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Use the BODMAS rule to determine the order in which to evaluate some arithmetic expressions. 

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Wrong order of solving operations can often lead to incorrect answers.  Therefore, the order in which we carry out a calculation is important.

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BODMAS is a mnemonic which tells us the correct order in which operations should be carried out:

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Brackets ⇒ Ordinals ⇒ Division/Multiplication ⇒ Addition/Subtraction

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Apply BODMAS and try to solve these calculations.

", "advice": "

The correct order of carrying out operations can be remembered by the mnemonic BODMAS:

\n
\n

Brackets ⇒ Ordinals ⇒ Division/Multiplication ⇒ Addition/Subtraction

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It is important to notice that division and multiplication have the same priority - division does not have a priority over multiplication. Similarly, adition and subtraction also have the same priority. When the order is unclear, we work from left to right.

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Note that brackets have the highest priority, but when we evaluate them, we still need to follow BODMAS inside them.

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Sometimes, an alternative acronym BIDMAS (Brackets, Indices, ...) is also used.

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a)

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Division and multiplication have the same priority, so we just work from left to right. $\\var{int*int} ÷ \\var{int}  = \\var{int}$ and hence

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\\[\\begin{align} \\var{int*int} ÷ \\var{int} \\times \\var{int} &= \\var{int} \\times \\var{int} \\\\&= \\var{int*int} \\text{.}   \\end{align}\\]

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b)

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Similarly, $\\var{eint*2} ÷ \\var{eint/2}  = 4 $ and hence

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\\[\\begin{align} \\var{eint*2} ÷ \\var{eint/2} \\times \\var{eint} &= 4 \\times \\var{eint} \\\\&= \\var{4*eint}\\text{.} \\end{align}\\]

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c)

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Applying BODMAS, multiplication has a priority over addition. $\\var{sint + 2} \\times \\var{sint} = \\var{(sint + 2)*sint}$ and hence

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\\[\\begin{align} \\var{sint} + \\var{sint + 2} \\times \\var{sint} &= \\var{sint} + \\var{(sint + 2)*sint} \\\\&= \\var{sint + (sint + 2)*sint}\\text{.} \\end{align}\\]

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d)

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Applying BODMAS, multiplication and division have priority over addition and subtraction. $1 \\times 0 = 0$ and $\\var{bint}\\div\\var{bint} = 1$ so

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\\[\\begin{align} \\var{bint - 15} - 1 \\times 0 + \\var{bint}\\div\\var{bint} &= \\var{bint - 15} - 0 + 1 \\\\&= \\var{bint - 14}\\text{.} \\end{align}\\]

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e)

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Roots can be considered as powers, while fractions can be considered as a bracket divided by a bracket.

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\\[\\displaystyle \\text{Numerator is considered as a bracket } (\\var{oint}^2+ \\sqrt{\\var{eint*eint}}) \\text{ and the denominator as } (3 \\times 2 - 2 \\times 2)\\text{.}\\]

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Before we evaluate numerator, we calculate powers:

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\\[\\begin{align} \\sqrt{\\var{eint*eint}} &= \\var{eint} \\text{,}
\\\\\\var{oint}^2 &= \\var{oint*oint} \\text{.} \\end{align}\\]

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Before we evaluate denominator we calculate multiplications:

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\\[\\begin{align} 3 \\times 2 &= 6 \\text{ and } \\\\ 2 \\times 2 &= 4\\text{.} \\end{align}\\]

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Performing addition/subtraction as the last step in evaluating numerator/denominator we get:

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\\[ \\begin{align} (\\var{oint}^2+ \\sqrt{\\var{eint*eint}}) &= \\var{oint*oint} + \\var{eint}
\\\\&= \\var{oint*oint + eint}
\\\\\\text{and}
\\\\(3 \\times 2 - 2 \\times 2) &= 6 - 4
\\\\&= 2 \\end{align} \\]

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So the fraction

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\\[\\begin{align} \\displaystyle \\frac{(\\var{oint}^2+ \\var{eint})}{(3 \\times 2 - 2 \\times 2)} &= \\frac{\\var{(oint*oint + eint)}}{2}\\text{.} \\end{align}\\]

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Evaluating the final bracket we get:

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\\[(10 - 2) = 8\\text{.}\\]

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As we evaluated all brackets, we can continue with:

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\\[\\displaystyle \\frac{\\var{oint}^2+ \\sqrt{\\var{eint*eint}}}{3 \\times 2 - 2 \\times 2} + (10 - 2) \\div \\var{pint} = \\frac{\\var{(oint*oint + eint)}}{2} + 8 \\div \\var{pint} \\]

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Now, division has a priority over addition so since $\\frac{\\var{(oint*oint + eint)}}{2} = \\var{(oint*oint + eint)/2}$ and $8 \\div \\var{pint} = \\var{8/pint}$:

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\\[\\begin{align} \\frac{\\var{(oint*oint + eint)}}{2} + 8 \\div \\var{pint} &= \\var{(oint*oint + eint)/2} + \\var{8/pint} \\\\&= \\var{(oint*oint + eint)/2 + 8/pint}\\text{.} \\end{align}\\]

\n

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Random integer from 2 to 10.

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Random odd integer from 1 to 9.

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A random slightly bigger integer.

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1, 2 or 4.

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Random even integer from 2 to 10.

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Random integer from 1 to 5.

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$\\var{bint} -15 - 1 \\times 0 + \\displaystyle \\frac{ \\var{bint} } {\\var{bint}}   =$

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{bint -18}

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{bint -14}

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{random(2..40 except bint-14)}

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This is a set of questions designed to help you practice adding, subtracting, multiplying and dividing fractions.

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All of these can be done without a calculator.

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

These are basic questions to help you practice adding, subtracting, multiplying and dividing fractions.

\n

Attempt the questions without a calculator.

\n

Give your answer as a fraction.

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When adding/subtracting fractions, you must first find a common denominator between the fractions. If they already have the same denominator then you only need to worry about adding/subtracting the numerators and dividing the result by the common denominator.

\n

For example:
To find a common denominator of $\\frac{2}{5} + \\frac{7}{15}$, the most obvious would be $15$, because $5\\times3=15$. Therefore, you must multiply both sides of the fraction $\\frac{2}{5}$ by $3$ to obtain a new fraction $\\frac{6}{15}$. This is known as 'scaling up'.
Now you can add the two fractions together (by adding the numerators) because they have the same denominator:
$\\frac{6}{15}+\\frac{7}{15}=\\frac{13}{15}$.

\n

The same applies with subtraction as well as addition.

\n

\n

When multiplying fractions, you can simply multiply the two numerators and divide this by the multiplication of the two denominators.

\n

For example:
$\\frac{a}{b}\\times\\frac{c}{d}$ = $\\frac{a\\times{c}}{b\\times{d}}$

\n

\n

When dividing fractions, you firstly need to reciprocate (flip) the second fraction, then multiply the numerators and denominators as you would a normal multiplication question.

\n

For example:
$\\frac{a}{b} \\div \\frac{c}{d}$ would be flipped to become $\\frac{a}{b} \\div \\frac{d}{c}$ and then treated as a normal multiplication question (as explained above).

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"random(1..bb-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "jj": {"name": "jj", "group": "Ungrouped variables", "definition": "random(2..20) //random denominator", "description": "", "templateType": "anything"}, "dd": {"name": "dd", "group": "Ungrouped variables", "definition": "random(2..20) //random denominator", "description": "", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(1..b-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "ll": {"name": "ll", "group": "Ungrouped variables", "definition": "random(2..20) //random denominator", "description": "", "templateType": "anything"}, "k": {"name": "k", "group": "Ungrouped variables", "definition": "random(1..j-h) //random numerator less than denominator", "description": "", "templateType": "anything"}, "y": {"name": "y", "group": "Ungrouped variables", "definition": "random(1..x-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "ss": {"name": "ss", "group": "Ungrouped variables", "definition": "random(1..tt-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "r": {"name": "r", "group": "Ungrouped variables", "definition": "random(2..20) //random denominator", "description": "", "templateType": "anything"}, "h": {"name": "h", "group": "Ungrouped variables", "definition": "random(1..j-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "qq": {"name": "qq", "group": "Ungrouped variables", "definition": "random(1..rr-1) //random numerator less than denominator", "description": "", "templateType": "anything"}, "rr": {"name": "rr", "group": "Ungrouped variables", "definition": "random(2..20) //random denominator", "description": "", "templateType": "anything"}, "w": {"name": "w", "group": "Ungrouped variables", "definition": "random(1..x-1) //random 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"ungrouped_variables": ["a", "b", "c", "d", "f", "g", "j", "m", "p", "r", "t", "v", "x", "z", "bb", "dd", "ff", "hh", "jj", "ll", "nn", "pp", "rr", "tt", "h", "k", "l", "n", "o", "q", "s", "u", "w", "y", "aa", "cc", "ee", "gg", "ii", "kk", "mm", "oo", "qq", "ss", "cf", "ker1", "ker2", "ker3", "ker4"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "1", "showCorrectAnswer": false, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

What is the answer to $\\frac{\\var{h}}{\\var{j}} + \\frac{\\var{k}}{\\var{j}}$?

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Check to see if the denominators are the same. If they are, you only need to add the numerators together and leave the denominator as it is for the final answer.

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Add the numerators

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Write as a fraction over the similar denominator; cancel down if you can.

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What is the answer to $\\frac{\\var{l}}{\\var{m}} - \\frac{\\var{n}}{\\var{m}}$?

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Check to see if the denominators are the same. If they are, you only need to subtract the numerators and leave the denominator as it is.

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Subtract the numerators

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Create a fraction with the original denominator; cancel down if you can.

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What is the answer to $\\frac{1}{\\var{x}} + \\frac{1}{\\var{z}}$?

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Check to see if the denominators are the same

\n

If they are not - multiply each fraction up to equivalent fractions with equal denominators

\n

Once they are equal add the numerators and put over the equal denominator

\n

Cancel down if needed

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What is the answer to $\\frac{\\var{ker2}}{\\var{ker1}} + \\frac{\\var{ker4}}{\\var{cf*ker1}}$?

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What is the answer to $\\frac{\\var{ker2}}{\\var{ker1}} - \\frac{\\var{ker4}}{\\var{(cf+1)*ker1}}$?

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What is the answer to $\\frac{\\var{ker2}}{\\var{ker1+1}} + \\frac{\\var{ker4+1}}{\\var{(cf)*(ker1+1)}}$?

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Eight expressions, of increasing complexity. The student must simplify them by expanding brackets and collecting like terms.

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

For each expression below, collect like terms and expand brackets.

\n

The * symbol is required between algebraic symbols, e.g. $5ab^2$ should be written 5*a*b^2.

", "advice": "

When simplifying expressions, only terms of the same type or like terms can be added together.

\n

Algebraic symbols or letters can be added together provided that they are raised to the same power. For example, we can add $x^2+x^2=2x^2$, but we cannot collect both $x^2$ and $x$ into one term.

\n

a)

\n

\\[ 
\\begin{align}
\\var{c[0]}x+\\var{c[1]}x+\\var{c[2]}x&=(\\var{c[0]}+\\var{c[1]}+\\var{c[2]})x\\\\
&=\\simplify{({c[0]}+{c[1]}+{c[2]})}x
\\end{align}
\\]

\n

b)

\n

\\[ 
\\begin{align}
\\var{a[1]}x^2+\\var{a[2]}x^2+\\var{a[3]}x+\\var{a[4]}x +\\var{a[0]}&=(\\var{a[1]}+\\var{a[2]})x^2+(\\var{a[3]}+\\var{a[4]})x +\\var{a[0]}\\\\
&=\\simplify{({a[1]}+{a[2]})}x^2+\\simplify{({a[3]}+{a[4]})}x+\\var{a[0]}
\\end{align}
\\]

\n

c)

\n

\\[ 
\\begin{align}
\\var{b[0]}y^5+\\var{b[1]}y^5+\\var{b[2]}y^5+\\var{b[4]}y^5+\\var{b[3]}y^5&=(\\var{b[0]}+\\var{b[1]}+\\var{b[2]}+\\var{b[4]}+\\var{b[3]})y^5\\\\
&=\\simplify{({b[1]}+{b[2]}+{b[3]}+{b[4]}+{b[0]})}y^5
\\end{align}
\\]

\n

d)

\n

\\[ 
\\begin{align}
\\var{d[0]}ab+\\var{d[1]}abc+\\var{d[2]}a+\\var{d[3]}b+\\var{d[4]}c+\\var{d[5]}abc
&=(\\var{d[1]}+\\var{d[5]})abc+\\var{d[0]}ab+\\var{d[2]}a+\\var{d[3]}b+\\var{d[4]}c\\\\
&=\\simplify{{d[1]}+{d[5]}}abc+\\var{d[0]}ab+\\var{d[2]}a+\\var{d[3]}b+\\var{d[4]}c
\\end{align}
\\]

\n

e)

\n

\\[ 
\\begin{align}
\\var{f[0]}a^2b+\\var{f[1]}ab^2+\\var{f[2]}ab+\\var{f[3]}a^2b+\\var{f[4]}ab^2
&=(\\var{f[0]}+\\var{f[3]})a^2b+(\\var{f[1]}+\\var{f[4]})ab^2+\\var{f[2]}ab\\\\
&=\\simplify{{f[0]}+{f[3]}}a^2b+\\simplify{{f[1]}+{f[4]}}ab^2+\\var{f[2]}ab
\\end{align}
\\]

\n

f)

\n

\\[
\\begin{align}
\\var{g[0]}(\\var{g[1]}x+\\var{g[2]}y)+\\var{g[4]}x+\\var{g[5]}y
&=(\\var{g[0]}\\times \\var{g[1]}+\\var{g[4]})x+(\\var{g[0]} \\times\\var{g[2]}+\\var{g[5]})y\\\\
&=(\\simplify{{g[0]}*{g[1]}}+\\var{g[4]})x+(\\simplify{{g[0]}*{g[2]}}+\\var{g[5]})y\\\\
&=\\simplify{{g[0]}*{g[1]}+{g[4]}}x+\\simplify{{g[0]}*{g[2]}+{g[5]}}y
\\end{align}
\\]

\n

g)

\n

\\[ 
\\begin{align}
\\var{h[0]}x(\\var{h[1]}x+\\var{h[2]}z)+\\var{h[3]}x+\\var{h[6]}z+\\var{h[4]}x^2+\\var{h[5]}z^2
&=(\\simplify[]{{h[0]}{h[1]}}+\\var{h[4]})x^2+(\\simplify[]{{h[0]}{h[2]}})zx+\\var{h[3]}x+\\var{h[5]}z^2+\\var{h[6]}z\\\\
&=(\\simplify{{h[0]}{h[1]}}+\\var{h[4]})x^2+(\\simplify[]{{h[0]}{h[2]}})zx+\\var{h[3]}x+\\var{h[5]}z^2+\\var{h[6]}z\\\\
&=\\simplify{{h[0]}*{h[1]}+{h[4]}}x^2+\\simplify{{h[0]}*{h[2]}}zx+\\simplify{{h[3]}x+{h[5]}}z^2+\\var{h[6]}z
\\end{align}
\\]

\n

h)

\n

\\[ 
\\begin{align}
\\var{j[0]}(\\var{j[1]}x-\\var{j[2]}y)+\\var{j[3]}(\\var{j[4]}x-\\var{j[5]}y)+\\var{j[6]}(\\var{j[7]}x-\\var{j[8]}y)
&= (\\simplify[]{{j[0]}{j[1]}}+\\simplify[]{{j[3]}{j[4]}}+\\simplify[]{{j[6]}{j[7]}})x-(\\simplify[]{{j[0]}{j[2]}}+\\simplify[]{{j[3]}{j[5]}}+\\simplify[]{{j[6]}{j[8]}})y\\\\
&= (\\simplify{{j[0]}{j[1]}}+\\simplify{{j[3]}{j[4]}}+\\simplify{{j[6]}{j[7]}})x-(\\simplify{{j[0]}{j[2]}}+\\simplify{{j[3]}{j[5]}}+\\simplify{{j[6]}{j[8]}})y\\\\
&= \\simplify{({j[0]}*{j[1]}+{j[4]*j[3]}+{j[6]}*{j[7]})x}-\\simplify{({j[0]}*{j[2]}+{j[5]}{j[3]}+{j[6]}*{j[8]})y}
\\end{align}
\\]

\n

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random variables for part 1

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$\\var{c[0]}x+\\var{c[1]}x+\\var{c[2]}x=$ [[0]]

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You must collect like terms to fully simplify.

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$\\var{a[1]}x^2+\\var{a[2]}x^2+\\var{a[3]}x+\\var{a[4]}x +\\var{a[0]}=$ [[0]]

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$\\var{g[0]}(\\var{g[1]}x+\\var{g[2]}y)=$ [[0]]

\n

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You must condense your answer to fully simplify. *'s are not needed to indicate multiplication here.

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Solve for  $x$:

\n

$\\var{j[6]+1} x  + \\var{j[7]}= \\var{j[0]}$ [[0]]

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You must condense your answer to fully simplify. *'s are not needed to indicate multiplication here.

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Solve for  $x$:

\n

$\\var{j[5]+1} x  + \\var{j[6]}a= \\var{j[2]}$ [[0]]

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You must condense your answer to fully simplify. *'s are not needed to indicate multiplication here.

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