# 2.5: Dividing Decimals

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Pole Vault Mat

The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.

The track and field team at Harrison Middle school had a special visitor after practice. Jody, a pole vaulter from the nearby college, visited to share his experiences with the students. He brought some pictures of himself in different events and after showing them to the students, spent the rest of the time answering student questions.

“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”

Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.

9009 cubic feet

16.5 feet in length

21 feet in width

Justin knows that he is missing the height of the mat.

“How high is the mat?” Justin asked Kara, showing her his notes.

“Who cares?” Kara whispered looking back at Jody.

“I do,” Justin said turning away.

Justin began to complete the math, but he couldn't remember how to work the equation and the division.

This is where you come in. Pay attention to this lesson. By the end of it, you will help Justin with his dilemma.

What You Will Learn

By the end of this lesson, you will be able to:

• Divide decimals with and without rounding.
• Estimate and confirm decimal quotients by dividing leading digits.
• Identify and apply the Inverse Property of Multiplication in decimal operations, using numerical and variable expressions.
• Solve real-world problems using formulas involving decimal quotients.

Teaching Time

I. Divide Decimals With and Without Rounding

Division is useful any time that you want to split up an object or a group of objects. We divide all the time. Division is very important. While you have been dividing whole numbers since elementary school, now it is time for you to learn how to divide decimals.

Dividing decimals is like dividing whole numbers. Dividing a divisor into a dividend gives a quotient.

divisor)dividend¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯quotient\begin{align*}\overset{ \qquad \quad quotient}{divisor \overline{ ) {dividend \;}}}\end{align*}

Division of decimals has special rules for how to deal with the decimal point.

Our place-value system is based on 10. Anytime we multiply by 10, we can move the decimal point to the right the number of places per multiple of 10.

\begin{align*}4.55 \times 10 = 45.5\end{align*}

Notice that we multiplied by 10 so we moved the decimal point one place to the right.

\begin{align*}4.55 \times 100 = 455\end{align*}.

Notice that here we multiplied by 100, so we moved the decimal point two places to the right because \begin{align*}10 \times 10\end{align*} is 100. Two tens = two decimal places.

The first step to dividing decimals is to move the divisor’s decimal point all the way to the right.

When we move the decimal point in the divisor to the right until it is a whole number, we also move a decimal point the same number of places in the dividend.

Example

\begin{align*}{2.3 \overline{ ) {4.6 \;}}}\end{align*}

2.3 is our divisor and 4.6 is the dividend. We need to make the divisor into a whole number. To do this, we only have to move the decimal point one place to the right. That is the same as multiplying by 10.

\begin{align*}2.3 \times 10 = 23\end{align*}

If we do this to the divisor, we have to do it to the dividend too. We multiply the dividend by 10 to move the decimal one place to the right.

\begin{align*}4.6 \times 10 = 46\end{align*}

Now we can divide.

\begin{align*}\overset{ \quad 2}{23 \overline{ ) {46 \;}}}\end{align*}

What if there aren’t enough places in the dividend to move the decimal point?

If there aren’t enough places in the dividend, then we can add zeros. Here is an example of how that works.

Example

\begin{align*}{2.25 \overline{ ) {9 \;}}}\end{align*}

First, we need to make the divisor into a whole number. There are two places after the decimal point, so we need to multiply by 100.

\begin{align*}2.25 \times 100 = 225\end{align*}

We also have to do this with the dividend. To move the decimal point two places after the nine, we have to add two zero.

\begin{align*}9 \times 100 = 900\end{align*}

Now we can divide.

\begin{align*}\overset{ \quad \ \ 4}{225 \overline{ ) {900 \;}}}\end{align*}

Take a few notes on how to move the decimal point in the divisor and dividend and then continue the lesson.

You have already seen how rounding can help us when estimating in addition, subtraction and multiplication problems. Rounding can also be useful when dividing. Sometimes, you will need to round before dividing and sometimes, you will need to round after dividing.

Pay attention and be sure to round to the designated place.

Example

Round each number to the nearest tenth, then divide, \begin{align*}67.521 \div 2.243\end{align*}

This problem asks us to round each number to the tenths place before dividing.

The rounding steps: underline the number we’re rounding to and bold or circle the number directly to the right of it.

We’re rounding to the tenth place, so we’ll round to the place directly to the right of the decimal place.

The bolded number, the hundredths place, is the one to look at when deciding to round up or down.

67.521 \begin{align*}\rightarrow\end{align*} rounded to the tenth place \begin{align*}\rightarrow\end{align*} 67.5

2.243 \begin{align*}\rightarrow\end{align*} rounded to the tenth place \begin{align*}\rightarrow\end{align*} 2.2

\begin{align*}{2.2 \overline{ ) {67.5 \;}}}\end{align*} Place our rounded numbers in the long division format.

\begin{align*}{22 \overline{ ) {675.0 \;}}}\end{align*} Multiply both numbers by 10 (making the divisor a whole number).

To adjust the dividend, move the decimal to the right of 67.5 and add a zero.

Next, we divide just as we do with whole numbers. Notice, that we move the decimal point from the dividend right up into the quotient. You can draw an arrow to show how it moves from the dividend to the quotient.

\begin{align*}& \overset{ \ \ 30.68181}{22 \overline{ ) {675.00000 \;}}}\\ & \underline{\; \;-66 \;}\\ & \qquad 150\\ & \quad \underline{- \; 132 \;\;}\\ & \qquad \ \ 180\\ & \quad \ \ \underline{- \; 176 \;}\\ & \qquad \quad \ \ 40\\ & \qquad \ \ \underline{- \; 22 \;\;\;}\\ & \qquad \quad \ \ \ 180\\ & \qquad \ \ \ \underline{- \; 176 \;\;}\\ & \qquad \qquad \ \ \ 40\\ & \qquad \quad \ \ \ \underline{- \; 22}\\ & \qquad \qquad \ \ \ 18\end{align*}

Yes, there is. You can see how we added zeros to help us with the dividing well there is also a pattern that is repeating at the end of this quotient. Because of this, we can conclude that we could keep adding zeros and the pattern would continue to repeat indefinitely.

What we have here is called a repeating decimal, because the 8181 pattern will repeat over and over. The more zeros we add to the dividend, the longer the quotient will be. In this case, you can either round the quotient to 30.682 or you can notate the repeating decimal by putting a line over the repeating part.

\begin{align*}30.6\overline{81}\end{align*}

This line means that the digits under the line will repeat. For this problem, let’s use a rounded answer.

2L. Lesson Exercises

Divide.

1. \begin{align*}3.96 \div 1.2\end{align*}
2. \begin{align*}24.288 \div 9.6\end{align*}
3. Round to the nearest tenth, then divide, \begin{align*}4.7 \div 2.4\end{align*}

Now take a few minutes to check your work with a partner. Be sure that your work is accurate and correct any mistakes before continuing.

II. Estimate and Confirm Decimal Quotients by Dividing Leading Digits

Estimation is a process by which we approximate solutions. Estimating either before or after solving a problem helps to generalize or confirm a solution.

Rounding decimals before division is one way to find an estimate. You can also simply divide the leading digits.

Remember how we estimated products by multiplying the leading digits? Estimating quotients works the same way. Leading digits are the first two values in a decimal. To estimate a quotient, divide the leading digits exactly as you have been—move the decimal point in the divisor to make it a whole number, adjust the decimal point in the dividend accordingly, then divide, inserting the decimal point in the solution in line with its position in the dividend.

Example

Estimate the quotient. \begin{align*}7.882 \div .4563\end{align*}

To estimate, we are going to work with only the leading digits, or the first two.

Let’s examine the decimals and reduce them to their leading digits. \begin{align*}7.882 \rightarrow 7.8\end{align*} and \begin{align*}.4563 \rightarrow .45\end{align*}.

Now let’s move the decimal point in the divisor and dividend. In long-division form, we have \begin{align*}{.45 \overline{ ) {7.8 \;}}}\end{align*} so we’re going to multiply both numbers by 100 and move the decimal points two places to the right: \begin{align*}{45 \overline{ ) {780 \;}}}\end{align*}. Notice how we added a zero to the end of the dividend to make the move of the decimal point possible.

Now we can divide.

\begin{align*}& \overset{ \quad 17.33}{45 \overline{ ) {780.00 \;}}}\\ & \quad \underline{- \; 45 \;\;}\\ & \qquad 330\\ & \quad \underline{- \; 315\;\;}\\ & \qquad \ \ 150\\ & \quad \ \ \underline{- \; 135\;\;}\\ & \qquad \quad \ 150\\ & \qquad \ \underline{- \; 135}\\ & \qquad \qquad 15\end{align*}

Notice how we added 2 extra zeros to the dividend to facilitate our division? We could keep adding zeros and keep dividing and our quotient would get longer and longer. Do you notice a pattern in the quotient? Each time we add a zero to the dividend, we come up with another 3 in the quotient. This is another repeating decimal. We can either round our answer to 17.33 or notate the repeating decimal by putting a line over the repeating part \begin{align*}17.\overline{3}\end{align*}.

Our answer is 17.33 or \begin{align*}17.\overline{3}\end{align*}.

Let’s look at another example where we divide leading digits, but this time, we will use estimation to confirm our answer when we are finished.

Example

\begin{align*}4.819 \div 1.245\end{align*}

First, let’s simplify these numbers to their leading digits.

4.819 = 4.8

1.245 = 1.2

Now we can set up the problem as a division problem.

\begin{align*}{1.2 \overline{ ) {4.8 \;}}}\end{align*}

Next, we make the divisor into a whole number by multiplying by 10. We do the same thing in the dividend. Next, we can divide to find the quotient.

\begin{align*}\overset{ \quad 4}{12 \overline{ ) {48 \;}}}\end{align*}

Now, let’s go back to the original problem to confirm our answer using estimation. If we were to round these two numbers to the nearest tenth, here is our answer.

4.819 = 4.8

1.245 = 1.2

If we ignore the decimal points, we have twelve and forty-eight. Twelve times four is forty-eight. We see that our answer makes sense.

2M. Lesson Exercises

1. \begin{align*}15.934 \div 2.57\end{align*}
2. \begin{align*}4.368 \div 3.12\end{align*}
3. \begin{align*}6.16 \div 1.12\end{align*}

III. Identify and Apply the Inverse Property of Multiplication in Decimal Operations, Using Numerical and Variable Expressions

Sometimes division is called the inverse, or opposite, of multiplication. This means that division will “undo” multiplication.

\begin{align*}5 \times 6 = 30\!\\ 30 \div 6 = 5\end{align*}

See how that works? You can multiply two factors to get a product. Then when you divide the product by one factor, you get the other factor.

The Inverse Property of Multiplication states that for every number \begin{align*}x, x\left (\frac{1}{x} \right )= 1\end{align*}.

In other words, when you multiply \begin{align*}x\end{align*} by its opposite, \begin{align*}\frac{1}{x}\end{align*} (also known as the multiplicative inverse), the result is one.

Let’s put whole numbers in place of \begin{align*}x\end{align*} to make the property clear: \begin{align*}3 \times \left ( \frac{1}{3} \right ) = \frac{3}{3} = 1\end{align*}. The Inverse Property of Multiplication may seem obvious, but it has important implications for our ability to solve variable equations which aren’t easily solved using mental math—such as variable equations involving decimals.

How does this apply to our work with equations?

With equations, the two expressions on either side of the equal sign must be equal at all times. The Inverse Property of Multiplication lets us multiply or divide the same number to both sides of the equation without changing the solution to the equation. This technique is called an inverse operation and it lets us get the variable \begin{align*}x\end{align*} alone on one side of the equation so that we can find its value. Take a look at how it’s done.

Example

\begin{align*}5x & = 15\\ \frac{5x}{5} &= \frac{15}{5} \rightarrow \text{inverse operation } = \text{divide} \ 5 \ \text{from both sides}\\ x &= 3\end{align*}

Remember how we said that division can “undo” multiplication? Well, this is an example where that has happened.

Notice how, on the left side of the equation, \begin{align*}\frac{5}{5} = 1\end{align*}, leaving the \begin{align*}x\end{align*} alone on the left side. You will see this often in Algebra when solving for unknowns in equations. When you divide a number by itself the answer is 1. 1 times any number is just that number. Knowing this rule helps us see why we would want to divide numbers around a variable by themselves. Our goal is to "isolate" the variable. That means we want to be left with just the variable by itself on one side of the equals sign. After a while, this will become so natural that you won't even think about why you are doing it.

Example

Solve \begin{align*}2.7x = 3.78\end{align*}

We need to find a value of \begin{align*}x\end{align*} that, when multiplied by 2.7, results in 3.78. Let’s begin by using inverse operations to get \begin{align*}x\end{align*} alone on the left side of the equation.

\begin{align*}2.7x & = 3.78\\ \frac{2.7}{2.7} x&= \frac{3.78}{2.7} \rightarrow \text{Inverse Operations} = \text{divide both sides by} \ 2.7\end{align*}

Now, to find the value of \begin{align*}x\end{align*}, we complete the decimal division. First we multiply by ten and move the decimal places accordingly. \begin{align*}2.7 \rightarrow 27\end{align*} and \begin{align*}3.78 \rightarrow 37.8\end{align*}

\begin{align*}& \overset{ \quad \ \ 1.4}{27 \overline{ ) {37.8 \;}}}\\ & \quad \ \underline{27\;\;}\\ & \qquad 108\\ & \quad \underline{- \; 108}\\ & \qquad \quad 0\end{align*}

2N. Lesson Exercises

Use the inverse operation to solve each equation.

1. \begin{align*}2.3x = 5.06\end{align*}
2. \begin{align*}1.6x= 5.76\end{align*}
3. \begin{align*}4.7x = 10.81\end{align*}

Take a few minutes to check your work with a friend.

IV. Solve Real-World Problems Using Formulas Involving Decimal Quotients

Decimals are all around us. Anytime that you work with money you are working with decimals. We can use what we have learned about equations and division to help us with solving real-world problems.

Equations help define the relationships between different elements in a problem. To write an equation, you read the problem carefully, define the unknown as a variable, and write an equation that describes the mathematical relationship between that variable and the other information in the problem. The mathematical relationship may involve one or more operations. Once you have the equation, you can use one or more inverse operations to solve for the value of the variable.

Example

The Landry’s living room has an area of \begin{align*}97.92 \ m^2\end{align*} and a length of 13.6 meters. What is the width of the living room?

Remember the formula for finding the area of a rectangle? \begin{align*}A = lw\end{align*}. In this problem, we are given the area and the length. We need to substitute those values into the formula and solve for the width.

\begin{align*}A & = lw\\ 91.92 & = 13.6w\\ \frac{91.92}{13.6} & = \frac{13.6}{13.6} w \rightarrow \text{inverse operations:} \ \text{divide both sides by} \ 13.6\\ w & = \frac{91.92}{13.6}\end{align*}

Multiply both the divisor and dividend by 10 to make the divisor a whole number; then divide.

\begin{align*}& \overset{ \qquad \ \ 7.2}{136 \overline{ ) {979.2 \;}}}\\ & \ \ \underline{- \; 952\;\;}\\ & \qquad \ 272\\ & \quad \ \underline{- \; 272}\\ & \qquad \quad \ 0\end{align*}

\begin{align*}w = 7.2\end{align*}

The answer is width = 7.2 meters

Now let’s apply this information in solving our introductory problem!

## Real Life Example Completed

The Pole Vault Mat

Here is the original problem once again. Reread it and underline any important information.

The pole vault is a track and field event where a student uses a pole to launch themselves over a bar. Then the student lands on a large mat underneath the pole. Each part of the pole vault event is very specific. The height of the bar is specific. The length of the pole is specific, and the dimensions of the mat are specific as well.

The track and field team at Harrison Middle school had a special visitor after practice. Jody, a pole vaulter from the nearby college, visited to share his experiences with the students. He brought some pictures of himself in different events and after showing them to the students, spent the rest of the time answering student questions.

“Even the mat has specific dimensions,” Jody explained. “They measure the length, width and height of the mat to be sure that it has an accurate volume. The mat that we are using has a volume of 9009 cubic feet. The length of the mat is 16.5 feet and the width is 21 feet.”

Justin and Kara were listening intently to Jody’s explanation of the pole vault event. Justin, who loves numbers, began jotting down the dimensions of the mat on a piece of paper.

9009 cubic feet

16.5 feet in length

21 feet in width

Justin knows that he is missing the height of the mat.

“How high is the mat?” Justin asks Kara, showing her his notes.

“Who cares?” Kara whispered, looking back at Jody.

“I do, some things are worth figuring out,” Justin said turning away.

Justin began to complete the math. But he couldn't remember how to work the equation and the division.

To solve this problem of height, we need to remember that the formula for volume is length times width times height. Justin has the measurements for the volume and for the length and the width. He is missing the height. Justin can write the following formula.

\begin{align*}V &= lwh\\ 9009 & = 16.5 \times 21 \times h\end{align*}

Next, we multiply \begin{align*}16.5 \times 21\end{align*} to begin solving our math problem.

\begin{align*}16.5 \times 21 = 346.5\end{align*}

Now can write this equation.

\begin{align*}9009 = 346.5h\end{align*}

Next we divide 9009 by 346.5.

\begin{align*}h = 26 \ feet\end{align*}

The height of the mat is equal to 26 feet.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Divisor
the number outside the division box. This is the number that is doing the dividing.
Dividend
the number being divided. It is the number inside the division box.
Quotient
the answer in a division problem.
Estimation
Inverse
the opposite
Inverse Property of Multiplication
when you multiply a value by its opposite, the answer is one.
Inverse Operation
the opposite operation. The opposite operation of division is multiplication.
Equation
a statement where the value on one side of the equals sign must be the same as the value on the other side of the equals sign.

## Technology Integration

Other Videos:

http://www.mathplayground.com/howto_dividedecimals.html – This is a video on how to divide decimals.

http://www.mathplayground.com/howto_dividedecimalspower10.html – This is a video on how to divide decimals by powers of ten.

## Time to Practice

Directions: Find the quotient.

1. \begin{align*}5.4 \div 4.5\end{align*}

2. \begin{align*}8.71 \div 6.7\end{align*}

3. \begin{align*}3.375 \div 2.25\end{align*}

4. \begin{align*}11.2 \div 5.6\end{align*}

5. \begin{align*}19.11 \div 1.3\end{align*}

6. \begin{align*}28.992 \div 18.12\end{align*}

7. \begin{align*}113.52 \div 12.9\end{align*}

8. \begin{align*}31.93 \div 3.1\end{align*}

9. \begin{align*}46.125 \div 6.15\end{align*}

10. \begin{align*}84.28 \div 17.2\end{align*}

Directions: Find the quotient after rounding each to the nearest tenth.

11. \begin{align*}113.409 \div 25.2157\end{align*}

12. \begin{align*}81.862 \div 6.453\end{align*}

13. \begin{align*}377.151 \div 11.54269\end{align*}

14. \begin{align*}9.177 \div 4.5712\end{align*}

Directions: Estimate the quotient by dividing the leading digits.

15. \begin{align*}4.992 \div .07123\end{align*}

16. \begin{align*}1.8921 \div 6.0341\end{align*}

17. \begin{align*}3,026.2129 \div 1.5612\end{align*}

18. \begin{align*}1.00765 \div .33\end{align*}

Directions: Solve the following problems using what you have learned about dividing decimals and equations. Write an equation when necessary.

19. Solve \begin{align*}13.7x = 63.02\end{align*}

20. Solve \begin{align*}.55x = .31955\end{align*}

21. Solve \begin{align*}9.8x = 114.66\end{align*}

22. In a week of track practice, Rose ran 3.12 times more than Jamie. If Rose ran 17.16 kilometers, how many kilometers did Jamie run? Write an equation and solve.

23. Ling’s flower bed has an area of \begin{align*}23.12 \ m^2\end{align*} and a width of 3.4 meters. What is the length of Ling’s flower bed? Write an equation and solve.

24. A jet airplane travels 6.5 times faster than a car. If the jet travels at 627.51 kilometers per hour, how fast is the car? Write an equation and solve.

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