// Numbas version: exam_results_page_options {"name": "Division, two digit divisor results in a remainder, answer expressed as a decimal rounded to 1 decimal place - long or short division", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Division, two digit divisor results in a remainder, answer expressed as a decimal rounded to 1 decimal place - long or short division", "tags": [], "metadata": {"description": "

Divisor is a two-digit number. There is a remainder which we express as a decimal by continuing the 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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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.

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

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qd2

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$\\var{dividend1}\\div\\var{divisor1}=$[[0]] (1 decimal place)

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Short Division

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We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. By the way, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$. Both of these expressions mean \"how many $\\var{divisor1}$s go into $\\var{dividend1}$?\"

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The short 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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We need to know the $\\var{divisor1}$ times tables or write the $\\var{divisor1}$ times tables out (by 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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Step 1: 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 $\\var{dd3*10}$) 

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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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$\\begin{array}{r} \\color{red}{\\var{qd3}\\phantom{^\\var{diff3}\\var{qd2}}.\\,\\phantom{^{\\var{diff2}}\\var{qd1}^{\\var{diff1}}\\var{qd0}}} \\\\[-.5cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\color{green}{\\var{dd3}}^\\phantom{\\var{diff3}}\\var{dd2}\\,.\\,^\\phantom{\\var{diff2}}\\var{dd1}^\\phantom{\\var{diff1}}\\var{dd0}} \\end{array}$

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Step 2: We only made it to $\\var{prod3}$ but we were aiming for $\\var{dd3}$, so we were $\\var{diff3}$ away. This is the remainder (what remains to be divided) in the tens column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff3}}\\var{dd2}$ in the ones column. We made it to exactly $\\var{dd3}$, so there is no remainder in the tens column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd2}$ in the ones column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}\\var{qd2}}\\,.\\,\\phantom{{^\\var{diff2}}\\var{qd1}\\,{^\\var{diff1}}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^\\color{red}{\\var{diff3}}\\var{dd2}\\,.\\,\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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The ones column

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Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b2}}$?\"

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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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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}\\color{red}{\\var{qd2}}\\,.\\,\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^\\color{green}{\\var{diff3}}\\color{green}{\\var{dd2}}\\,.\\phantom{^\\color{red}{\\var{diff2}}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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Step 2: We only made it to $\\var{prod2}$ but we were aiming for $\\var{b2}$, so we were $\\var{diff2}$ away. This is the remainder (what remains to be divided) in the ones column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff2}}\\var{dd1}$ in the tenths column. We made it to exactly $\\var{b2}$, so there is no remainder in the ones column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd1}$ in the tenths column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,.\\,\\phantom{^\\var{diff2}\\var{qd1}\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,.\\,^\\color{red}{\\var{diff2}}\\var{dd1}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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The tenths column

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Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b1}}$?\" (note this $\\var{b1}$ actually represents $\\var{b1*0.1}$)

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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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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,.\\,\\phantom{^\\var{diff2}}\\color{red}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,.\\,^\\color{green}{\\var{diff2}}\\color{green}{\\var{dd1}}\\,\\phantom{^\\color{red}{\\var{diff1}}}\\var{dd0}}\\end{array}$

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Step 2: We only made it to $\\var{prod1}$ but we were aiming for $\\var{b1}$, so we were $\\var{diff1}$ away. This is the remainder (what remains to be divided) in the tenths column and we write it in front of the next digit as a superscript in order to represent $\\color{red}{\\var{diff1}}\\var{dd0}$ in the hundredths column. We made it to exactly $\\var{b1}$, so there is no remainder in the tenths column and you can continue to the next column. However, for the sake of my own sanity (writing this solution), I will put a remainder of $\\color{red}{0}$ in front of the next digit as a superscript in order to represent $\\color{red}{0}\\var{dd0}$ in the hundredths column. You do not need to do this but you can if you wish.

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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,.\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}\\var{qd0}}} \\\\[-.7cm] {\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,.\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{red}{\\var{diff1}}\\var{dd0}}\\end{array}$

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The hundredths column

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Step 1: Now we ask ourselves, \"How many $\\color{green}{\\var{divisor1}}$s go into $\\color{green}{\\var{b0}}$?\" (note this $\\var{b0}$ actually represents $\\var{b0*0.01}$)

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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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$\\begin{array}{r} {\\var{qd3}\\,\\phantom{^\\var{diff3}}{\\var{qd2}}\\,.\\,\\phantom{^\\var{diff2}}{\\var{qd1}}\\phantom{\\,^\\var{diff1}}\\color{red}{\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}\\var{dd3} \\,^{\\var{diff3}}{\\var{dd2}}\\,.\\,^{\\var{diff2}}{\\var{dd1}}\\,^\\color{green}{\\var{diff1}}\\color{green}{\\var{dd0}}}\\end{array}$

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Step 2: We could work out the remainder, 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 updown to $\\var{ans}$. Therefore, $\\var{dividend1}\\div\\var{divisor1}=\\var{ans}$ (1 dec. pl.).

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Long Division

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We want to calculate $\\var{dividend1}\\div\\var{divisor1}$. By the way, this is the same thing as $\\frac{\\var{dividend1}}{\\var{divisor1}}$. 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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    \n
  1. Divide
  2. \n
  3. Multiply
  4. \n
  5. Subtract
  6. \n
  7. Bring down 
  8. \n
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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 (by 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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$\\begin{array}{r} {\\var{qd3}\\color{red}{\\var{qd2}}\\phantom{.\\!\\var{qd1}\\var{qd0}}} \\\\[-.7cm] \\color{green}{\\var{divisor1}} \\strut \\overline{\\smash{\\raise.09ex{)}}{\\var{dd3}}{\\var{dd2}}.\\!\\var{dd1}\\var{dd0}} \\\\[-0.5cm] \\underline{{\\var{prod3}}}\\phantom{5.55}\\\\[-.7cm]\\color{green}{\\var{diff3}\\var{dd2}}\\phantom{.55}\\end{array}$

\n

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:

\n

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

\n

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. 

\n

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

\n

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

\n

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

\n

\n

The tenths column

\n

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)

\n

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: 

\n

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

\n

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:

\n

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

\n

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. 

\n

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

\n

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

\n

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

\n

\n

The hundredths column

\n

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)

\n

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: 

\n

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

\n

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:

\n

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

\n

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. 

\n

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

\n


\n

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.

\n

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

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