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iGCSE (2021 Edition)

3.06 Dividing with decimals

Lesson

In this lesson we will look at dividing with decimals. We will look at methods for solving problems including dividing decimal numbers by whole numbers, dividing whole numbers by decimals, and dividing decimals by decimals.

Why do we need to divide decimals?

Sometimes, before you can really understand how to perform some number problems, it helps to understand why you need to do them.

Worked example

Example 1

Evaluate: Matilda works at her local supermarket for four hours every Saturday, and earns a total of $\$84.80$$84.80. How much does she earn per hour?

Think: If she works for $4$4 hours, we can divide Matilda's total earnings by $4$4 to find the amount she earns per hour. So we want to find the value of $84.80\div4$84.80÷​4.

Do: We can solve this in a number of ways, but let's start by partitioning $84.80$84.80 into tens, units, tenths and hundredths.

We can see that $84.80=80+4+0.80$84.80=80+4+0.80, that is, it is made up of $8$8 tens, $4$4 units, and $8$8 tenths (or $80$80 hundredths). We can now divide each of these parts by $4$4.

  • $80\div4$80÷​4, or $8$8 tens divided by $4$4, is equal to $20$20 (or two tens).
  • $4\div4$4÷​4, or $4$4 units divided by $4$4, is equal to $1$1.
  • $8$8 tenths divided by $4$4 is equal to $2$2 tenths. We can also think of this as $80$80 hundredths divided by $4$4, which is $20$20 hundredths.

Notice that as we are dealing with currency, it makes sense to refer to the decimals as hundredths, as this relates to cents.

The table below shows each step in the division.

$84.80\div4$84.80÷​4 $=$= $\left(80+4+0.80\right)\div4$(80+4+0.80)÷​4

Partitioning the number $84.80$84.80

  $=$= $80\div4+4\div4+0.80\div4$80÷​4+4÷​4+0.80÷​4

Dividing each part by $4$4 separately

  $=$= $20+1+0.20$20+1+0.20

Evaluating each division

  $=$= $21.20$21.20

Evaluating the addition

This shows that Matilda earns $\$21.20$$21.20 per hour. 

Reflect: By partitioning or breaking up the number $84.80$84.80 into smaller pieces we were able to easily divide each part by $4$4, and then combine them back together at the end to get our final answer.

Remember!

We use the same approach we use for dividing with whole numbers, it's just that we work on places to the right of the decimal point.

Now we’ll look at using short division as a way to solve a division with decimals. In this example we need to rename some of our digits, just like we would when dividing whole numbers.

Practice question

Question 1

We want to find $92.1\div3$92.1÷​3

  1. Choose the most reasonable estimate for $92.1\div3$92.1÷​3

    $31$31

    A

    $310$310

    B

    $3.1$3.1

    C

    $0.31$0.31

    D
  2. Complete the short division to find $92.1\div3$92.1÷​3

      $\editable{}$ $\editable{}$ $.$. $\editable{}$
    $3$3   $9$9 $2$2 $.$. $\editable{}$ $1$1
     

When we need to rename

Consider the quotient $54.05\div5$54.05÷​5. We can break this up by partitioning the decimal into $5$5 tens, $4$4 units, $0$0 tenths and $5$5 hundredths. The $5$5 tens and $5$5 hundredths parts are easily divisible by $5$5, but how shall we try to divide the $4$4 units by $5$5?

We can use the fact that $4$4 units is the same as $40$40 tenths, which is now divisible by $5$5. Now the number $54.05$54.05 is partitioned into $5$5 tens, $0$0 units, $40$40 tenths and $5$5 hundredths. When we renamed $4$4 units to $40$40 tenths, we made use of a zero placeholder.

The process we went through in partitioning and renaming is what gets used behind the scenes when we perform short division, as shown in the video below.

Practice question

Question 2

Evaluate $11.54\div2$11.54÷​2 using short division.

  1.     $\editable{}$ $\cdot$· $\editable{}$ $\editable{}$
    $2$2   $1$1 $1$1 $\cdot$· $\editable{}$ $5$5 $\editable{}$ $4$4
       

Dividing by a decimal number

But what about when we want to divide by a decimal number?

We've already seen how to divide decimal numbers by whole numbers, so it would be great if we could just keep using this strategy. Using our knowledge of place value, we can!

We know that $10\div5=2$10÷​5=2, but what is $100\div50$100÷​50? Well, obviously that is just $2$2 as well. We can see that even though both numbers were ten times larger, we ended up with exactly the same answer. We can use the same strategy but with numbers ten times smaller. This means $1\div0.5=2$1÷​0.5=2 as well!

Let's look at some examples to show how this strategy works.

Worked examples

Example 2

Evaluate$5.6\div0.8$5.6÷​0.8

Think: This will be much easier to solve if both numbers were whole numbers, so what can we do to make this happen?

Do: Each decimal in the expression has one decimal place, so we can rewrite the expression with whole numbers by multiplying both decimals by $10$10. Since $5.6\times10=56$5.6×10=56, and $0.8\times10=8$0.8×10=8, then $5.6\div0.8=56\div8$5.6÷​0.8=56÷​8.

Now $56\div8=7$56÷​8=7, so we can return to the original expression and say that $5.6\div0.8=7$5.6÷​0.8=7.

Reflect: By multiplying both numbers by ten, we were able to turn this in to a division using whole numbers, which could then solve easily.

 

EXAMPLE 3

Evaluate$5.28\div0.04$5.28÷​0.04

Think: How can we write this expression with whole numbers?

DoNow each decimal in the expression has two decimal places, so we can rewrite this expression with whole numbers by multiplying both decimals by $100$100. Since $5.28\times100=528$5.28×100=528, and $0.04\times100=4$0.04×100=4, then $5.28\div0.04=528\div4$5.28÷​0.04=528÷​4.

We can evaluate $528\div4$528÷​4 to get $132$132, which means that the original expression is $5.28\div0.04=132$5.28÷​0.04=132.

Remember!

To divide a decimal by a decimal, we can first multiply both numbers by a suitable power of $10$10 to make them whole numbers. This can make the division easier.

Practice question

Question 3

Evaluate the quotient $1.2\div0.3$1.2÷​0.3

Outcomes

0580C1.8B

Use the four rules for calculations with decimals, including correct ordering of operations and use of brackets.

0580E1.8B

Use the four rules for calculations with decimals, including correct ordering of operations and use of brackets.

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