}\\times\\frac{\\var{j_coprime}}{\\var{h_coprime}} \\right) = \\frac{\\var{fj}}{\\var{gh}} \\]

Then, simplify by finding the highest common divisor in the numerator and denominator which in this case is $\\var{gcd1}$.

This gives a final answer of $\\displaystyle\\simplify{{fj}/{gh}}$.

Use this link to find some resources which will help you revise this topic

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variable f4 times h4.

Used in part c)

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PART C

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Random number but not the same number as variable g4.

Used in part c.

", "templateType": "anything", "can_override": false}, "h3_coprime": {"name": "h3_coprime", "group": "Ungrouped variables", "definition": "h3/gcd(g3,h3)", "description": "

PART C

", "templateType": "anything", "can_override": false}, "f_coprime": {"name": "f_coprime", "group": "part a", "definition": "f/gcd(f,g)", "description": "

PART A

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PART A

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PART B

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greatest common divisor of variables f1j1 and g1h1.

Used in part b).

", "templateType": "anything", "can_override": false}, "g1_coprime": {"name": "g1_coprime", "group": "Ungrouped variables", "definition": "g1/gcd(f1,g1)", "description": "

PART B

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PART B

", "templateType": "anything", "can_override": false}, "gcd3": {"name": "gcd3", "group": "Ungrouped variables", "definition": "gcd(num,denom)", "description": "

greatest common denominator for part c.

", "templateType": "anything", "can_override": false}, "j1": {"name": "j1", "group": "Ungrouped variables", "definition": "random(h1..11 except h1)", "description": "

Random number between 2 and 20 and not the same value as variable h1.

Used in part b).

", "templateType": "anything", "can_override": false}, "g1h1": {"name": "g1h1", "group": "Ungrouped variables", "definition": "g1_coprime*h1_coprime", "description": "

variable g1 times h1.

Used in part b).

", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "part a", "definition": "random(2..10)", "description": "

Random number between 2 and 10.

Used in part a).

", "templateType": "anything", "can_override": false}, "f4": {"name": "f4", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "

Random number.

Used in part c).

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Random number between 2 and 20.

Used in part b)

", "templateType": "anything", "can_override": false}, "g3": {"name": "g3", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "

Random number.

Used in part c).

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variable f3 times h3.

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Random number from 2 to 10.

Used in part a).

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variable g times variable h.

Used in part a).

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PART A

", "templateType": "anything", "can_override": false}, "denom": {"name": "denom", "group": "Ungrouped variables", "definition": "h3_coprime*(f4h4+g4_coprime)", "description": "

Unsimplified denominator of part c.

", "templateType": "anything", "can_override": false}, "j": {"name": "j", "group": "part a", "definition": "random(h..12 except h)", "description": "

Random number between 2 and 10 and not the same value as h.

Used in part a).

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variable f1 times j1.

Used in part b).

", "templateType": "anything", "can_override": false}, "h4_coprime": {"name": "h4_coprime", "group": "Ungrouped variables", "definition": "h4/gcd(g4,h4)", "description": "

PART C

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Random number between 2 and 30 and not the same value as variable f1.

Used in part b).

", "templateType": "anything", "can_override": false}, "fj": {"name": "fj", "group": "part a", "definition": "f_coprime*j_coprime", "description": "

variable f times variable j.

Used in part a).

", "templateType": "anything", "can_override": false}, "f3": {"name": "f3", "group": "Ungrouped variables", "definition": "random(1 .. 3#1)", "description": "

Random number between 2 and 6.

Used in part c).

", "templateType": "randrange", "can_override": false}, "f1_coprime": {"name": "f1_coprime", "group": "Ungrouped variables", "definition": "f1/gcd(f1,g1)", "description": "

PART B

", "templateType": "anything", "can_override": false}, "h3": {"name": "h3", "group": "Ungrouped variables", "definition": "random(5..8)", "description": "

Random number and not the same value as variable g3.

Used in part c).

", "templateType": "anything", "can_override": false}, "gcd1": {"name": "gcd1", "group": "part a", "definition": "gcd(fj,gh)", "description": "

greatest common divisor of variable fj and gh.

Used in part a).

", "templateType": "anything", "can_override": false}, "g3_coprime": {"name": "g3_coprime", "group": "Ungrouped variables", "definition": "g3/gcd(g3,h3)", "description": "

PART C

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PART A

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Random number.

Used in part c).

", "templateType": "anything", "can_override": false}, "h1": {"name": "h1", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "

Random number between 2 and 20.

Used in part b).

", "templateType": "anything", "can_override": false}, "num": {"name": "num", "group": "Ungrouped variables", "definition": "h4_coprime*(f3h3+g3_coprime)", "description": "

numerator of the improper fraction in part c. Unsimplified.

", "templateType": "anything", "can_override": false}, "g": {"name": "g", "group": "part a", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["f1", "g1", "f1_coprime", "g1_coprime", "h1", "j1", "h1_coprime", "j1_coprime", "f1j1", "g1h1", "gcd2", "f3", "g3", "h3", "g3_coprime", "h3_coprime", "f4", "g4", "h4", "g4_coprime", "h4_coprime", "f3h3", "f4h4", "num", "denom", "gcd3"], "variable_groups": [{"name": "part a", "variables": ["g", "f", "f_coprime", "g_coprime", "h", "j", "h_coprime", "j_coprime", "fj", "gh", "gcd1"]}], "functions": {}, "preamble": {"js": "", "css": "fraction {\n display: inline-block;\n vertical-align: middle;\n}\nfraction > numerator, fraction > denominator {\n float: left;\n width: 100%;\n text-align: center;\n line-height: 2.5em;\n}\nfraction > numerator {\n border-bottom: 1px solid;\n padding-bottom: 5px;\n}\nfraction > denominator {\n padding-top: 5px;\n}\nfraction input {\n line-height: 1em;\n}\n\nfraction .part {\n margin: 0;\n}\n\n.table-responsive, .fractiontable {\n display:inline-block;\n}\n.fractiontable {\n padding: 0; \n border: 0;\n}\n\n.fractiontable .tddenom \n{\n text-align: center;\n}\n\n.fractiontable .tdnum \n{\n border-bottom: 1px solid black; \n text-align: center;\n}\n\n\n.fractiontable tr {\n height: 3em;\n}\n"}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$\\displaystyle\\frac{\\var{f_coprime}}{\\var{g_coprime}}\\div\\frac{\\var{h_coprime}}{\\var{j_coprime}}=$ [[0]] [[1]]

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

To convert an improper fraction into a mixed number, find out how many times the denominator \\var{h_coprime/gcdb} goes into the numerator \\var{num/gcdb}. You can do this by dividing the numerator by the denominator and taking the whole number part or you can just add the denominator to itself until one more addition would make it bigger. This gives us a whole number part of our mixed fraction of \\var{f}.

The numerator of our mixed fraction is what is left from dividing out the whole number. For this question that is $\\var{num/gcdb}-\\var{f*h_coprime}.

Finally the denominator of our mixed fraction is just the denominator of the improper fraction.

\\[
\\frac{\\var{num/gcdb}}{\\var{h_coprime/gcdb}} = {\\var{f}\\frac{\\var{g_coprime}}{\\var{h_coprime}}}\\text{.}
\\]

Use this link to find some resources which will help you revise this topic.

Random number between 1 and 15

", "templateType": "anything", "can_override": false}, "g_coprime": {"name": "g_coprime", "group": "Ungrouped variables", "definition": "g/gcd_gh", "description": "

PART C

", "templateType": "anything", "can_override": false}, "gcd_gh": {"name": "gcd_gh", "group": "Ungrouped variables", "definition": "gcd(g,h)", "description": "

PART C

", "templateType": "anything", "can_override": false}, "num": {"name": "num", "group": "Ungrouped variables", "definition": "((h_coprime*f)+g_coprime)", "description": "

numerator for the improper fraction c(i)

", "templateType": "anything", "can_override": false}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(10 .. 24#1)", "description": "

Random number between 1 and 24

", "templateType": "randrange", "can_override": false}, "h_coprime": {"name": "h_coprime", "group": "Ungrouped variables", "definition": "h/gcd_gh", "description": "

PART C

", "templateType": "anything", "can_override": false}, "gcdb": {"name": "gcdb", "group": "Ungrouped variables", "definition": "gcd({num},{h_coprime})", "description": "

gcd of num and h

", "templateType": "anything", "can_override": false}, "f": {"name": "f", "group": "Ungrouped variables", "definition": "random(2 .. 5#1)", "description": "

Random number between 1 and 5

Write the improper fraction as a mixed number and reduce it down to its simplest form.

$\\displaystyle{\\frac{\\var{num/gcdb}}{\\var{h_coprime/gcdb}}} = $ [[2]] [[0]] [[1]] .

Put fractions in size order.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of these two fractions is the largest?

", "advice": "

To find which is bigger of $\\frac{\\var{top1}}{\\var{bot1}}$ and $\\frac{\\var{top2}}{\\var{bot2}}$ we need them to have the same denominator. A way to do this is to multiply the top and bottom of $\\frac{\\var{top1}}{\\var{bot1}}$ by $\\var{bot2}$ and multiply the top and bottom of $\\frac{\\var{top2}}{\\var{bot2}}$ by {bot1}. This doesn't change the the value of the fractions as this is just like multiplying by one.

\\[\\frac{\\var{top1}}{\\var{bot1}}=\\frac{\\var{top1*bot2}}{\\var{bot1*bot2}},\\quad \\frac{\\var{top2}}{\\var{bot2}}=\\frac{\\var{top2*bot1}}{\\var{bot1*bot2}}\\]

Now we can easily see which is bigger by comparing the numerator.

Use this link to find some resources which will help you revise this topic.

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Convert a percentage to a fraction.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Convert $\\var{perc}$% into its equivalent fraction expressed in its simplest form.

", "advice": "

Percentages can be converted to fractions by treating them as fractions out of $100$:

\\[\\frac{\\var{perc}}{100},\\]

and then simplifying. In this case giving:

\\[\\frac{\\var{ansn}}{\\var{ansd}}\\]

Use this link to find some resources which will help you revise this topic.

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[[0]] numerator

[[1]] denominator

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Convert a decimal to a fraction.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express $\\var{dec}$ as a fraction in simplest form.

", "advice": "

{advice}

Use this link to find some resources which will help you revise this topic.

Decimals can be converted to fractions using place value. This decimal only has 1 decimal place and therefore finishes in the \\\"tenths\\\" column. Hence, we can write it as:

\\n

\\\\[\\\\frac{\\\\var{dec*10}}{10},\\\\]

\\n

and then simplifying (if necessary). In this case no simplification is needed.

\"", "description": "", "templateType": "long string", "can_override": false}, "adviceyes": {"name": "adviceyes", "group": "Ungrouped variables", "definition": "\"

Decimals can be converted to fractions using place value. This decimal only has 1 decimal place and therefore finishes in the \\\"tenths\\\" column. Hence, we can write it as:

\\n

\\\\[\\\\frac{\\\\var{dec*10}}{10},\\\\]

\\n

and then simplifying (if necessary). In this case giving:

\\n

\\\\[\\\\frac{\\\\var{ansn}}{\\\\var{ansd}}\\\\]

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[[0]] Numerator

--------------

[[1]] Denominator

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "Numerator", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "ansn", "maxValue": "ansn", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "Denominator", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "ansd", "maxValue": "ansd", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF22 FDP convert 3 - Fraction into decimal", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Andrew Neate", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21832/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Convert a fraction into a decimal.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write $\\frac{\\var{x}}{\\var{y}}$ as a decimal. Round your answer to 3 decimal places.

", "advice": "

You can calculate the decimals by hand using long division of $\\var{x}.000$ divided by $\\var{y}$.

In some cases you may be able to simplify the fraction to something that you know the decimal for.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"x": {"name": "x", "group": "Ungrouped variables", "definition": "random(1..(y-1))", "description": "", "templateType": "anything", "can_override": false}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(2..10)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["x", "y"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "x/y", "maxValue": "x/y", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NF23 Equivalent Ratios", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Picking a ratio out of a list that is not equivalent to the others.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of the following ratios is not equivalent to the others?

", "advice": "

The key to this question is understanding how to simplify ratios. In this case all the ratios simplify to $\\var{a}:\\var{b}$ except for $\\var{6*a}:\\var{7*b}$.

For more help with this topic have a look at the resources here.

Dividing amounts in ratios

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The ratio of ethanol to water is {a}:{b} for an experiment. If I have {volWater}ml of water, how much ethanol do I need?

", "advice": "

If there is a ratio of {a}:{b} for ethanol:water then that means for every {b}ml of water we need {a}ml of ethanol.

In our experiment there is {volwater}ml of water so to find the amount of ethanol we divide by {b} and then multiply by {a}.

\\[\\var{volwater}\\text{ml}\\times\\frac{\\var{a}}{\\var{b}}=\\var{volwater*a/b}\\text{ml}\\]

Use this link to find some resources which will help you revise this topic.

[[0]]ml

Convert numbers greater than 1 into standard form/scientific notation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write the following numbers in scientific notation.

", "advice": "

Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. The decimal point is currently to the right of the last digit in $\\var{q2}$ and needs to be between the first and second digits, i.e $\\var{dec2}$. Count the places that the digits must move and you get $\\var{pow2}$ places. That is,

\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]

We have a positive $\\var{pow2}$ as the power because we need to make the number $\\var{dec2}$ bigger to get to $\\var{q2}$.

Use this link to find some resources which will help you revise this topic.

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$\\var{q2} =$ [[0]]$\\times 10$ ^[[1]]

Convert numbers between 0 and 1 intro standard form/scientific notation.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Write the following numbers in scientific notation.

", "advice": "

Suppose we have the number $\\var{q2}$. In scientific notation, this number would start with $\\var{dec2}$ since we only want one digit in front of the decimal point. Count the places that the digits must move and you get $\\var{-pow2}$ places to the right. That is,

\\[\\var{q2}=\\var{dec2}\\times 10^{\\var{pow2}}\\]

We have a negative $\\var{-pow2}$ as the power because we need to make the number $\\var{dec2}$ smaller to get to $\\var{q2}$.

Use this link to find some resources which will help you revise this topic.

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$\\var{q2}$ = [[0]]$\\times 10$ ^[[1]]

Calculations involving Standard form.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

To divide two numbers in standard form we can calculate the division of each part of the standard form number separately. In general we have,

\\[\\frac{x\\times10^j}{y\\times10^k}=\\frac xy\\times\\frac{10^j}{10^k}=\\frac xy\\times 10^{j-k}\\]

In this question we therefore have,

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=\\frac{\\var{a}}{\\var{b}}\\times\\frac{10^{\\var{n}}}{10^{\\var{m}}}=\\var{aDivBRound}\\times10^\\var{n-m}.\\]

Since {aDivBRound} is less than 1 then our answer isn't in standard form. In this case we need to reduce the exponent by 1 so the final answer is

\\[\\var{MantAnsRound}\\times10^{\\var{ExponentAns}}.\\]

Use this link to find some resources which will help you revise this topic.

For the equation

\\[\\frac{\\var{a}\\times10^{\\var{n}}}{\\var{b}\\times10^{\\var{m}}}=a\\times10^n\\]

find the values of $a$ and $n$ which keep the answer in standard form.

Give $a$ to two decimal places.

$a=$[[0]]

$n=$[[1]]

Exchange rates.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

On a particular day, 1 Pound Sterling (£) is equivalent to {exchanget} United States Dollars (\\$). If you have \\$ {amount}, what does this equate to in Pound Sterling?

", "advice": "

To convert Dollars (\\$) into Pounds (£) you will need to caculate:

\\begin{align}
\\frac{\\var{amount}}{\\var{exchange}} &= \\var{answer}\\\\
&= \\var{answert},
\\end{align}

ensuring to round off to the nearest penny.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"exchange": {"name": "exchange", "group": "Ungrouped variables", "definition": "random(0.9 .. 1.2#0.01)", "description": "

Pound to dollar exchange rate.

", "templateType": "randrange", "can_override": false}, "amount": {"name": "amount", "group": "Ungrouped variables", "definition": "random(100 .. 1000#5)", "description": "

Amount of dollars.

", "templateType": "randrange", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "amount/exchange", "description": "", "templateType": "anything", "can_override": false}, "answert": {"name": "answert", "group": "Ungrouped variables", "definition": "currency(answer,\"\u00a3\",\"p\")", "description": "", "templateType": "anything", "can_override": false}, "exchanget": {"name": "exchanget", "group": "Ungrouped variables", "definition": "dpformat(exchange,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["exchange", "amount", "answer", "answert", "exchanget"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]]

Unit conversion between two compound units.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

A Cheetah runs at a speed of {speedkm} kilometres per hour (km/h). What is the Cheetah's speed in metres per second (m/s)?

Give your ansswer to 2 decimal places where appropriate.

", "advice": "

There are a number of ways to work out the conversion. Here are a couple of suggestions.

METHOD 1

Since there are $1000$m in $1$km we first multiply by $1000$ to get the speed in metres per hour:

\\begin{equation} 1000*\\var{speedkm} = \\var{step1}\\end{equation}

Then we divide by $3600$ since that is the number of seconds in an hour to get the speed in metres per second:

\\begin{equation} \\frac{\\var{step1}}{3600} = \\var{speedms} \\end{equation}

Finally we round off to 2 decimal places as required, $\\var{roundanswer}$m/s.

METHOD 2

We can actually do all of the above in one step of working by using a single conversion factor. Since there are $1000$m in a km and $3600$ seconds in an hour, we can calaculate the conversion factor:

\\begin{equation} \\frac{3600}{1000} = 3.6 \\end{equation}

and then simply divide by that conversion factor:

\\begin{equation} \\frac{\\var{speedkm}}{3.6} = \\var{speedms} \\end{equation}

finally rounding off as before, $\\var{roundanswer}$m/s.

Use this link to find some resources which will help you revise this topic.

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[[0]]m/s

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "speedms", "maxValue": "speedms", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "speedms", "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU03 Convert Units - volume - l to ml", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Simple unit conversion with metric units.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {liquid} litres ($l$) in millilitres ($ml$).

", "advice": "

There are $1000ml$ in $1l$. To work out the conversion: $\\var{liquid}*1000 = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"liquid": {"name": "liquid", "group": "Ungrouped variables", "definition": "random(1 .. 6#0.01)", "description": "", "templateType": "randrange", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "liquid*1000", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["liquid", "answer"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]]$ml$

Metric Unit conversion - division by 1000.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {liquid} millilitres ($ml$) in litres ($l$). Give your answer to 3 decimal places.

", "advice": "

There are $1000ml$ in $1l$. To work out the conversion: $\\frac{\\var{liquid}}{1000} = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"liquid": {"name": "liquid", "group": "Ungrouped variables", "definition": "random(100 .. 5200#1)", "description": "", "templateType": "randrange", "can_override": false}, "answer": {"name": "answer", "group": "Ungrouped variables", "definition": "liquid/1000", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["liquid", "answer"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

[[0]]$l$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU05 Convert Units - metric prefixes - milligrams to grams", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Using prefixes (milli) in this case.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {x} milligrams ($mg$) in grams ($g$). Give your answer to 3 decimal places.

", "advice": "

There are $1000mg$ in $1g$. To work out the conversion: $\\frac{\\var{x}}{1000} = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

[[0]]$g$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU06 Convert Units - metric prefixes - milligrams to micrograms", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

Using prefixes - milli and micro.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {x} milligrams ($mg$) in micrograms ($\\mu g$).

", "advice": "

There are $1000\\mu g$ in $1mg$. To work out the conversion: $\\var{x}*1000 = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

[[0]]$\\mu g$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU07 Convert Units - metric prefixes - micrograms to milligrams", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

convert from micrograms to milligrams.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {x} micrograms ($\\mu g$) in milligrams ($mg$). Give your answer to 3 decimal places.

", "advice": "

There are $1000\\mu g$ in $1mg$. To work out the conversion: $\\frac{\\var{x}}{1000} = \\var{answer}$.

Use this link to find some resources which will help you revise this topic.

[[0]]$mg$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "precisionType": "dp", "precision": "3", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU08 Convert Units - Volume - ml to cubic cm", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

\"Convert\" from millilitres to cubic centimeters.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Express {x} millilitres ($ml$) in cubic centimetres ($cm^3$).

", "advice": "

$1 ml$ is the same measurement of volume as $1 cm^3$ so there is nothing to do to convert except change the units.

Use this link to find some resources which will help you revise this topic.

[[0]]$cm^3$

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "NU09 Convert Units - Length - Inches to cm", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}], "tags": [], "metadata": {"description": "

convert from inches to cm.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Given that $1$ $inch$ is approximately $2.54$ $cm$. How long is a $\\var{x}$ inch ruler in $cm$?

", "advice": "

The information given in the question tells you to use a conversion factor of $2.54$. So you must multiply $\\var{x}$ by $2.54$ to find the length of the ruler in $cm$.

Use this link to find some resources which will help you revise this topic.

[[0]]$cm$.

Give your answer to 2 decimal places.

", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "answer", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "answer", "maxValue": "answer", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "answer", "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Stats - Data Display", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], []], "questions": [{"name": "SD01 Choosing a suitable chart", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

This question is about identifying what types of charts or visual representations of data you can use for different data sets.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

This question is about recognising what types of charts or visual representations of data you can use with what types of data sets.

", "advice": "

There are many different types of visual representations of data and sometimes there will be a choice of what you use.

Start by looking at these resources to build up your understanding of data display options and methods.

The table shows different names of charts on the left hand side and different descriptions of data sets along the top.

Pair up each description with the chart that would be most suitable.

", "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": false, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["Scatter plot", "Histogram", "Bar Chart"], "matrix": [["1", 0, 0], [0, "0", "1"], [0, "1", "0"]], "layout": {"type": "all", "expression": ""}, "answers": ["Two continuous variables plotted against each other to investigate their relationship.", "Non-numerical categories and the frequencies of each category.", "A continuous variable such as \"height in $cm$\" grouped into intervals showingthe frequency of the data in each interval."]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SD02 Interpret Pie Charts", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

This question is about correctly interpreting pie charts.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

{geogebra_applet{\"https://www.geogebra.org/calculator/pmvumdrv\",[C: C,M: M]}}

The Pie Chart above shows the responses to a question asked by someone trying to plan a social event for their workplace. It shows answers given to the question \"Where would you like to go for a staff social?\" with the options \"Meal\", \"Cinema\" and \"Games Cafe\".

", "advice": "

A Pie chart of this type can only be used to make statements about the proportions of data in each category and does not provide information about the actual frequencies.

For more information on Pie Charts follow this link.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true, "j": false}, "constants": [], "variables": {"C": {"name": "C", "group": "Ungrouped variables", "definition": "random(60 .. 70#5)", "description": "", "templateType": "randrange", "can_override": false}, "M": {"name": "M", "group": "Ungrouped variables", "definition": "random(5 .. 25#5)", "description": "", "templateType": "randrange", "can_override": false}}, "variablesTest": {"condition": "C<>M", "maxRuns": 100}, "ungrouped_variables": ["C", "M"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "m_n_x", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

From the following comments which can you say are definitely true, definitely false and which do you not have enough information to know?

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Reading a value from a simple bar chart.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "

The bar heights give the values of the spend.

Each company has two bars, the left one for last year (in red) and the right one for this year (in purple).
Isolate last years spend by looking at the the bars on the right side, and choose the tallest bar, corresponding to the highest value.

Use this link to find some resources which will help you revise this topic.

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new y values after the transformation

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the (random) original y values which relate to the x values

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vertical shift

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horizontal shift

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new transformed x values

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original x values

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\n var yo8 = scope.variables.yo8.value;\n var yo9 = scope.variables.yo9.value; \n \n var div = Numbas.extensions.jsxgraph.makeBoard('550px','550px',{boundingBox:[-0.8,82,16,-8], axis:false, grid:true});\n \n question.display.html.querySelector('#dragpoint').append(div);\n \n var board = div.board;\n \n// board.suspendUpdate(); \n\n \n var dataArr = [yo0,yo5,0,yo1,yo6,0,yo2,yo7,0,yo3,yo8,0,yo4,yo9]; \n \n var xaxis = board.create('axis', [[0, 0], [12, 0]], {withLabel: true, name: \"Bank\", label: {offset: [250,-30]}});\n \n xaxis.removeAllTicks(); \n \n board.create('axis', [[0, 0], [0, 10]], {hideTicks:true, withLabel: false, name: \"\", label: {offset: [-110,300]}});\n \n var pop0 = board.create('point', [1.5,0],{name:'Morgan',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop1 = board.create('point',[4.5,0],{name:'Strome',fixed:true,size:0,color:'black',face:'diamond', label:{offset:[-20,-8]}});\n var pop2 = board.create('point',[7.5,0],{name:'Bentley',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop3 = board.create('point',[10.5,0],{name:'Sand',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var pop4 = board.create('point',[13.5,0],{name:'Karchen',fixed:true,size:0,color:'black', face:'diamond', label:{offset:[-15,-8]}});\n var leg1 = board.create('point',[12,75],{name:'last year',fixed:true,size:6,color:'#DA2228', face:'square', label:{offset:[9,0]}});\n var leg2 = board.create('point',[12,72],{name:'this year',fixed:true,size:6,color:'#6F1B75', face:'square', label:{offset:[9,0]}});\n \n \n// var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', fillOpacity:1, width:1,\n // colorArray:['#8E1B77','#8E1B77','Red','Red','blue','red','blue','red','red','blue', 'red','blue','red','red'], shadow:false});\n \n//var chart = board.createElement('chart', dataArr, \n // {chartStyle:'bar', width:1,fillOpacity:1, fillColor:'red', shadow:false}); \n \n \n var a = board.create('chart', [[1,2,3],[yo0,yo5,0]], {chartStyle:'bar',colors:['#DA2228','#6F1B75','#6F1B75'],width:1,fillOpacity:1});\n var b = board.create('chart', [[4,5,6],[yo1,yo6,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var c = board.create('chart', [[7,8,9],[yo2,yo7,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var d = board.create('chart', [[10,11,12],[yo3,yo8,0]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75','#6F1B75'],fillOpacity:1});\n var e = board.create('chart', [[13,14],[yo4,yo9]], {chartStyle:'bar',width:1,colors:['#DA2228','#6F1B75'],fillOpacity:1});\n \n board.unsuspendUpdate();\n \n var txt1 = board.create('text',[-0.3,30, 'Investment \u00a3(m)'], {fontColor:'black', fontSize:14, rotate:90});\n \n // var txt = board.create('text',[0.5,75, 'Investment (m)'], {fontSize:14, rotate:90});\n \n // var txt1 = board.create('text',[8,76, 'red bars represent 2010'], {fontColor:'red', fontSize:14, rotate:90});\n \n // var txt2 = board.create('text',[8,73, 'blue bars represents 2011'], {fontSize:14, rotate:90});\n\n // var myColors = new Array('red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue', 'white','red', 'blue');\n \n \n \n //board.unsuspendUpdate();\n\n // Rotate text around the lower left corner (-2,-1) by 30 degrees.\n // var tRot = board.create('transform', [90.0*Math.PI/180.0, -1,40], {type:'rotate'}); \n // tRot.bindTo(txt);\n // board.update();\n\n \n//var chart2 = board.createElement('chart', dataArr, {chartStyle:'line,point'});\n//chart2[0].setProperty('strokeColor:black','strokeWidth:2','shadow:true');\n//for(var i=0; i<11;i++) {\n // chart2[1][i].setProperty({strokeColor:'black',fillColor:'white',face:'[]', size:4, strokeWidth:2});\n//}\n//board.unsuspendUpdate(); \n \n //board.unsuspendUpdate();\n\n}\n\nquestion.signals.on('HTMLAttached',function() {\n dragpoint_board();\n});", "css": "table#values th {\n background: none;\n text-align: center;\n}"}, "parts": [{"type": "1_n_2", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Banking Sector: IT Infrastructure Spending

What was the maximum spend by a single company last year?

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£{answer} m

", "

£{fakeanswer1} m

", "

£{fakeanswer2} m

", "

£{fakeanswer3} m

", "

£{fakeanswer4} m

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Interpreting the elements of a box plot

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The diagram below shows a box plot of some data.

{geogebra_applet{\"https://www.geogebra.org/m/aj2hcbhg\",[lv: lv,lq: lq,m: m,uq: uq,hv: hv]}}

", "advice": "

A boxplot (also known as a box-and-whisker diagram or plot) is a convenient way of graphically depicting groups of numerical data through their five-number summaries: the smallest observation (sample minimum), lower quartile (Q1), median (Q2), upper quartile (Q3), and largest observation (sample maximum). A boxplot may also indicate which observations, if any, might be considered outliers.

For more information on box plots follow this link.

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Which of these statements are true and which are false?

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The median of the data is $\\var{m}$.

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Calculate an intersection probability given a two way table.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

a) Each row and column must sum to the 'total'.

b) Look for the column containing '$\\var{q1a}$' and the row containing '$\\var{q1b}$'. The entry where they intersect, $\\var{q1*total}$, is the value we are interested in.

Since we require a probability, this is $\\var{q1*total}$ out of $\\var{total}$, i.e.

\\[ \\frac{\\var{q1*total}}{\\var{total}} \\]

Use this link to find some resources which will help you revise this topic

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$\\var{total}$ items are sampled. Complete the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	$\\var{nB}$	not $\\var{nB}$	TOTAL
$\\var{nA}$	[[0]]	$\\var{AnB'}$	$\\var{A}$
not $\\var{nA}$	$\\var{notAnB}$	[[1]]	[[2]]
TOTAL	[[3]]	[[4]]	$\\var{total}$

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If one item is picked at random, use the table to calculate the probability that the item is '{q1a}' and '{q1b}'.

Give your answer as a fraction, or a decimal correct to 2dp.

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Reading a value from a histogram.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

The histogram shows information about distances run on a Sunday by some randomly asked people in a park.

", "advice": "

To calculate the frequencies using the following formula:

frequency = class width $\\times$ frequency density

Hence our table becomes:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Distance, $km$	Frequency
$0 < t \\leq 10$	$10 \\times \\var{yo0} = \\var{freq0}$
$10 < t \\leq 15$	$5 \\times \\var{yo1} = \\var{freq1}$
$15 < t \\leq 20$	$5 \\times \\var{yo2} = \\var{freq2}$
$20 < t \\leq 30$	$10 \\times \\var{yo3} = \\var{freq3}$

Hence, to find the total number of people that ran that day:

$\\var{freq0} + \\var{freq1} + \\var{freq2} + \\var{freq3} = \\var{total}.$

The frequency of runners that ran less than $10km$ is found by calculating the frequency from the first bar on the histogram:

class width $\\times$ frequency density $= 10 \\times \\var{yo0} = \\var{freq0}$.

The frequency of runners that ran between $15km$ and $20km$ is found by calculating the frequency from the third bar on the histogram:

class width $\\times$ frequency density $= 5 \\times \\var{yo2} = \\var{freq2}$.

To estimate how many runners ran more than $25km$ we need again need to use the frequency = class width \\times frequency density formula.

Here the class width is $5$ because we are looking for the frequency of runners that ran between $25km$ and $30km$ from the college.

Frequency $= 5 \\times \\var{yo3} = \\var{5*yo3}.$

If this number is a decimal we round up to get $\\var{m25}$.

As in part d),to estimate how many runners ran less than $7km$ we use the frequency = class width \\times frequency density formula.

Here the class width is $7$ because we are looking for the frequency of runners that ran between 0 and 7 km.

Frequency $= 7 \\times \\var{yo0} = \\var{7*yo0}.$

If this number is a decimal we round up to get $\\var{l7}$.

To estimate how many runners ran between $5km$ and $12.5km$ we use a method similar to that in part d) and e) but, this time, we need to use information from both the first and second bar on the histogram.

Let's first calculate how many runners ran between $5km$ and $10km$:

frequency $=$ class width $\\times$ frequency density $= 5 \\times \\var{yo0} = \\var{5*yo0}$.

Now we need to calculate how many runners ran between $10km$ and $12.5km$:

frequency $=$ class width $\\times$ frequency density $= 2.5 \\times \\var{yo1} = \\var{2.5*yo1}$.

Putting this together the number of runners that ran between $5km$ and $12.5km$ is $\\var{5*yo0} + \\var{2.5*yo1} = \\var{5*yo0 + 2.5*yo1}.$

If this number is a decimal we round up to get $\\var{ef}.$

The number of runners that ran between $10km$ and $14km$ is $4 \\times \\var{yo1} = \\var{4*yo1}.$

If this number is a decimal we round up to get $\\var{eh}.$

The number of runners that ran further than $18km$ is:

$2 \\times \\var{yo2} + 10 \\times \\var{yo3} = \\var{2*yo2} + \\var{freq3} = \\var{2*yo2+freq3}.$

If this number is a decimal we round up to get $\\var{ej}.$

Did you get a decimal answer? Were you surprised to see a whole number answer when you got a decimal on one of the estimate questions? Look at the context of the question, you cannot have $0.5$ of a student so we round our answers up to the next whole number!

Use this link to find some resources which will help you revise this topic.

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Frequency Density

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Frequency Density

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Frequency Density

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Frequency Density

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Sum of frequencies

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Frequency

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Frequency

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Frequency

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Frequency

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How many people ran more than 25km

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How many runners ran less than 7km

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Number of people running between 5 & 12.5 km

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Number of runners running between 10 & 14km

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Number of runners running more than 18km

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How many people were asked about the distance they ran that day?

", "minValue": "total", "maxValue": "total", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "b)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

How many runners ran less than 10 km that day?

", "minValue": "freq0", "maxValue": "freq0", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "c)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

How many people ran between 10 and 15 km that day?

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Estimate how many runners ran more than 25 km that day?

", "minValue": "m25", "maxValue": "m25", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": true, "customName": "e)", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Estimate how many runners ran less than 7 km that day?

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Estimate how many people ran between 5 km and 12.5 km.

", "minValue": "Ef", "maxValue": "Ef", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SD07 Interpreting Line Graphs", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Francesca recorded the number of customers in a supermarket every two hours.

She began at 9 am and finished at 7 pm.

The line graph below shows her results.

{geogebra_applet{\"https://www.geogebra.org/classic/s4w7nmga\",[y1:y1,y3:y3,y4:y4,y6:y6]}}

", "advice": "

a) You want to find the point on the graph with the greatest frequency. You can see that this is at 1pm when there were $\\var{y3}$ cars in the car park.

b) We find 11am on the $x$-axis and look vertically upwards until we find the point. From here we go horiztonally across to the $y$-axis to read the frequency. We can see that at 11am there were $\\var{y2}$ cars in the car park.

c) We want to find $\\var{number}$ on the $y$-axis and then look horiztonally across until we find the point. From here we move down to the $x$-axis to see at which time there were $\\var{number}$ cars in the car park. We can see this occured at $\\var{answerc}\\var{time}$.

d) We must find 6pm on the $x$-axis. This isn't marked on like the other times but we know it sits half way between 5pm and 7pm. From here we look vertically upwards until we meet the red line on our graph. Notice we don't have a point for this time, hence why this is an estimate. From here we move hortizontally across to the $y$-axis to find the frequency. At 6pm we estimate that there were $\\var{y56}$ cars in the car park.

e) We must find the frequency of cars at 1pm and at 3pm using the same steps as in part b. At 3pm there were $\\var{y4}$ cars in the car park and we subtract this from $\\var{y3}$ which is the number of cars in the car park at 1pm. Hence, $\\var{y3}-\\var{y4}=\\var{y3-y4}$.

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Number of customers in supermarket at 7pm.

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Number of customers in supermarket at 5pm.

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Number of customers in supermarket at 1pm.

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Number of customers in supermarket at 3pm.

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Number of customers in supermarket at 11am.

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Number of customers in supermarket at 9am

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Number of customers in supermarket at 6pm.

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Randomly select which number of customers user needs to identify associated time for.

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Associated time to randomly selected frequency variable (number)

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For printing am or pm depending on time selelected - to appear in advice only

", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["y6", "y5", "y3", "y4", "y2", "y1", "y56", "number", "answerc", "time"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "1_n_2", "useCustomName": true, "customName": "a)", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

When were the most customers in the supermarker?

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How many customers were in the supermarket at 11 am?

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At what time were there {number} customers in the supermarket?

If your answer was 12pm you would just write 12 in the box.

Estimate the number of customers in the supermarket at 6pm.

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How many less customers were there in the supermarket at 3pm than 1pm?

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Decide whether each of the described sets of data is drawn from a discrete or continuous distribution.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Decide whether the following data sets are discrete or continuous.

", "advice": "

Data can either be discrete or continuous.

Discrete data consists of integers, units of measurement cannot be split. Only certain values have a meaning.
Continuous data can be shown on a number line, all points on this line have a meaning. In between two measurements there's always a valid \"middle\" measurement. So for example, between 156.3cm and 156.4cm there's 156.35cm.

a)

Height is a continuous variable. For example, 180.3cm and 180.4cm have a valid midpoint 180.35cm.Weight is a continuous variable. For example, 54.5kg and 54.6kg have a valid midpoint 54.55kg.Time is a continuous variable. For example, 54.2s and 54.3s have a valid midpoint 54.25s.Temperature is a continuous variable, it can take any value between -273.15°C (absolute zero) and positive infinity. For example, 25°C and 26°C have a valid midpoint 25.5°C. Hence, this data is continuous.

b)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5. The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

c)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not..The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy a half of a bar. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

d)

e)

When we round continuous variables to the nearest integer, this data becomes discrete, as there are no valid middle measurements between the integers. Therefore, the weight of a dog to the nearest kgthe height of Olympic medalists to the nearest cmthe time taken to run 10km to the nearest min is discrete and not continuous.

f)

The number of Stage 1 students will always be an integer. You cannot split one student into two, for example value 19.5 students does not make sense. Therefore, this is a discrete set of data.The result of rolling 3 dice can take values of integers from 3 up to 18. For example, values 3 and 4 do not have any valid middle measurement. Therefore, this is a discrete set of data.Shoe sizes are a discrete set of data. For example, sizes 39 and 40 mean something while the middle value 39.5 does not.The number of chocolate bars sold on Monday will always be an integer. There is no middle measurement between 1 and 2 bars sold. You cannot buy half of a bar of chocolate. Therefore, this is a discrete set of data.The number of movies downloaded will always be an integer. You can either download a movie successfully or unsuccessfuly, so this is a discrete set of data. It is impossible to split 0 and 1 movies downloaded into 0.5.The number of cinema tickets sold will always be a whole number. There is no middle measurement between 1 and 2 tickets sold. You simply cannot buy half of a ticket. Therefore, this is a discrete set of data.

Use this link to find some resources which will help you revise this topic

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{cont[ranc]}

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{disc[rand]}

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{disc[rand2]}

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{disc[rand3]}

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"], "matrix": ["1", 0], "distractors": ["", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Stats - Measure and Location of Spread", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], [], [], [], [], []], "questions": [{"name": "SM01 Calculate Range", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["mean", "measures of average and spread", "median", "mode", "range", "taxonomy"], "metadata": {"description": "

This question provides a list of data to the student. They are asked to find the \"range\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

Use this link to find some resources which will help you revise this topic.

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Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

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Sorted list.

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Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

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The vector of number of times of each value in the data.

The final list.

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Find the range.

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Calculating the Mean from a basic list of integers.

", "licence": "None specified"}, "statement": "

Calculate the Mean from a list

", "advice": "

The MEAN is the sum, divided by the number of values summed i.e.

$\\frac{\\var{list[0]} + \\var{list[1]} + \\var{list[2]} + \\var{list[3]} + \\var{list[4]}}{5}$

use your calculator to find

mean = $\\var{mean}$.

Use this link to find some resources which will help you revise this topic.

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Given a list of numbers:

{list}

Calculate the mean: [[0]]

This question provides a list of data to the student. They are asked to find the \"mode\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The mode is the value that occurs the most often in the data.

To find a mode, we can look at our sorted list:

$\\var{a_s[0]}, \\var{a_s[1]}, \\var{a_s[2]}, \\var{a_s[3]}, \\var{a_s[4]}, \\var{a_s[5]}, \\var{a_s[6]}, \\var{a_s[7]}, \\var{a_s[8]}, \\var{a_s[9]}, \\var{a_s[10]}, \\var{a_s[11]}, \\var{a_s[12]}, \\var{a_s[13]}, \\var{a_s[14]}, \\var{a_s[15]}, \\var{a_s[16]}, \\var{a_s[17]}, \\var{a_s[18]}, \\var{a_s[19]}$.

We notice that $\\var{mode1}$ occurs the most ($\\var{modetimes[mode1]}$ times) so $\\var{mode1}$ is the mode.

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

Option 1 for the list. Only used if there is only one mode.

", "templateType": "anything", "can_override": false}, "a_s": {"name": "a_s", "group": "final list", "definition": "sort(a)", "description": "

Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

", "templateType": "anything", "can_override": false}, "modetimes": {"name": "modetimes", "group": "final list", "definition": "map(\nlen(filter(x=j,x,a)),\nj, 0..8)", "description": "

The vector of number of times of each value in the data.

", "templateType": "anything", "can_override": false}, "mode1": {"name": "mode1", "group": "final list", "definition": "mode[0]", "description": "

Mode as a value.

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Mode as a vector.

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The final list.

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Find the mode.

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This question provides a list of data to the student. They are asked to find the \"median\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A random sample of 20 residents from Newcastle were asked about the number of times they went to see a play at the theatre last year.

Here is the list of their answers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

The median is the middle value. We need to sort the list in order:

\\[ \\var{a_s[0]}, \\quad \\var{a_s[1]}, \\quad \\var{a_s[2]}, \\quad \\var{a_s[3]}, \\quad \\var{a_s[4]}, \\quad \\var{a_s[5]}, \\quad \\var{a_s[6]}, \\quad \\var{a_s[7]}, \\quad \\var{a_s[8]}, \\quad \\var{a_s[9]}, \\quad \\var{a_s[10]}, \\quad \\var{a_s[11]}, \\quad \\var{a_s[12]}, \\quad \\var{a_s[13]}, \\quad \\var{a_s[14]}, \\quad \\var{a_s[15]}, \\quad \\var{a_s[16]}, \\quad \\var{a_s[17]}, \\quad \\var{a_s[18]}, \\quad \\var{a_s[19]} \\]

There is an even number of responses, so there are two numbers in the middle (10^th and 11^th place). To find the median, we need to find the mean of these two numbers $\\var{a_s[9]}$ and $\\var{a_s[10]}$:

\\begin{align}
\\frac{\\var{a_s[9]} + \\var{a_s[10]}}{2} &= \\frac{\\var{a_s[9] + a_s[10]}}{2} \\\\
&= \\var{median} \\text{.}
\\end{align}

Use this link to find some resources which will help you revise this topic.

Option 2 for the list. Only used if there is only one mode and option 1 was not used.

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Option 1 for the list. Only used if there is only one mode.

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Sorted list.

Option 3 for the list. Ensures there is only one mode (2) while still randomising the data.

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The vector of number of times of each value in the data.

Mode as a value.

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Mode as a vector.

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The final list.

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Find the median.

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Using a calculator, work out the mean for each of these frequency tables, give your answers to two decimal places.

", "advice": "

To calculate the mean shoe size we need to add a third column to our table where we multiply the grade by the number (frequency) of students who are that shoe size.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Shoe Size	Frequency	*Shoe Size Frequency**
3	{p1}	3 * {p1} = {3*p1}
4	{p2}	4 * {p2} = {4*p2}
5	{p3}	5 * {p3} = {5*p3}
6	{p4}	6 * {p4} = {6*p4}
7	{p5}	7 * {p5} = {7*p5}
8	{p6}	8 * {p6} = {8*p6}

Now we find the total sum of that third column:

$\\var{3*p1} + \\var{4*p2} + \\var{5*p3} + \\var{6*p4} + \\var{7*p5} + \\var{8*p6} = \\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}.$

To find the mean shoe size we must divide this total by the total number of students:

$\\frac{\\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}}{\\var{p1}+\\var{p2}+\\var{p3}+\\var{p4}+\\var{p5}+\\var{p6}} = \\frac{\\var{3*p1 + 4*p2 + 5*p3 + 6*p4 + 7*p5 + 8*p6}}{\\var{sum1}} = \\var{mean1a}.$

The question asks us for our answer to two decimal places so the last thing we need to do is round. Hence, the mean is $\\var{mean1}$.

You can use this same method to answer the other parts of this question!

Use this link to find some resources which will help you revise this topic.

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The mean - part a

", "templateType": "anything", "can_override": false}, "p1": {"name": "p1", "group": "Part a", "definition": "random(0 .. 50#1)", "description": "

Frequency 1 - part a

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Frequency 2 - part a

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Frequency 3 - part a

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Frequency 4 - part a

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Frequency 5 - part a

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Frequency 6 - part a

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Mean rounded to two decimal places - part a

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Sum of frequencies - part a

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Sum of frequencies - part b

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Frequency 1 - part b

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Frequency 2 - part b

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Frequency 3 - part b

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Frequency 4 - part b

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The mean - part b

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Mean rounded to two decimal places - part b

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Frequency 1 - part c

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Frequency 2 - part c

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Frequency 3 - part c

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Frequency 4 - part c

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Frequency 5 - part c

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Frequency 6 - part c

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Mean rounded to two decimal places - part c

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Sum of frequencies - part c

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The mean - part c

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The table below shows the distribution of shoe sizes amongst {sum1} students at a school in Sheffield.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Shoe size	Frequency
3	{p1}
4	{p2}
5	{p3}
6	{p4}
7	{p5}
8	{p6}

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The table below shows the distribution of number of pets that {sum2} randomly asked people have.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Number of Pets	Frequency
0	{p1b}
1	{p2b}
2	{p3b}
3	{p4b}

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The table below shows the distribution of the number of drinks per order at a coffee shop in Manchester.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Number of Drinks	Frequency
0	{p1c}
1	{p2c}
2	{p3c}
3	{p4c}
4	{p5c}
5	{p6c}

", "minValue": "mean3", "maxValue": "mean3", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM06 Choosing the appropriate average", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": [], "metadata": {"description": "

This question asks the student to choose the appropriate measure of average and spread for a data with outliers.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Which of the following measures would you choose if you were dealing with data which includes outliers? Select one measure of average and one measure of spread.

", "advice": "

The median is a more appropriate measure of average when your data contains outliers because outliers do not affect the median.

The interquartile range is the best measure of variability for skewed distributions or data sets with outliers. Because it’s based on values that come from the middle half of the distribution, it’s unlikely to be influenced by outliers.

Use this link to find some resources which will help you revise this topic.

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Identifying measures of spread or location (average)

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Match each of the following with what they measure.

", "advice": "

The mean is a measure of location or central tendancy. It is calcuated by summing all of the data values and dividing by the number of values.

The median is a measure of location or central tendancy. It is the middle value of an ordered data set.

The inter-quartile range is a measure of spread. The interquartile range is the difference between upper and lower quartiles.The lower quartile, or first quartile (Q1), is the value under which 25% of data points are found when they are arranged in increasing order. The upper quartile, or third quartile (Q3), is the value under which 75% of data points are found when arranged in increasing order. The inter-quartile range therefore gives us an idea of the middle 50% of the ordered data set.

The standard deviation is a measure of spread. It measures the dispersion of a data set relative to its mean.

The variance is a measure spread because it is the square of the standard deviation.

A p-value the probability that a particular statistical measure, such as the mean or standard deviation, of an assumed probability distribution will be greater than or equal to (or less than or equal to in some instances) observed results. A p-value is used to determine statistical significance, not measures of spread or location.

Use this link to find some resources which will help you revise this topic.

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\n // tRot.bindTo(txt);\n // board.update();\n\n \n//var chart2 = board.createElement('chart', dataArr, {chartStyle:'line,point'});\n//chart2[0].setProperty('strokeColor:black','strokeWidth:2','shadow:true');\n//for(var i=0; i<11;i++) {\n // chart2[1][i].setProperty({strokeColor:'black',fillColor:'white',face:'[]', size:4, strokeWidth:2});\n//}\n//board.unsuspendUpdate(); \n \n //board.unsuspendUpdate();\n\n}\n\nquestion.signals.on('HTMLAttached',function() {\n dragpoint_board();\n});", "css": "table#values th {\n background: none;\n text-align: center;\n}"}, "parts": [{"type": "m_n_x", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minMarks": 0, "maxMarks": 0, "minAnswers": 0, "maxAnswers": 0, "shuffleChoices": true, "shuffleAnswers": true, "displayType": "radiogroup", "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": ["Variance", "Mean", "Median", "Inter-quartile range", "P-value", "Standard deviation"], "matrix": [["1", 0, 0], [0, "1", 0], [0, "1", 0], ["1", 0, 0], [0, 0, "1"], ["1", 0, 0]], "layout": {"type": "all", "expression": ""}, "answers": ["Measure of Spread", "Measure of location (average)", "Neither measure of location nor measure of spread"]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "SM08 Calculate Range (Decimal)", "extensions": ["stats"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Chris Graham", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/369/"}, {"name": "Stanislav Duris", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1590/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}, {"name": "Lauren Desoysa", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21504/"}, {"name": "Will Morgan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/21933/"}, {"name": "Megan Oliver", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23526/"}, {"name": "Michael Pan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/23528/"}], "tags": ["mean", "measures of average and spread", "median", "mode", "range", "taxonomy"], "metadata": {"description": "

This question provides a list of data to the student. They are asked to find the \"range\".

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Here is a list of 20 random numbers:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$\\var{a[0]}$	$\\var{a[1]}$	$\\var{a[2]}$	$\\var{a[3]}$	$\\var{a[4]}$	$\\var{a[5]}$	$\\var{a[6]}$	$\\var{a[7]}$	$\\var{a[8]}$	$\\var{a[9]}$
$\\var{a[10]}$	$\\var{a[11]}$	$\\var{a[12]}$	$\\var{a[13]}$	$\\var{a[14]}$	$\\var{a[15]}$	$\\var{a[16]}$	$\\var{a[17]}$	$\\var{a[18]}$	$\\var{a[19]}$

", "advice": "

Range is the difference between the highest and the lowest value in the data.

To find this, we subtract the lowest value from the highest value:

\\[ \\var{max(a)} - \\var{min(a)} = \\var{range} \\text{.}\\]

Use this link to find some resources which will help you revise this topic.

Option 1 for the list. Only used if there is only one mode.

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Sorted list.

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Find the range.

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Here are the ages of 7 people:

{a1}, {a2}, {a3}, {a4}, {a5}, {a6}, {a7}.

", "advice": "

First we should order our list of ages from smallest to largest:

$\\var{list[0]}, \\var{list[1]}, \\var{list[2]}, \\var{list[3]}, \\var{list[4]}, \\var{list[5]}, \\var{list[6]}$.

To find the range of ages we must subtract the smallest age from the largest age,

$\\var{max(list)} - \\var{min(list)} = \\var{range}$.

Hence, the range is $\\var{range}$.

To calculate the interquartile range we subtract the lower quartile from the upper quartile.

To calculate the lower quartile:

Since we have an odd number of ages we had one to the total and divide by 4,

$\\frac{7+1}{4} = \\frac{8}{4} = 2.$

So we are looking for the $2^{nd}$ value in our list which is $\\var{lq}$.

Hence the lower quartile is $\\var{lq}$.

To find the upper quartile:

We still add one to the number of ages because this number is odd and then we find $75$% of it,

$\\frac{3\\times(7+1)}{4} = \\frac{3\\times8}{4} = \\frac{24}{4} = 6.$

So we are looking for the $6^{th}$ value in our list which is $\\var{uq}$.

Hence the lower quartile is $\\var{uq}$.

Now we can calculate the interquartile range by subtracting the lower quartile from the upper quartile,

$\\var{uq}-\\var{lq} = \\var{iqr}.$

Age 1

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Age 2

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Age 3

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Age 4

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Age 5

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Age 6

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Age 7

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Ages in ascending order

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Range

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Lower quartile

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Upper quartile

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Interquartile range

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Work out the range of ages.

Work out the interquartile range of the ages.

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Tina needs to take a train to work.

The table below shows information about how long the train is delayed for on each day.

The policy of the train company is that if your train is delayed for longer than 30 minutes, you will recieve a 50% discount on your train ticket.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Delay Time	Frequency
0 < t $\\leq$ 10	{f1}
10 < t $\\leq$ 20	{f2}
20 < t $\\leq$ 30	{f3}
30 < t $\\leq$ 40	{f4}
40 < t $\\leq$ 50	{f5}

", "advice": "

To find the estimated mean we must work out the midpoint of the delay time intervals and multiply this by the frequency to create the $fx$ column on the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Delay Time	Frequency	Midpoint	Frequency $\\times$ Midpoint, $fx$
0 < t $\\leq$ 10	{f1}	5	= 5$\\times\\var{f1} = \\var{f1m}$
10 < t $\\leq$ 20	{f2}	15	= 15$\\times\\var{f2} = \\var{f2m}$
20 < t $\\leq$ 30	{f3}	25	= 25$\\times\\var{f3} = \\var{f3m}$
30 < t $\\leq$ 40	{f4}	35	= 35$\\times\\var{f4} = \\var{f4m}$
40 < t $\\leq$ 50	{f5}	45	= 45$\\times\\var{f5} = \\var{f5m}$

Now to find the mean we calculate the sum of the $fx$ column and divide it by the sum of the frequencies:

Estimated Mean = $\\frac{\\var{f1m}+\\var{f2m}+\\var{f3m}+\\var{f4m}+\\var{f5m}}{5+15+25+35+45} = \\frac{\\var{freqmid}}{\\var{sum}} = \\var{meana}.$

Hence the estimated mean delay time is $\\var{mean}$ to two decimal places.

To work out what percentage of delays that were over 30 minutes we first need to calculate the number of days where the train was delayed for over 30 minutes.

The number of delays that over 30 minutes is $\\var{f4}+\\var{f5} = \\var{f4+f5}$.

We divide this by the total number of deliveries to find the fraction of delays that were over 30 minutes and then multiply this number by 100 to find the percentage:

$\\frac{\\var{f4+f5}}{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}+\\var{f5}} \\times 100\\% = \\frac{\\var{f4+f5}}{\\var{sum}} \\times 100\\% = \\var{percentagea} \\times 100\\%$.

Hence, the percentage of delays that were over 30 minutes is $\\var{percentage}\\%$ to two decimal places.

The percentage of delays that take over 40 minutes is:

$\\frac{\\var{f5}}{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}+\\var{f5}} \\times 100\\% = \\frac{\\var{f5}}{\\var{sum}} \\times 100\\% = \\var{percent40} \\times 100\\%$, to one decimal place.

So, only $\\var{percent40}\\%$ of the time, delays were longer than 40 minutes so the manager would probably wanto the policy changed to 40 minutes.

Use this link to find some resources which will help you revise this topic.

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Frequency 3

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Frequency 5

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Mean

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Sum of frequencies

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Frequency * midpoint 1

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Frequency * midpoint 2

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Frequency * midpoint 4

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Frequency * midpoint 5

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Sum of Frequency * midpoint

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Mean rounded to two decimal places

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Sum of 4th & 5th frequencies

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Percentage over 30 mins

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Percentage rounded to one decimal place

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Percentage over 40 mins

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Using a calculator, calculate an estimate for the mean delay time. Give your answer to two decimal places.

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What percentage of delays were over 30 minutes? Give your answer to one decimal place.

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The train company propose that they should only offer the 50% discount if the train is delayed for over 40 minutes.

If you were the manager of the train company would you agree? Type yes or no into the box and think about the reasons for your answer.

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This frequence table stores data about the lengths of some plants:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

Length ($x$ cm)	Frequency
10 < $x$ $\\leq$ 20	{f1}
20 < $x$ $\\leq$ 30	{f2}
30 < $x$ $\\leq$ 40	{f3}
40 < $x$ $\\leq$ 50	{f4}

", "advice": "

To find the median we must find the middle value. To do this we sum the frequencies and divide the result by 2:

$\\frac{\\var{f1}+\\var{f2}+\\var{f3}+\\var{f4}}{2} = \\frac{\\var{total}}{2} = \\var{middle1}$.

Hence we must find value number $\\var{middle1}$, we can do this by totalling the cumulative frequency in the table below.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

\n Length $x$ cm \n	Frequency	Cumulative Frequency
1. 10 < $x$ $\\leq$ 20	{f1}	$\\var{f1}$
2. 20 < $x$ $\\leq$ 30	{f2}	$\\var{f1}+\\var{f2}=\\var{cf2}$
3. 30 < $x$ $\\leq$ 40	{f3}	$\\var{f1}+\\var{f2} +\\var{f3} =\\var{cf3}$
4. 40 < $x$ $\\leq$ 50	{f4}	$\\var{f1}+\\var{f2} +\\var{f3} + \\var{f4} =\\var{total}$

You can see value number $\\var{middle1}$ lies in interval number $\\var{interval1}$ hence this is the median class interval for the time taken.

We can estimate the median by using interpolation:

$$
\\text{Estimate of median} = \\text{class start value} + \\frac{\\text{position in class}}{\\text{frequency in class}}\\times \\text{class width}
$$

In this case this is $\\var{lower1}+\\frac{\\var{middle1}-\\var{cf11}}{\\var{fint1}}\\times 10 =\\var{lower1}+\\var{interpolation1}=\\var{median1}.$

(Note: that sometimes it is convenient to use $(\\var{total} + 1)/2 = \\var{middle2}$ as the way to work out the median position. The distinction does not really matter as this is an estimated value and often the decision is made based on which is the most convenient to calculate). This question has been written so that both answers will be marked correctly.

To calculate the $i^{th}$ quartile ($Q_i$) we must use the following formula:

$Q_i = l + \\frac{\\frac{iN}{4}-F}{f}\\times h,$

where:

$l$ = lower limit of the interval in which $Q_i$ lies;

$N$ = Total number of observations;

$F$ = Cumulative frequency of class previous to the $i^{th}$ quartile class;

$f$ = Frequency of $i^{th}$ quartile class.

Since we want to calculate the lower quartile, $i=1$.

If we calculate $\\frac{iN}{4} = \\frac{\\var{total}}{4} = \\var{quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervallq}$. This means that $F = \\var{fprev}$ and $f = \\var{frequency}$.

Hence,

$Q_1 = \\var{lowerlq} + \\frac{\\var{quarter} - \\var{Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\frac{\\var{quarter-Fprev}}{\\var{frequency}} \\times 10 = \\var{lowerlq} + \\var{((quarter-Fprev)/frequency)*10} = \\var{lq}.$

So, The lower quartile is $\\var{lqround}$ to two decimal places.

To calculate the upper quartile we use the same formula as in part b) but this time $i=3$. So, $\\frac{iN}{4} = \\frac{\\var{3*total}}{4} = \\var{3*quarter}$, we can see in the table from part a) that the lower quartile will be in interval number $\\var{intervaluq}$. This means that $F = \\var{ufprev}$ and $f = \\var{ufrequency}$.

Hence,

$Q_3 = \\var{upperlq} + \\frac{\\var{3*quarter} - \\var{uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\frac{\\var{3*quarter-uFprev}}{\\var{ufrequency}} \\times 10 = \\var{upperlq} + \\var{((3*quarter-uFprev)/ufrequency)*10} = \\var{uq}.$

So, The lower quartile is $\\var{uqround}$ to two decimal places.

To calculate the interquartile range we subtract the lower quartile from the upper quartile.

Hence,

$IQR = Q_3 - Q_1 = \\var{uq}-\\var{lq}=\\var{iqr}$.

So the interquartile range rounded to two decimal places is $\\var{iqrround}$.

Use this link to find some resources which will help you revise this topic.

Frequency 1

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Frequency 2

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Frequency 3

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Frequency 4

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Sum of frequencies

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Middle Value

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Cumulative Frequency - first and second frequencies

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Cumulative freqeuncy - sum of first, second, third frequencies

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Lower bound of interval which median sits

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Formula to find lower quartile

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Interval which lower quartile sits in

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Quarter value

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Cumulative frequency of class previous to one which lower quartile sits in

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Frequency of class which lower quartile sits in

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Lower quartile rounded to 2 decimal places

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Upper quartile rounded to 2 decimal places

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Upper quartile

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cumulative frequency of class previous to the one which the upper quartile sits in

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Frequency of class upper quartile sits in

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class which upper quartile sits in

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Interquartile range

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width of class which lower quartile sits in

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width of class upper quartile sits in

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Round interquartle range to two decimal places

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Calculate the lower quartile, give your answer to 2 decimal places.

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Calculate an estimate for the upper quartile using linear interpolation, give your answer to two decimal places.

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Estimate the interquartile range, give your answer to two decimal places.

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Tests understanding of scatter plots and related concepts.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

The scatter plot below shows the relationship between an employee’s height in centimetres and how long it takes them to walk to work in minutes.

\n\n\n\n\n\n\n\n\n\n\n\n

time (mins)	{drawgraph()}
	height (cm)

", "advice": "

The graph shows that there is a positive correlation between a person's height and how long it takes them to walk to work.

A postive correlation is a relationship between two variables where both variables move in the same diection.

This tells us that as a person's height increases, the time it takes to walk to work increases.

Use this link to find some resources which will help you revise this topic

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p6y

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Mark the statement that best describes what this scatter plot shows.

", "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "radiogroup", "displayColumns": 0, "showCellAnswerState": true, "choices": ["

In general, there is a positive correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a negative correlation between a person's height and how long it takes them to walk to work.

", "

In general, there is a no correlation between a person's height and how long it takes them to walk to work.

"], "matrix": ["1", 0, 0], "distractors": ["", "", ""]}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}, {"name": "Stats - Probability", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "questions": [{"name": "SP01 Probability - \"sample space\"", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ruth Hand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3228/"}, {"name": "Mash Sheffield", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4679/"}], "tags": [], "metadata": {"description": "

Calculate probability of selecting coloured counters from a bag.

", "licence": "None specified"}, "statement": "

A bag contains:

$\\var{srn}$ small, red tokens,
$\\var{sbn}$ small, blue tokens,
$\\var{brn}$ large, red tokens, and
$\\var{bbn}$ large, blue tokens.

", "advice": "

part a)

A probability is a fraction. You can give your answer as a fraction, decimal or percentage as these are all equivalent.

The formula for probability is:

\\[ P(A) = \\frac{\\text{number of possibilities for A}}{\\text{number of total possible outcomes}} \\]

For this question the total possible outcomes are $\\var{srn}+\\var{sbn}+\\var{brn}+\\var{bbn} = \\var{total}$.

Therefore

\\[ P(\\text{A large red token}) = \\frac{\\var{brn}}{\\var{total}} = \\var[fractionnumbers]{brn/total}\\]

part b)

For this question we need to know the total number of small tokens, i.e. $\\var{srn}+\\var{sbn} = \\var{srn+sbn}$.

Therefore

\\[ P(\\text{A small token}) = \\frac{\\var{srn+sbn}}{\\var{total}} = \\var[fractionnumbers]{(srn+sbn)/total}\\]

Use this link to find some resources which will help you revise this topic.

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You take a token at random.

What is the probability that it is a large, red token?

Give your answer as a fraction, or a decimal correct to 2dp.

You take a token at random.

What is the probability that it is a small token?

Give your answer as a fraction, or a decimal correct to 2dp.

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Predicting the probability of an unbiased coin landing on heads based on the results of previous throws.

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When we flip an unbiased coin there are two possible events that we could measure: the coin lands on heads or the coin lands on tails.

Each toss of the coin is independent; if we flip a coin once and it lands on heads then the next time we flip the coin it is still equally likely to land on either heads or tails.

It doesn't matter what the coin landed on previously as this outcome does not affect the outcome of the next flip of the coin.

Even when we flip an unbiased coin $\\var{no_flips}$ times and it lands on heads each time; the next time we flip the coin, it is still equally likely to land on either heads or tails.

So the probability that the coin lands on heads the next time that the coin is flipped is still $\\displaystyle\\frac{1}{2}$.

Use this link to find some resources which will help you revise this topic.

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Number of flips of the coin

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An unbiased coin is flipped $\\var{no_flips}$ times. Given that the coin landed on heads each time, what is the probability of the coin landing on heads the next time it is flipped?

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Complete the two way table.

", "advice": "\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

	Football	Rugby	Tennis	Total
Year 7	$\\var{f7}$	$\\var{r7}$	$\\var{total7}-(\\var{f7}+\\var{r7}) = \\var{total7}-\\var{f7+r7} = \\var{t7}$	$\\var{total7}$
Year 8	$\\var{totalf}-\\var{f7} = \\var{f8}$	$\\var{r8}$	$\\var{t8}$	$\\var{total}-\\var{total7} = \\var{total8}$
Total	$\\var{totalf}$	$\\var{r7}+\\var{r8}=\\var{totalr}$	$\\var{t7}+\\var{t8}=\\var{totalt}$	$\\var{total}$

Use this link to find some resources which will help you revise this topic.

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number of year 7's playing football

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Number of year 7's playing rugby

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Number of year 7's playing tennis

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Number of year 8's play8ng football

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Number of year 8's playing rugby

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Number of year 8's playing tennis

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Total number of year 8 students

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Total number of students playing football

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Total number of students playing rugby

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Total number of students playing tennis

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Total number of students in year 7 and year 8

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Total number of students playing a sport

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Total number of students

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	Football	Rugby	Tennis	Total
Year 7	{f7}	{r7}	[[2]]	{total7}
Year 8	[[0]]	{r8}	{t8}	[[4]]
Total	{totalf}	[[1]]	[[3]]	{total}

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The answer is a comma-separated list of numbers.

The list is marked correct if each number occurs the same number of times as in the expected answer, and no extra numbers are present.

You can optionally treat the answer as a set, so the number of occurrences doesn't matter, only whether each number is included or not.

Is every number in the student's list valid?

Are the student's answers in ascending order?

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Is each number in the expected answer present in the student's list the correct number of times?

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True if the student's list doesn't contain any numbers that aren't in the expected answer.

Should the answer be considered as a set, so the number of times an element occurs doesn't matter?

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Numbers included in the student's answer that are not in the expected list.

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The answer is either 'True' or 'False'

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