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A quick practice set of problems for education students to take in preparation for their numeracy test.

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Note: I wasted waaay too long trying to use map() to make this neater and I could never get it to work.

Couldn't get this to work with map()

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Couldn't get this to work with map() :/

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This question is intended to test the recall of basic facts.

Match the fractions with their decimal or percentage by selecting the corresponding radio buttons.

Repeat until you can answer them correctly without too much thought or working out.

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The following common fractions should be memorised to avoid working them out each time. Note $0.33\\ldots$ is more correctly written $0.\\dot{3}$ and is read \"zero point three repeater\".

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

Fraction	Decimal	Percentage
$1$	$1$	$100\\%$
\n $\\frac{1}{2}$ \n	$0.5$	$50\\%$
\n $\\frac{1}{3}$ \n	$0.33\\ldots$	$33.3\\ldots\\%$
\n $\\frac{2}{3}$ \n	$0.66\\ldots$	$66.6\\ldots\\%$
\n $\\frac{1}{4}$ \n	$0.25$	$25\\%$
\n $\\frac{3}{4}$ \n	$0.75$	$75\\%$
\n $\\frac{1}{5}$ \n	$0.2$	$20\\%$
\n $\\frac{2}{5}$ \n	$0.4$	$40\\%$
\n $\\frac{3}{5}$ \n	$0.6$	$60\\%$
\n $\\frac{4}{5}$ \n	$0.8$	$80\\%$
\n $\\frac{1}{8}$ \n	$0.125$	$12.5\\%$
\n $\\frac{1}{10}$ \n	$0.1$	$10\\%$
\n $\\frac{1}{20}$ \n	$0.05$	$5\\%$
\n $\\frac{1}{100}$ \n	$0.01$	$1\\%$

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Basics, percentage of an amount, converting to fractions and decimals.

{percent}% means {percent} out of [[0]].

This means we can write {percent}% as the fraction [[1]] .

And so we can write {percent}% as the decimal [[2]].

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'Percent' means 'out of 100'. So we can write $\\var{percent}\\%$ as $\\frac{\\var{percent}}{100}$. Since this is the same as $\\var{percent}\\div 100$ we can write this as a decimal by moving the decimal point two places (the number of zeroes in $100$). So $\\var{percent}$ (which has a decimal point after the last digit) becomes $\\var{percent/100}$. In summary:

\\[\\var{percent}\\%=\\frac{\\var{percent}}{100}=\\var{percent/100}\\]

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Basics, percentage of an amount, converting to fractions and decimals.

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Rewrite the following decimal as a percentage and a simplified fraction:

$\\var{dec} =$ [[0]]$\\% =$ [[1]]

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To go from a decimal to a percentage, move the decimal point to the right two places.

To get from a percentage to a decimal we ultimately moved the decimal place twice to the left. For instance

\\[\\var{dec*100}\\% = \\frac{\\var{dec*100}}{100}=\\var{dec*100}\\div 100 =\\var{dec}\\]

That means to go the other way (from a decimal to a percentage) we need to move the decimal place twice to the right. In particular,

\\[\\var{dec}=\\var{dec*100}\\div 100 = \\frac{\\var{dec*100}}{100}=\\var{dec*100}\\%\\]

Often it is useful to remember that 1 represents 100%, this can help you to check if your answer makes sense.

As we saw above, the decimal $\\var{dec}$ can be written as a fraction $\\frac{\\var{dec*100}}{100}$. If we want to use a fraction, it is often best if we simplify the fraction by cancelling any common factors in the numerator and denominator. If there are no common factors (other than $1$) the fraction is said to be in reduced or lowest form. Since $\\var{dec*100}$ and $100$ both contain a common factor of $\\var{cf}$, we can remove that factor from both. That is,

\\[\\frac{\\var{dec*100}}{\\var{100}}=\\frac{\\var{dec*100}\\div \\var{cf}}{\\var{100}\\div \\var{cf}}=\\frac{\\var{dec*100/cf}}{\\var{100/cf}}.\\]

As we saw above, the decimal $\\var{dec}$ can be written as $\\frac{\\var{dec*100}}{100}$, however, we don't normally like decimals in fractions, so we can multiply top and bottom by ten to move the decimal point to the right one more spot. That is,

\\[\\frac{\\var{dec*100}}{100}=\\frac{\\var{dec*100}\\times 10}{100\\times 10}=\\frac{\\var{dec*1000}}{1000}.\\]

If we want to use a fraction, it is best if we simplify the fraction by cancelling any common factors in the numerator and denominator. If there are no common factors (other than $1$) the fraction is said to be in reduced or lowest form. Since $\\var{dec*1000}$ and $1000$ both contain a common factor of $\\var{cf}$, we can remove that factor from both. That is,

\\[\\frac{\\var{dec*1000}}{\\var{1000}}=\\frac{\\var{dec*1000}\\div \\var{cf}}{\\var{1000}\\div \\var{cf}}=\\frac{\\var{dec*1000/cf}}{\\var{1000/cf}}.\\]

Note, instead of writing something like $0.123$ as $\\frac{12.3}{100}$ we can write it as $\\frac{123}{1000}$. In fact, one could read both the decimal and the fraction as \"one hundred and twenty-three thousandths\".

Note, when the denominator is a power of ten e.g., $10$, $100$, $1000$ etc, we can get away with only checking for common factors of $2$ and $5$ (repeatedly) since these are the prime factors of $10$.

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Basics, percentage of an amount, converting to fractions and decimals.

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{a} out of {b} as a fraction is [[0]].

What is this as a percentage? [[1]]$\\%$

What is this as a decimal? [[2]]

Recall that $\\frac{\\var{a}}{\\var{b}}$ means '$\\var{a}$ out of $\\var{b}$', and so going the other way, expressions like '$\\var{a}$ out of $\\var{b}$' are equivalent to $\\frac{\\var{a}}{\\var{b}}$.

For this fraction, since the denominator is $\\var{b}$, we can multiply the top and bottom by $\\var{m}$ to create an equivalent fraction over $100$. Then the numerator will be the number we want for the percentage. In particular,

$\\frac{\\var{a}}{\\var{b}}=\\frac{\\var{a}\\times \\var{m}}{\\var{b}\\times \\var{m}}=\\frac{\\var{a*m}}{100}=\\var{a*m}\\%$

To determine the decimal, we recall that dividing by $100$ moves the decimal point to the left two places:

$\\frac{\\var{a*m}}{100}=\\var{a*m}\\div {100}=\\var{a*m/100}$

In general, to convert a fraction into a percentage, we can try the following:

make an equivalent fraction with a denominator of 100, then the numerator will be the percentage. For example, given $\\frac{4}{5}$ we can multiply the top and bottom by 20 to get it over 100 (we could have multiplied by 2 and then 10 as well since this is the same thing). Our working could look like this:

\\[\\frac{4}{5}=\\frac{4\\times 20}{5\\times 20}=\\frac{80}{100}=80\\%=0.8\\]

Cancel any common factors and see if you can get the fraction to be over $100$. For example, given $\\frac{44}{55}$, since there is a common factor of $11$, we can divide the numerator and denominator by $11$ to get the fraction $\\frac{4}{5}$. Now we can multiply the numerator and denominator by $20$ to get $\\frac{80}{100}$, which is $80\\%$ or $0.8$.
If after simplifying the fraction, the numerator isn't just made up of factors of $2$ and $5$, the above approaches will not work, so we will need to do the division (either with a calculator or by hand using the division algorithm). The result will be a decimal that continues on forever. You can change that to a percentage by multiplying by $100\\%$. For example, $\\frac{4}{99}=\\var{4/99}\\ldots=\\var{400/99}\\ldots\\%$ which we can round as needed.

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{c} out of {d} as a fraction is [[0]].

What is this as a percentage? [[1]]%

What is this as a decimal? [[2]]

Recall that $\\frac{\\var{c}}{\\var{d}}$ means '$\\var{c}$ out of $\\var{d}$', and so going the other way, expressions like '$\\var{c}$ out of $\\var{d}$' are equivalent to $\\frac{\\var{c}}{\\var{d}}$.

Since $\\var{c}$ and $\\var{d}$ both contain a common factor of $\\var{cf_b}$, we can remove that factor from both. That is,

\\[\\frac{\\var{c}}{\\var{d}}=\\frac{\\var{c}\\div \\var{cf_b}}{\\var{d}\\div \\var{cf_b}}=\\frac{\\var{c/cf_b}}{\\var{d/cf_b}}.\\]

For this simplified fraction, since the denominator is $\\var{d/cf_b}$, we can multiply the top and bottom by $\\var{n}$ to create an equivalent fraction over $100$. Then the numerator will be the number we want for the percentage. In particular,

\\[\\frac{\\var{c/cf_b}}{\\var{d/cf_b}}=\\frac{\\var{c/cf_b}\\times \\var{n}}{\\var{d/cf_b}\\times \\var{n}}=\\frac{\\var{c*n/cf_b}}{100}=\\var{c*n/cf_b}\\%\\]

To determine the decimal, we recall that dividing by $100$ moves the decimal point to the left two places:

\\[\\frac{\\var{c*n/cf_b}}{100}=\\var{c*n/cf_b}\\div {100}=\\var{c*n/cf_b/100}\\]

In general, to convert a fraction into a percentage, we can try the following:

make an equivalent fraction with a denominator of 100, then the numerator will be the percentage. For example, given $\\frac{4}{5}$ we can multiply the top and bottom by 20 to get it over 100 (we could have multiplied by 2 and then 10 as well since this is the same thing). Our working could look like this:

\\[\\frac{4}{5}=\\frac{4\\times 20}{5\\times 20}=\\frac{80}{100}=80\\%=0.8\\]

Cancel any common factors and see if you can get the fraction to be over $100$. For example, given $\\frac{44}{55}$, since there is a common factor of $11$, we can divide the numerator and denominator by $11$ to get the fraction $\\frac{4}{5}$. Now we can multiply the numerator and denominator by $20$ to get $\\frac{80}{100}$, which is $80\\%$ or $0.8$.
If after simplifying the fraction, the numerator isn't just made up of factors of $2$ and $5$, the above approaches will not work, so we will need to do the division (either with a calculator or by hand using the division algorithm). The result will be a decimal that continues on forever. You can change that to a percentage by multiplying by $100\\%$. For example, $\\frac{4}{99}=\\var{4/99}\\ldots=\\var{400/99}\\ldots\\%$ which we can round as needed.

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Divisor is single digit. 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.

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

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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 $\\var{dividend1}$?\"

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:

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

Note the positions of the numbers!

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!

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

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

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

Divide
Multiply
Subtract
Bring down

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\".

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.

\\[\\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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Short Division

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 $\\var{dividend1}$?\"

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:

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

Note the positions of the numbers!

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!

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

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

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.

The tens column

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 $\\var{dd3*10}$)

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:

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

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.

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

The ones column

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

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:

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

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.

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

The tenths column

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 $\\var{b1*0.1}$)

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:

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

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.

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

The hundredths column

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 $\\var{b0*0.01}$)

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:

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

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.

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.).

Long Division

The tens column

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)

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

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:

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

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.

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

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

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

The ones column

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)

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

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:

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

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.

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

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

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

The tenths column

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)

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

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:

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

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.

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

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

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

The hundredths column

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)

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

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:

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

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.

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

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.

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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$\\frac{1}{2}$

$\\frac{1}{3}$

$\\frac{2}{3}$

$\\frac{1}{4}$

$\\frac{3}{4}$

$\\frac{1}{5}$

$\\frac{2}{5}$

$\\frac{3}{5}$

$\\frac{4}{5}$

$\\frac{1}{8}$

$\\frac{1}{10}$

$\\frac{1}{20}$

$\\frac{1}{100}$

Short Division

Long Division