When we divide a whole number by a non-unit fraction, the answer is often not a whole number.

In this lesson, we'll learn how to solve division problems in these cases using models. The process is similar to cases when the answer is a whole number (we saw this in a previous lesson). The difference now is that we have to deal with a remainder.

To demonstrate, let's use a model to solve the following division problem:

{\color{blue}{1}} \div \dfrac{2}{3}

We start with a model that represents \color{blue}1 whole:

We need to determine how many times \dfrac{\bbox[2pt, lightgray]{2}}3 fits into \color{blue}1 whole. So, we proceed as follows:

  • First, we split our whole into thirds (the denominator of our fraction):

  • Next, we break this into groups of \bbox[2pt, lightgray]{2} :

Note the following:

  • We have \fbox{1} full group. This is the whole number part of our answer.

  • We see that there is \bbox[2pt, Moccasin]{1} piece left over out of a possible group of \bbox[2pt, lightgray]{2}. So, the fractional part of our answer is \dfrac{\bbox[2pt, Moccasin]{1}}{\bbox[2pt, lightgray]{2}}.

Therefore, the solution to our division problem is as follows: 1 \div \dfrac{2}{3} = \fbox{1} \, \dfrac{\bbox[2pt, Moccasin]{1}}{\bbox[2pt, lightgray]{2}}

Let's see another example.

FLAG

Use the model above to determine the missing fraction in the following division problem. 2 \div \dfrac{5}{6} = 2 \, \dfrac{ \fbox{[math]\,\phantom{0}\,[/math]} }{ \fbox{[math]\,\phantom{0}\,[/math]} }

EXPLANATION

Let's start by interpreting our model:

  • The model shows 2 wholes split into sixths.

  • There are \bbox[2pt, lightgray]{5} sixths in each group, with \bbox[2pt, Moccasin]{2} remaining.

Note the following:

  • There are \color{red}2 full groups.

  • There are \bbox[2pt, Moccasin]{2} sixths left over out of a possible group of \bbox[2pt, lightgray]{5}. So the fractional part of our answer is \dfrac{\bbox[2pt, Moccasin]{2}}{\bbox[2pt, lightgray]{5}}.

Therefore, 2 \div \dfrac{5}{6} = {\color{red}{2}} \, \dfrac{\bbox[2pt, Moccasin]{2}}{\bbox[2pt, lightgray]{5}}.

FLAG

Use the model above to determine the missing fraction in the following division problem. \[ 2 \div \dfrac{3}{4} = 2 \, \dfrac{ \fbox{$\,\phantom{0}\,$} }{ \fbox{$\,\phantom{0}\,$} } \]

a
$\dfrac13$
b
$\dfrac23$
c
$\dfrac12$
d
$\dfrac14$
e
$\dfrac34$

Use the model above to determine the missing fraction in the following division problem. \[ 3 \div \dfrac{2}{5} = 7 \, \dfrac{ \fbox{$\,\phantom{0}\,$} }{ \fbox{$\,\phantom{0}\,$} } \]

a
$\dfrac23$
b
$\dfrac15$
c
$\dfrac13$
d
$\dfrac35$
e
$\dfrac12$

Use the model above to calculate the value of 2 \div \dfrac{3}{7}.

EXPLANATION

To calculate {\color{red}2} \div \dfrac{\bbox[2pt, lightgray]{3}}{\color{blue}7} , we first need to divide each of the \color{red}2 wholes into \color{blue}7 equal parts.

Next, we break this into groups of \bbox[2pt, lightgray]{3} :

Note the following:

  • We got \fbox{4} full groups.

  • There are \bbox[2pt, Moccasin]{2} pieces left over out of a possible group of \bbox[2pt, lightgray]{3}. So the fractional part of our answer is \dfrac{\bbox[2pt, Moccasin]{2}}{\bbox[2pt, lightgray]{3}}.

Therefore, 2 \div \dfrac{3}{7} = \fbox{4} \, \dfrac{\bbox[2pt, Moccasin]{2}}{\bbox[2pt, lightgray]{3}}.

FLAG

If the model above represents $1$ whole, use the model to solve the following division problem. Express your answer as a mixed number in its simplest form.

a
b
c
d
e

If the model above represents $2$ wholes, use the model to solve the following division problem. Express your answer as a mixed number in its simplest form.

a
b
c
d
e

If the model above represents $4$ wholes, use the model to solve the following division problem. Express your answer as a mixed number in its simplest form.

a
b
c
d
e
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