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4.19: Division of Decimals by Decimals using Zero Placeholders

Difficulty Level: At Grade Created by: CK-12
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Practice Division of Decimals by Decimals using Zero Placeholders
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Have you ever tried to divide two numbers but it didn't work perfectly? In the last Concept, you saw Miles divide a decimal by a decimal and it worked out perfectly in the end. What if it didn't? What would Miles do then? Look at this problem.

Jessie decided to try her hand at the sand and the hourglass too. But she used a different bucket. Jessie's bucket would hold 6.75 pounds of sand. 1.25 pounds of sand could go through the hourglass at one time.

Jessie thinks that she can run the sand through the hourglass six times before needing to refill the bucket.

Is she correct?

To solve this problem, you will need to divide once again. Pay attention to this Concept because you will need a new skill to complete the problem.

Guidance

The decimals that we divided in the last Concept were all evenly divisible. This means that we had whole number quotients. We didn’t have any decimal quotients.

What can we do if a decimal is not evenly divisible by another decimal?

If you think back, we worked on some of these when we divided decimals by whole numbers. When a decimal was not evenly divisible by a whole number, we had to use a zero placeholder to complete the division.

$5 \overline{)13.6 \;}$

When we divided 13.6 by 5, we ended up with a 1 at the end of the division. Then we were able to add a zero placeholder and finish finding a decimal quotient. Here is what this looked like.

$& \overset{ \quad 2.72}{5 \overline{ ) {13.60 \;}}}\\& \underline{-10 \;\;}\\& \quad \ 36\\& \ \ \underline{-35\;\;}\\& \qquad 1 - \ \text{here is where we added the zero placeholder}\\& \qquad 10\\& \quad \ \underline{-10}\\& \qquad \ \ 0$

We add zero placeholders when we divide decimals by decimals too.

$1.2 \overline{)2.79 \;}$

The first thing that we need to do is to multiply the divisor and the dividend by a multiple of ten to make the divisor a whole number. We can multiply both by 10 to accomplish this goal.

$12 \overline{)27.9 \;}$

Now we can divide.

$& \overset{ \quad \ \ 2.3}{12 \overline{ ) {27.9 \;}}}\\& \ \underline{-24 \;\;}\\& \quad \ \ 39\\& \quad \underline{-36}\\& \qquad \ 3$

Here is where we have a problem. We have a remainder of 3. We don’t want to have a remainder, so we have to add a zero placeholder to the problem so that we can divide it evenly.

$& \overset{ \quad \ \ 2.32}{12 \overline{ ) {27.90 \;}}}\\& \ \underline{-24\;\;}\\& \quad \ \ 39\\& \quad \underline{-36\;\;}\\& \qquad \ 30\\& \quad \ \ \underline{-24}\\& \qquad \quad 6$

Uh Oh! We still have a remainder, so we can add another zero placeholder.

$& \overset{ \quad \ 2.325}{12 \overline{ ) {27.900 \;}}}\\& \ \ \underline{-24\;\;}\\& \ \quad \ \ 39\\& \ \quad \underline{-36\;\;}\\& \ \qquad \ 30\\& \ \quad \ \ \underline{-24\;\;}\\& \ \qquad \quad 60\\& \ \qquad \ \underline{-60\;\;}\\& \ \qquad \quad \ \ 0$

Sometimes, you will need to add more than one zero. The key is to use the zero placeholders to find a quotient that is even without a remainder.

Example A

$1.2 \overline{)2.76 \;}$

Solution: 2.3

Example B

$8.7 \overline{)53.94 \;}$

Solution: 6.2

Example C

$5.4 \overline{)18.9 \;}$

Solution: 3.5

Now back to Jessie and the sand. Did you figure out that you will need a zero placeholder to figure out if Jessie's estimate is correct? Let's look at the original problem once again.

Jessie decided to try her hand at the sand and the hourglass too. But she used a different bucket. Jessie's bucket would hold 6.75 pounds of sand. 1.25 pounds of sand could go through the hourglass at one time.

Jessie thinks that she can run the sand through the hourglass six times before needing to refill the bucket.

To figure out if a 6.75 pound bucket of sand can be divided into 1.25 pounds six times, we will need to divide.

$1.25 \overline{)6.25 \;}$

Next, we can divide.

Our answer is $5.4$ .

Jessie isn't correct. She will need to refill the bucket after pouring 1.25 pounds of sand through the hourglass five times.

Vocabulary

Here are the vocabulary words in this Concept.

Divisor
the number doing the dividing, it is found outside of the division box.
Dividend
the number being divided. It is found inside the division box.
Quotient
the answer in a division problem

Guided Practice

Here is one for you to try on your own.

$3.2 \overline{)28.52 \;}$

To complete this problem, we must first move the decimal point one place in the divisor which makes 3.2 into 32. Then we can simply divide. Notice that you will need to use zero placeholders and keep adding them until the division is complete.

Our answer is $8.9125$ .

Practice

Directions: Divide the following decimals. Use zero placeholders when necessary.

1. $1.3 \overline{)5.2 \;}$
2. $6.8 \overline{)13.6 \;}$
3. $4.5 \overline{)13.5 \;}$
4. $2.5 \overline{)10 \;}$
5. $3.3 \overline{)19.8 \;}$
6. $8.5 \overline{)17 \;}$
7. $9.3 \overline{)27.9 \;}$
8. $1.2 \overline{)7.2 \;}$
9. $5.3 \overline{)26.5 \;}$
10. $6.5 \overline{)13 \;}$
11. $1.25 \overline{)7.5 \;}$
12. $3.36 \overline{)20.16 \;}$
13. $5.87 \overline{)52.83 \;}$
14. $2.5 \overline{)3 \;}$
15. $3.2 \overline{)8 \;}$
16. $4.6 \overline{)10.58 \;}$
17. $8.1 \overline{)17.82 \;}$

Vocabulary Language: English

Dividend

Dividend

In a division problem, the dividend is the number or expression that is being divided.
divisor

divisor

In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Quotient

Quotient

The quotient is the result after two amounts have been divided.

Oct 29, 2012

Jul 08, 2015