LESSON: Dividing by Decimals

READ: Using Additional Zero Placeholders

Find Quotients of Decimals by Using Additional Zero Placeholders

The decimals that we divided in the last section 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.

Here is a blast from the past problem.

Example

$5 \overline{)13.6 \;}$

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.

Example

$1.2 \overline{)2.79 \;}$

The first thing that we need to do is to multiply the divisor and the dividend by a base ten number 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.