// Numbas version: exam_results_page_options {"name": "Fractions are division: Fraction to decimal", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"advice": "", "tags": [], "functions": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "metadata": {"description": "

Divisor is single digit. There is a remainder which we express as a decimal by continuing the long division process. Rounding is required to one decimal place. The working suggests determining the second decimal place so the student knows whether to round up or down.

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We want to calculate $\\frac{\\var{dividend1}}{\\var{divisor1}}$, which is just the same as $\\var{dividend1}\\div\\var{divisor1}$, since both of these expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

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The long division algorithm allows you to work this out by working from the left to the right of $\\var{dividend1}$ whilst respecting place value. We normally set up the division in the following way:

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$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dividend1}}$

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Note the positions of the numbers!

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Actually, since we want the answer to one decimal place we add as many zeroes after the decimal place to ensure we have two decimal places!

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$\\var{divisor1} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}\\var{dd2}.\\!\\var{dd1}\\var{dd0}}$

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Why two? We use that extra digit to determine whether to round up or down.

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The algorithm (or procedure) seems complicated at first but you might find a mnemonic helps to remember the steps. We work left to right doing the following steps

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1. Divide
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3. Multiply
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5. Subtract
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7. Bring down
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and repeating until we run out of digits. The steps form the acronym DMSB. Popular mnemonics include \"Does McDonalds Sell Burgers?\", \"Dracula Must Suck Blood\" and \"Dead Mice Smell Bad\".

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We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (be repeatedly adding $\\var{divisor1}$) so that we can refer to them.

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\\\boxed{\\begin{align}1\\times\\var{divisor1}&=\\var{divisor1}\\\\2\\times\\var{divisor1}&=\\var{2*divisor1}\\\\3\\times\\var{divisor1}&=\\var{3*divisor1}\\\\4\\times\\var{divisor1}&=\\var{4*divisor1}\\\\5\\times\\var{divisor1}&=\\var{5*divisor1}\\\\6\\times\\var{divisor1}&=\\var{6*divisor1}\\\\7\\times\\var{divisor1}&=\\var{7*divisor1}\\\\8\\times\\var{divisor1}&=\\var{8*divisor1}\\\\9\\times\\var{divisor1}&=\\var{9*divisor1}\\end{align}}\

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The tens column

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D: The first thing we ask ourselves is, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{dd3}}$?\" (note this $\\var{dd3}$ actually represents $\\var{dd3*10}$ since it is in the tens column)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{dd3}$. So we write $\\color{red}0$ above the $\\var{dd3}$ in the tens column: Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fit and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the tens column:  Well, $\\var{qd3}\\times \\var{divisor1}=\\var{prod3}$ so $\\var{qd3}$ fits and we write $\\color{red}{\\var{qd3}}$ above the $\\var{dd3}$ in the tens column:

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$\\begin{array}{r} \\color{red}{\\var{qd3}\\phantom{\\var{qd2}.\\!\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}\\var{dd2}.\\!\\var{dd1}\\var{dd0}} \\end{array}$

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M: Now since $\\color{green}{\\var{qd3}}\\times \\color{green}{\\var{divisor1}}=\\var{prod3}$ we write $\\color{red}{\\var{prod3}}$ underneath in the tens column:

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$\\begin{array}{r} \\color{green}{\\var{qd3}\\phantom{\\var{qd2}.\\!\\var{qd1}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}\\var{dd2}.\\!\\var{dd1}\\var{dd0}} \\\\[-.5cm] \\color{red}{\\var{prod3}}\\phantom{5.55}\\end{array}$

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S: We now do the subtraction, $\\color{green}{\\var{dd3}-\\var{prod3}}$, to determine the remainder (what remains to be divided) in the tens column.

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$\\begin{array}{r} {\\var{qd3}\\phantom{\\var{qd2}.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}\\var{dd2}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{\\color{green}{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]\\color{red}{\\var{diff3}}\\phantom{5.55}\\end{array}$

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B: Now we bring the $\\color{green}{\\var{dd2}}$ in the ones column down next to the remainder so that it forms $\\var{diff3}\\var{dd2}$.

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$\\begin{array}{r} {\\var{qd3}\\phantom{\\var{qd2}.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}\\color{green}{\\var{dd2}}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}}\\color{red}{\\var{dd2}}\\phantom{.55}\\end{array}$

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The ones column

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D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\" (note this $\\var{b2}$ does actually represent $\\var{b2}$ since it is in the ones column)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b2}$. So we write $\\color{red}0$ above the $\\var{dd2}$ in the ones column: Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fit and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the ones column:  Well, $\\var{qd2}\\times \\var{divisor1}=\\var{prod2}$ so $\\var{qd2}$ fits and we write $\\color{red}{\\var{qd2}}$ above the $\\var{dd2}$ in the ones column:

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$\\begin{array}{r} {\\var{qd3}\\color{red}{\\var{qd2}}\\phantom{.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]\\color{green}{\\var{diff3}\\var{dd2}}\\phantom{.55}\\end{array}$

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M: Now since $\\color{green}{\\var{qd2}}\\times \\color{green}{\\var{divisor1}}=\\var{prod2}$ we write $\\color{red}{\\var{prod2}}$ underneath in the ones column:

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$\\begin{array}{r} {\\var{qd3}\\color{green}{\\var{qd2}}\\phantom{.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\color{red}{\\var{prod2}}\\phantom{.55}\\end{array}$

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S: We now do the subtraction, $\\color{green}{\\var{b2}-\\var{prod2}}$, to determine the remainder (what remains to be divided) in the ones column.

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\phantom{.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]\\color{green}{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]\\color{red}{\\var{diff2}}\\phantom{.55}\\end{array}$

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B: Now we bring the $\\color{green}{\\var{dd1}}$ in the tenths column down next to the remainder so that it forms $\\var{diff2}\\var{dd1}$.

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}\\phantom{.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!\\color{green}{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]\\var{diff2}\\phantom{.}\\color{red}{\\var{dd1}}\\phantom{5}\\end{array}$

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The tenths column

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D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1/10}$ since it is in the tenths column)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b1}$. So we write $\\color{red}0$ above the $\\var{dd1}$ in the tenths column: Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fit and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tenths column:  Well, $\\var{qd1}\\times \\var{divisor1}=\\var{prod1}$ so $\\var{qd1}$ fits and we write $\\color{red}{\\var{qd1}}$ above the $\\var{dd1}$ in the tenths column:

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!\\color{red}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\end{array}$

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M: Now since $\\color{green}{\\var{qd1}}\\times \\color{green}{\\var{divisor1}}=\\var{prod1}$ we write $\\color{red}{\\var{prod1}}$ underneath in the tenths column:

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!\\color{green}{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\color{red}{\\var{prod1}}\\phantom{5}\\end{array}$

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S: We now do the subtraction, $\\color{green}{\\var{b1}-\\var{prod1}}$, to determine the remainder (what remains to be divided) in the tenths column.

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]\\color{green}{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] \\color{red}{\\var{diff1}}\\phantom{5}\\end{array}$

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B: Now we bring the $\\color{green}{\\var{dd0}}$ in the hundredths column down next to the remainder so that it forms $\\var{diff1}\\var{dd0}$.

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!{\\var{qd1}}\\phantom{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}\\color{green}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] {\\var{diff1}}\\color{red}{\\var{dd0}}\\end{array}$

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The hundredths column

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D: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" (note this $\\var{b0}$ actually represents $\\var{b0/100}$ since it is in the hundredths column)

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Well, none! $\\var{divisor1}$ is too big to fit into $\\var{b0}$. So we write $\\color{red}0$ above the $\\var{dd0}$ in the hundredths column: Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fit and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the hundredths column:  Well, $\\var{qd0}\\times \\var{divisor1}=\\var{prod0}$ so $\\var{qd0}$ fits and we write $\\color{red}{\\var{qd0}}$ above the $\\var{dd0}$ in the hundredths column:

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!{\\var{qd1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm] \\color{green}{\\var{diff1}\\var{dd0}}\\end{array}$

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M: Now since $\\color{green}{\\var{qd0}}\\times \\color{green}{\\var{divisor1}}=\\var{prod0}$ we write $\\color{red}{\\var{prod0}}$ underneath in the hundredths column:

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!{\\var{qd1}}\\color{green}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm]{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\color{red}{\\var{prod0}}\\end{array}$

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S: We now do the subtraction, $\\color{green}{\\var{b0}-\\var{prod0}}$, to determine the remainder (what remains to be divided) in the hundredths column.

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$\\begin{array}{r} {\\var{qd3}{\\var{qd2}}.\\!{\\var{qd1}}{\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!{\\var{dd1}}{\\var{dd0}}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]{\\var{diff3}\\var{dd2}}\\phantom{.55}\\\\[-0.5cm]\\underline{{\\var{prod2}}}\\phantom{.55}\\\\[-0.5cm]{\\var{diff2}\\phantom{.}\\var{dd1}}\\phantom{5}\\\\[-0.5cm]\\underline{{\\var{prod1}}}\\phantom{5}\\\\[-0.5cm]\\color{green}{\\var{diff1}\\var{dd0}}\\\\[-0.5cm]\\underline{\\color{green}{\\var{prod0}}}\\\\[-0.5cm]\\color{red}{\\var{diff0}}\\end{array}$

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Now we could keep adding zeros and continue the procedure but we only needed to determine the second decimal place in order to correctly round to one decimal place and so we now stop the procedure.

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Since the second decimal place was $\\var{qd0}$ we round up down to $\\var{ans}$. Therefore, $\\frac{\\var{dividend1}}{\\var{divisor1}}=\\var{ans}$ (1 dec. pl.).

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$\\displaystyle\\frac{\\var{dividend1}}{\\var{divisor1}}=$[] (1 decimal place)

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Write the following question down on paper and evaluate it without using a calculator.

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If you are unsure of how to do a question, click on Show steps to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again.

", "preamble": {"js": "", "css": ""}, "variables": {"prod0": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "prod0", "definition": "divisor1*qd0"}, "diff1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "diff1", "definition": "b1-prod1"}, "b1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "b1", "definition": "10*diff2+dd1"}, "diff0": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "diff0", "definition": "b0-prod0"}, "ans": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "ans", "definition": "precround(quotient1,1)"}, "dd2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "dd2", "definition": "mod(floor(dividend1),10)"}, "b2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "b2", "definition": "10*diff3+dd2"}, "remainder": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "remainder", "definition": "mod(dividend1,divisor1)"}, "dividend1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "dividend1", "definition": "random(25..99 except list(divisor1/2..99#divisor1/2))\n//divisor1*quotient1+remainder/100"}, "diff3": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "diff3", "definition": "dd3-prod3"}, "qd0": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "qd0", "definition": "mod(floor(quotient1*100),10)"}, "divisor1": {"group": "Ungrouped variables", "description": "

excluded 5 so that the decimal part is longer than 1 place.

", "templateType": "anything", "name": "divisor1", "definition": "random(3..9 except 5)"}, "prod1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "prod1", "definition": "divisor1*qd1"}, "dd3": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "dd3", "definition": "mod(floor(dividend1/10),10)"}, "qd3": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "qd3", "definition": "mod(floor(quotient1/10),10)"}, "diff2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "diff2", "definition": "b2-prod2"}, "b0": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "b0", "definition": "10*diff1+dd0"}, "prod3": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "prod3", "definition": "divisor1*qd3"}, "quotient1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "quotient1", "definition": "dividend1/divisor1\n//random(ceil(1001/divisor1)..floor(9999/divisor1) except list(100..10000#10))/100"}, "qd1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "qd1", "definition": "mod(floor(quotient1*10),10)"}, "dd0": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "dd0", "definition": "mod(dividend1*100,10)"}, "dd1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "dd1", "definition": "mod(floor(dividend1*10),10)"}, "prod2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "name": "prod2", "definition": "divisor1*qd2"}, "qd2": {"group": "Ungrouped variables", "description": "

qd2

", "templateType": "anything", "name": "qd2", "definition": "mod(floor(quotient1),10)"}}, "type": "question", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}