Numbas Worksheet
", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", "", "", "", ""], "questions": [{"name": "Transposing formula", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "David Goulding", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2365/"}], "preamble": {"js": "", "css": ""}, "advice": "", "tags": [], "parts": [{"unitTests": [], "showFeedbackIcon": true, "useCustomName": false, "showCorrectAnswer": true, "sortAnswers": false, "extendBaseMarkingAlgorithm": true, "scripts": {}, "gaps": [{"checkingType": "absdiff", "useCustomName": false, "variableReplacements": [], "vsetRangePoints": 5, "answerSimplification": "std", "variableReplacementStrategy": "originalfirst", "customName": "", "vsetRange": [0, 1], "answer": "({c} - {b}x)/({a} - {d}x)", "marks": "5", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showFeedbackIcon": true, "checkVariableNames": false, "failureRate": 1, "checkingAccuracy": 0.001, "valuegenerators": [{"name": "x", "value": ""}], "showPreview": true, "showCorrectAnswer": true, "scripts": {}, "type": "jme", "customMarkingAlgorithm": ""}], "prompt": "$\\simplify[std]{{a}y + {b}x = {c} + {d}xy}\\;$
\n$y =$ [[0]]
\nYou can click on \"Show steps\" for more information, but you will lose one mark if you do so.
", "customName": "", "steps": [{"unitTests": [], "showFeedbackIcon": true, "useCustomName": false, "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "scripts": {}, "prompt": "To re-arrange $ay + bx = c + dxy$ we should first collect all of the terms involving $y$ to the one side
\n$ay - dxy = c - bx$
\nwe should then factorize out $y$ to find
\n$y(a-dx) = c - bx$
\nand then divide by $a-dx$ to get $y$ on its own
\n$y = \\frac{c - bx}{a - dx}$
", "customName": "", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "information", "customMarkingAlgorithm": ""}], "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "marks": 0, "type": "gapfill", "customMarkingAlgorithm": "", "stepsPenalty": 1}], "metadata": {"description": "Another transposition question.
\nrebalmaths
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Rearrange the following equation to make $y$ the subject.
", "variablesTest": {"maxRuns": 100, "condition": ""}, "ungrouped_variables": ["a", "c", "b", "d"], "rulesets": {"std": ["all", "!noLeadingMinus", "!collectNumbers"]}, "variables": {"a": {"group": "Ungrouped variables", "name": "a", "templateType": "anything", "definition": "random(-10..10 except 0)", "description": ""}, "b": {"group": "Ungrouped variables", "name": "b", "templateType": "anything", "definition": "random(-10..10)", "description": ""}, "c": {"group": "Ungrouped variables", "name": "c", "templateType": "anything", "definition": "random(-10..10)", "description": ""}, "d": {"group": "Ungrouped variables", "name": "d", "templateType": "anything", "definition": "random(-10..10)", "description": ""}}, "functions": {}, "variable_groups": [], "type": "question"}, {"name": "TP6", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "parts": [{"showFeedbackIcon": true, "prompt": "\n$y = $ [[0]]
", "variableReplacements": [], "sortAnswers": false, "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true, "scripts": {}, "customMarkingAlgorithm": "", "steps": [{"showFeedbackIcon": true, "scripts": {}, "prompt": "It may be helpful to factor out y. For example:
\n\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y}=y(\\simplify{x^{{n1}} + {n2}*x - {n3}})\\]
", "showCorrectAnswer": true, "variableReplacements": [], "customMarkingAlgorithm": "", "marks": 0, "type": "information", "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true}], "type": "gapfill", "stepsPenalty": 0, "gaps": [{"showFeedbackIcon": true, "checkVariableNames": false, "variableReplacements": [], "showCorrectAnswer": true, "vsetRangePoints": 5, "expectedVariableNames": [], "variableReplacementStrategy": "originalfirst", "unitTests": [], "extendBaseMarkingAlgorithm": true, "failureRate": 1, "scripts": {}, "customMarkingAlgorithm": "", "showPreview": true, "vsetRange": [0, 1], "type": "jme", "answer": "{n4}/(x^{{n1}} + {n2}*x-{n3})", "marks": "5", "checkingType": "absdiff", "checkingAccuracy": 0.001}]}], "tags": [], "functions": {}, "advice": "Here, we first have to collect all terms involving $y$ on the same side. Hence, we get:
\n\\[\\simplify{x^{{n1}}*y+{n2}*y*x - {n3}*y} = \\var{n4}\\]
\nWe then spot that $y$ appears exactly once in each term on the left, so factorise:
\n\\[y(\\simplify{x^{{n1}} + {n2}*x - {n3}}) = \\var{n4}\\]
\nand simple division gives the answer.
", "rulesets": {}, "ungrouped_variables": ["n1", "n2", "n3", "n4"], "statement": "Consider the equation:
\n\\[\\simplify{x^{{n1}}*y + {n2}*y*x} = \\simplify{{n3}*y} + \\var{n4}\\]
\nRe-arrange this equation to make $y$ the subject.
", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"js": "", "css": ""}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "Another transposition question, which requires (basic) factorisation.
\nrebelmaths
"}, "variables": {"n1": {"definition": "random(2..6)", "group": "Ungrouped variables", "templateType": "anything", "name": "n1", "description": ""}, "n4": {"definition": "random(1..10)", "group": "Ungrouped variables", "templateType": "anything", "name": "n4", "description": ""}, "n3": {"definition": "random(-7..7 except 0 n2)", "group": "Ungrouped variables", "templateType": "anything", "name": "n3", "description": ""}, "n2": {"definition": "random(-7..7 except 0)", "group": "Ungrouped variables", "templateType": "anything", "name": "n2", "description": ""}}, "type": "question"}, {"name": "TP7", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "preamble": {"js": "", "css": ""}, "parts": [{"variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "prompt": "$S = $ [[0]]
\n", "scripts": {}, "unitTests": [], "showFeedbackIcon": true, "variableReplacements": [], "marks": 0, "gaps": [{"answer": "({b}*V)/({a}-V)", "variableReplacementStrategy": "originalfirst", "expectedVariableNames": [], "showCorrectAnswer": true, "checkingAccuracy": 0.001, "scripts": {}, "failureRate": 1, "vsetRange": [0, 1], "vsetRangePoints": 5, "unitTests": [], "showFeedbackIcon": true, "variableReplacements": [], "marks": "5", "showPreview": true, "checkingType": "absdiff", "extendBaseMarkingAlgorithm": true, "checkVariableNames": false, "customMarkingAlgorithm": "", "type": "jme"}], "sortAnswers": false, "extendBaseMarkingAlgorithm": true, "customMarkingAlgorithm": "", "type": "gapfill"}], "tags": [], "functions": {}, "advice": "Start by multiplying both sides by the denominator.
\nFor example if you have $V=\\frac{5S}{S+12}$ then multiply both sides by $(S+12)$.
\nThis gives: $V \\color{red}{(S+12)}=\\frac{5S}{S+12} \\color{red}{(S+12)} $
\nThe $(S+12)$ terms on the right hand side cancel out to give: $V(S+12)=5S$
\nNow expand out the brackets on the left hand side: $\\color{red}{VS+12V}=5S$
\nNext collect the like terms, you want to get all the terms with $S$ in them onto one side, so subtract $VS$ from both sides:
\n$VS \\color{red}{-VS} + 12V=5S \\color{red}{-VS}$
\nThis becomes $12V=5S-VS$
\nNow you can factorise the right hand side: $12V=\\color{red}{S(5-V)}$
\nFinally divide both sides by $(5-V)$ to leave $S$ on its own: $\\frac{12V}{\\color{red}{5-V}}=S$
\n", "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "rearranging the Michelas-Menten equation to make the substrate the subject.
\nrebelmaths
"}, "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"name": "a", "templateType": "anything", "definition": "random(2..8)", "description": "", "group": "Ungrouped variables"}, "b": {"name": "b", "templateType": "anything", "definition": "random(5..16)", "description": "", "group": "Ungrouped variables"}}, "statement": "Rearrange the following equation to make $S$ the subject:
\n\n\\[ V=\\frac{\\var{a}S}{S+\\var{b}}\\]
\n\nNote: To write a fraction you type (numerator)/(denominator).
", "variable_groups": [], "ungrouped_variables": ["a", "b"], "type": "question"}, {"name": "TP8", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "variables": {"c": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "c", "definition": "random(2..30 except([a,b]))"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "a", "definition": "random(2..15)"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "description": "", "name": "b", "definition": "random([3,5,7,11,13,17,19] except(a))"}}, "ungrouped_variables": ["a", "b", "c"], "rulesets": {}, "parts": [{"sortAnswers": false, "variableReplacements": [], "prompt": "$x = $ [[0]]
", "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "marks": 0, "scripts": {}, "unitTests": [], "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "gaps": [{"variableReplacements": [], "vsetRangePoints": 5, "expectedVariableNames": [], "customMarkingAlgorithm": "", "variableReplacementStrategy": "originalfirst", "showPreview": true, "marks": 1, "scripts": {}, "unitTests": [], "showCorrectAnswer": true, "extendBaseMarkingAlgorithm": true, "answer": "({c} p^2 - {a})/{b}", "checkingType": "absdiff", "checkingAccuracy": 0.001, "checkVariableNames": false, "failureRate": 1, "showFeedbackIcon": true, "type": "jme", "vsetRange": [0, 1]}], "showFeedbackIcon": true, "type": "gapfill"}], "advice": "We wish to make $x$ the subject, so we must get $x$ on its own. At the moment, the only $x$ in the equation is in the fraction under the square root. Until the square root is gone, everything under the square root is trapped together.
\nHow can we get rid of the square root? Ans: Apply the inverse function of the square root (i.e. do the opposite of taking the square root) which is to square.
\nRemember, if we wish to keep the equation balanced i.e. keep both sides equal, we must always do the same thing to both sides. So, if we square the right hand side in order to get rid of the square root, we must also square the left hand side:
\n\\[ \\begin{align*} p^{\\color{red}{2}} & = \\left( \\sqrt{\\frac{\\var{a}+\\var{b}x}{\\var{c}}} \\right)^{\\color{red}{2}} \\\\ \\Rightarrow \\ p^2 & = \\frac{\\var{a}+\\var{b}x}{\\var{c}}\\end{align*}\\]
\nNow looking at the right hand side, we have a fraction. Everything on top of the line of a fraction is trapped together and everything under the line of a fraction is trapped together until the fraction is gone. So we need to get rid of the fraction. We do this by multiplying both sides by the denominator, $\\var{c}$ (since multiplication is the opposite of division). This gives:
\n\\[ \\begin{align*} p^2 \\color{red}{\\times \\var{c}} & =\\frac{\\var{a}+\\var{b}x}{\\var{c}} \\color{red}{\\times \\var{c}}\\\\ \\Rightarrow \\ \\var{c} p^2 & = \\var{a} + \\var{b}x \\end{align*} \\]
\n\nNext, we subtract $\\var{a}$ from both sides to get rid of it from the right hand side:
\n\\[ \\begin{align*} \\var{c} p^2 \\color{red}{- \\var{a}} & = \\var{a} + \\var{b}x \\color{red}{- \\var{a}}\\\\ \\Rightarrow \\ \\var{c} p^2 - \\var{a} & = \\var{b} x \\end{align*} \\]
\nFinally, we divide both sides by $\\var{b}$ to get $x$ on its own on the right hand side (since division is the opposite of multiplication). This gives:
\n\\[ \\begin{align*} \\frac{\\var{c} p^2 - \\var{a}}{\\color{red}{\\var{b}}} & = \\frac{\\var{b} x}{\\color{red}{\\var{b}}}\\\\ \\Rightarrow \\ \\frac{\\var{c} p^2 - \\var{a}}{\\var{b}} & = x\\end{align*} \\]
", "statement": "Make $x$ the subject of the following formula:
\n\\[ p=\\sqrt{\\frac{\\var{a}+\\var{b}x}{\\var{c}}} \\]
", "metadata": {"licence": "None specified", "description": ""}, "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"css": "", "js": ""}, "variable_groups": [], "tags": [], "functions": {}, "type": "question"}, {"name": "TP9", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}], "statement": "Transpose the following equation to find $B$
\n\\[ \\frac{1}{A} = \\frac{1}{B} + \\frac{1}{C} \\]
\nNote: When inputting an expression like $AB$, you need to input $A * B$.
", "rulesets": {}, "tags": [], "parts": [{"scripts": {}, "marks": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "sortAnswers": false, "showCorrectAnswer": true, "variableReplacements": [], "prompt": "$B = $ [[0]]
", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "unitTests": [], "type": "gapfill", "gaps": [{"checkingType": "absdiff", "scripts": {}, "marks": "2", "customMarkingAlgorithm": "", "checkingAccuracy": 0.001, "expectedVariableNames": [], "extendBaseMarkingAlgorithm": true, "vsetRangePoints": 5, "checkVariableNames": false, "showCorrectAnswer": true, "variableReplacements": [], "showPreview": true, "answer": "A*C/(C-A)", "showFeedbackIcon": true, "unitTests": [], "vsetRange": [0, 1], "failureRate": 1, "type": "jme", "variableReplacementStrategy": "originalfirst"}]}], "variablesTest": {"condition": "", "maxRuns": 100}, "preamble": {"css": "", "js": ""}, "variables": {}, "functions": {}, "metadata": {"licence": "None specified", "description": ""}, "ungrouped_variables": [], "variable_groups": [], "advice": "We need to get $B$ on its own. There are a number of different ways you might do this. This is just one possible solution:
\nAt the moment, there is a single $B$ in the first fraction on the right hand side. Therefore we need to get rid of that fraction. Very often it can make calculations easier if there are no fractions at all, and it is often as easy to get rid of all fractions as to get rid of just one fraction. That is the case here. If we multiply both sides by $ABC$, we can get rid of all fractions in our equation:
\n\\[ \\begin{align*} \\frac{1}{A} \\color{red}{\\times ABC} & = \\frac{1}{B} \\color{red}{\\times ABC} + \\frac{1}{C} \\color{red}{\\times ABC}\\\\ BC & = AC + AB \\end{align*} \\]
\nSince we wish to get $B$ on its own, we now make sure every term that has a $B$ is on the one side of the equals sign and every term without a $B$ is on the other side. If we subtract $AB$ from both sides we get:
\n\\[ \\begin{align*} BC \\color{red}{- AB} & = AC + AB \\color{red}{- AB}\\\\ BC - AB & = AC \\end{align*} \\]
\nNext, factorising the left hand side:
\n\\[ \\color{red}{B(C - A)} = AC \\]
\nFinally, dividing both sides by $C - A$:
\n\\[ \\begin{align*} \\frac{B(C - A)}{\\color{red}{C - A}} & = \\frac{AC}{\\color{red}{C - A}}\\\\ B & = \\frac{AC}{C - A} \\end{align*} \\]
", "type": "question"}, {"name": "Simple linear equations", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "David Goulding", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2365/"}], "tags": [], "metadata": {"description": "", "licence": "None specified"}, "statement": "Solve the following equations for x. Enter your answer as a fraction or an integer, do not enter a decimal.
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\n$x=\\;$ [[0]]
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\n$x=\\;$ [[0]]
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\n$x=\\;$ [[0]]
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\n$x=\\;$[[0]]
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Using the laws of indices, simplify each expression down to its simplest form. Recall that $a^{0} = 1$ for any number $a$.
\n\nTo write $a^{2}$, type a^2. For fractional powers, e.g. $a^{2/3}$, write a^(2/3).
", "advice": "Here we are using the rule of indices: $a^m \\times a^n = a^{m+n}$.
\nUsing this rule,
\n\\[
\\begin{align}
a^\\var{x} \\times a^\\var{y}\\ &= a^\\simplify[all, !collectNumbers]{{x}+{y}}\\\\
&= a^\\var{x+y}.
\\end{align}
\\]
We are asked to find $\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}$.
\nNotice there is a constant in front of each of the terms.
\nTo do this, write the product out explicitly, as
\n\\[\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q} = \\var{c} \\times \\var{d} \\times a^\\var{p} \\times a^\\var{q}.\\]
\nWe know that $\\var{c} \\times \\var{d} = \\var{c*d}$, and using the rule of indices: $a^\\var{p} \\times a^\\var{q} = a^\\var{p+q}$.
\nTherefore:
\n\\begin{align}
\\var{c}a^\\var{p} \\times \\var{d}a^\\var{q}&= \\var{c*d} \\times a^\\var{p+q} \\\\
&= \\simplify{{c*d}*a^{p+q}}.
\\end{align}
Here we are using: $a^m \\div a^n = a^{m-n}$.
\nWe are asked to simplify the expression, $\\displaystyle\\simplify{{b}*a^{x}/({g}*a^{y})}$.
\nTo do this, we just have to use the previously mentioned rule of indices. We write this out explicity as
\n\\[\\simplify{{b}*a^{x}/({g}*a^{y})} = \\simplify{{b}/{g}} \\times \\simplify{a^{x}/(a^{y})}.\\]
\nUsing rules of indices,
\n\\begin{align} \\frac{a^\\var{x}}{a^\\var{y}} &= a^\\var{x} \\div a^\\var{y}\\\\
&= a^\\simplify[all, !collectNumbers]{{x}-{y}}\\\\
&= a^\\var{x-y}.
\\end{align}
Therefore,
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\simplify{a^{{x}-{y}}}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
Alternatively,
\nUsing the rule of indices: $a^{-m} = \\displaystyle\\frac{1}{a^{m}}$, we can rewrite the question as:
\n\\begin{align}
\\frac{\\var{b}a^\\var{x}}{\\var{g}a^\\var{y}} &= \\simplify{{b}/{g}} \\times \\frac{a^\\var{x}}{a^\\var{y}}\\\\
&= \\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}}.
\\end{align}
And then using the rule: $a^m \\times a^n = a^{m+n}$, this becomes:
\n\\begin{align}
\\simplify{{b}/{g}} \\times a^\\var{x} \\times a^{-\\var{y}} &= \\simplify{{b}/{g}} \\times a^\\simplify[all,!collectNumbers]{{x}+(-{y})}\\\\
&= \\simplify{{b}/{g}*a^{x-y}}.
\\end{align}
The question asks us to simplify $(\\simplify{{c}*a^{p}})^{\\var{q}}$.
\nTo do this we use the rules:
\n\\[(a^{m})^{n} = a^{mn},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can then expand the equation as
\n\\[(\\simplify{{c}*a^{p}})^{\\var{q}}= \\var{c}^{\\var{q}} \\times (a^{\\var{p}})^{\\var{q}}.\\]
\nThen using the rule of indices mentioned previously,
\n\\[
\\begin{align}
(\\simplify{{c}*a^{p}})^{\\var{q}}&= \\simplify{{c}^{q}} \\times a^\\var{p*q}\\\\
&= \\simplify{{c}^{q}*a^{p*q}}.
\\end{align}
\\]
The question asks us to simplify $\\sqrt[\\var{d}]{\\var{x}^\\var{d}a}$.
\nTo do this we use the rules:
\n\\[a^\\frac{1}{m} = \\sqrt[m]{a},\\]
\n\\[(ab)^m = a^mb^m.\\]
\nWe can expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{d}]{a} &= (\\simplify{a})^\\frac{1}{\\var{d}}\\\\
&= a^\\frac{1}{\\var{d}}.
\\end{align}
\\]
The question requires us to simplify $\\sqrt[\\var{c}]{a^\\var{q}}$.
\nHere, we use the rule of indices: $a^\\frac{n}{m} = \\sqrt[m]{a^n}$, allowing us to expand the expression as follows:
\n\\[
\\begin{align}
\\sqrt[\\var{c}]{\\simplify{a^{q}}} &= \\simplify[fractionnumbers,all]{(a^{q})^{{1}/{{c}}}}\\\\
&= \\simplify[fractionnumbers,all]{a^{{q}/{c}}}.
\\end{align}
\\]
Used in part c
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\n$I=\\var{initial}\\times \\var{n}\\times \\var{p}/100$
\n$I=\\var{interest}$
\nAdding the interest to the initial amount gives €$\\var{answer1}$.
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\nGive your answer to $2$ decimal places.
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\nrebelmaths
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\nPart (a)
\nP represents the principal sum invested , so in this example it is €$\\var{P}$.
\nPart (b)
\nA represents the amount in the deposit account after $\\var{n}$ years, so in this example it is €$\\var{A}$.
\nPart (c)
\nn represents the number of compounding periods , so in this example it is $\\var{n}$ years.
\nPart(d)
\nUsing the compound interest formula:
\n$A=P(1+i)^n$
\n$\\var{A}=\\var{P}(1+i)^\\var{n}$
\nWe need to rearrange the equation to find the value of $i$.
\n$\\frac{\\var{A}}{\\var{P}}=(1+i)^\\var{n}$
\n$\\var{ratio}=(1+i)^\\var{n}$
\n$\\sqrt[\\var{n}]{\\var{ratio}}=1+i$
\n$\\var{intplus}=1+i$
\n$i=\\var{int}$ so the annual interest rate is $\\var{perc}$%.
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\n€[[0]]
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\nPlease give your answer as a percentage correct to 2 decimal places.
\n\n[[0]]%
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\nThe compound interest formula is:
\n$\\ A = P(1+i)^n $
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\nWhat is the value of $n$?
\n[[0]]
\nWhat is the value of $i$?
\n[[1]]
\nWhat is the value of $A$? Please give your answer to the nearest cent.
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\nWhat is the value of $n$?
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\nWhat is the value of $i$? Please include all the decimal places in your answer.
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\n\nWhat is the value of $A$? Please give your answer to the nearest cent.
\n€[[2]]
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\nBank A offers an account that earns interest at a rate of $\\var{perc1}$% per annum where interest is compounded annually.
\nBank B offers an account that earns interest at a nominal rate of $\\var{perc2}$% per annum where interest is compounded daily.
\nSuppose that a potential customer has €$\\var{P}$ to invest for $\\var{n1}$ years.
\nThe compound interest formula is:
\n$\\ A = P(1+i)^n $
\n\nYou may assume that there are 365 days per annum.
\n", "advice": "
The compound interest formula is: $\\ A = P(1+i)^n $
\nPart (a)
\nn represents the number of compounding periods , so for Bank A it is $\\var{n1}$ years.
\ni represents the rate of compound interest, for Bank A, the annual interest rate is $\\var{perc1}$% so i is $\\frac {\\var{perc1}} {100}=\\var{int1}$.
\nThe total amount saved after $\\var{n1}$ years is denoted by A. Using the compound interest formula:
\n$A=P(1+i)^n$
\n$A=\\var{P} \\times(1+\\var{int1})^\\var{n1}=\\var{A1}$
\nPart(b)
\nn represents the number of compounding periods. For Bank B, interest is compounded daily for $\\var{n1}$ years so there are a total of $365 \\times \\var{n1} =\\var{n2}$ compounding periods.
\ni represents the rate of compound interest. For Bank B, the interest rate is ${\\var{perc2}}$% per annum compounded daily.
\nThe interest rate per day is $\\frac{\\var{perc2}}{365}=\\var{int3}$%
\nTherefore $i=\\frac{\\var{int3}}{100}=\\var{int2}$
\nThe amount saved after $\\var{n1}$ years is denoted by A. Using the compound interest formula:
\n$A=P(1+i)^n$
\n$A=\\var{P} \\times(1+\\var{int2})^\\var{n2}=\\var{A2}$
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