2.2: Adding Decimals
Introduction
The Relay
Last year, the girl’s middle school relay team ran the \begin{align*}4 \times 100\end{align*} meters in 53.87 seconds and won the regional championship. This year, there are four new girls on the team and they have set a goal of running the \begin{align*}4 \times 100\end{align*} in a faster time than the girls did the year before.
Jessica is the lead runner on the team. The second runner is Karin. The third runner is Tasha and the fourth, and fastest, runner is Uniqua. They didn’t know each other prior to making the team, so it took the girls most of the season to get to know each other’s strengths and weaknesses. Their coach, Ms. Sutter, worked hard to help the girls work through their differences while maintaining a good attitude. Ms. Sutter doesn’t put up with a lot of nonsense and all of the runners show respect to her and their team mates.
Because of their hard work, the relay team made it all the way to the Regional championship.
“Just be sure to run your best race,” Ms. Sutter told them prior to the start.
All four girls shook hands and prepared to run. Here are their individual times:
Jessica = 13.00
Karin = 14.10
Tasha = 14.23
Uniqua = 12.40
Did the girls accomplish their goal of running the relay faster than the previous team?
This lesson is about adding decimals. To figure out the total time of the relay team, you will need to know how to add decimals. Finish this lesson and you will be ready!
What You Will Learn
In this lesson you will the following skills.
- Add decimals with and without rounding.
- Estimate decimal sums using front-end estimation.
- Identify and apply the commutative and associative properties of addition in decimal operations, using numerical and variable expressions.
- Model and solve real-world problems using simple equations involving decimal addition.
Teaching Time
I. Add Decimals With and Without Rounding
Decimals are used in many aspects of life. Any time there is a part of a whole number—from money to measurement to percentages, you will see and work with decimals. Meghan finished 0.75 of her homework; Leo ran 1.34 kilometers; the class ate 5.6 pizzas.
You already know how to compare, order, and round decimals.
In this lesson we will learn how to add them. Once you know how to add decimals, you can apply your knowledge to a whole new dimension of real-world situations, some of which involve algebraic reasoning and equations.
How do we get started when we want to add decimals?
Well, first of all, it all comes back down to place value. To add decimals, we line up the digits according to place value. To keep the place values straight, we always align the decimal points of the numbers we are working with.
Example
\begin{align*}4.5 + 2.137 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
To add this problem, let’s remember to line up the decimal points and each digit according to place value. Notice that there is one digit after the decimal point in the first number and three in the second number. You will need to add some zeros on the end of the first number to help with the addition. Now let’s line up the problem.
\begin{align*}& \quad 4.500\\ & \underline{+ \; 2.137}\\ & \quad 6.637\end{align*}
Sometimes, particularly in real-world situations, decimal numbers are just too long to work with. In these cases, to find a quick and workable solution we might round the decimals before we add. For instance, rather than getting bogged down in the ten-millionths place, we can round to the tenths or hundredths place and then add. Look for clues in the problem to tell you whether or not to round before adding. If you are asked to round, make sure you round to the right place!
Let’s look at an example of a problem where rounding would be helpful.
Example
Round each number to the nearest hundredths then find the sum of 9.199, 11.353, 10.648.
This problem asks us to round before adding and it asks us to round to the hundredths place. To make the rounding steps clear, it helps to underline the number you’re rounding to and bolding or circling the number directly to the right of it. Because we’re rounding to the hundredth place, we’re going to round to the two places to the right of the decimal place. The bolded number, the thousandths place, is the one you’ll look at when deciding to round up or down.
9.199 \begin{align*}\rightarrow\end{align*} rounded to the hundredths place \begin{align*}\rightarrow\end{align*} 9.20
11.353 \begin{align*}\rightarrow\end{align*} rounded to the hundredths place \begin{align*}\rightarrow\end{align*} 11.35
10.648 \begin{align*}\rightarrow\end{align*} rounded to the hundredths place \begin{align*}\rightarrow\end{align*} 10.65
Now that the numbers are rounded, we add zeros where needed, line up the decimal places, and add.
\begin{align*}& \quad \ 09.20\\ & \quad \ 11.35\\ & \underline{+ \;\; 10.65}\\ & \quad \ 31.20\end{align*}
Our answer is 31.2.
2D. Lesson Exercises
- \begin{align*}4.56 + 1.2 + 37.89 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}14.2 + 56.78 + 123.4 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}189.34 + 123.5 + 7.2 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
Take a few minutes to check your work with a partner.
II. Estimate Decimal Sums Using Front-End Estimation
What is estimation?
Estimation is a method for finding an approximate solution to a problem. For example, the sum of 22 and 51 is exactly 73. We can estimate the sum by rounding to the tens place and adding 20 + 50. Then we can say the sum of 22 and 51 is “about 70.” Before we perform an operation, an estimate gives us a rough idea of what our answer will be.
Estimation is also useful when checking answers. If, after you solve a problem, you estimate to get a general idea of what an answer should be, you know if your answer is reasonable. You could say that an estimate is an “educated guess.”
By “educated guess” we don’t mean a guess that is out of nowhere. You have to have a method to figuring out the estimate. Rounding numbers before adding is one way to find an estimate.
Front-end estimation is another way. Like rounding, front-end estimation uses place value; unlike rounding, it doesn’t involve any changing of values.
There are three steps to front-end estimation.
First, you add the front end—or leftmost—digits to get a general estimate of the sum. Next, to refine the first estimate, you add the digits directly to the right of the front-end digits.
Finally you add the two sums together.
Front-End Estimation for Addition:
- Add the front (leftmost) digits.
- Add the digits directly to the right of the front-end.
- Add the estimates from steps 1 and 2.
Write these steps in your notebook and then continue with the lesson.
Example
Use front-end estimation to find the sum of 6.819 and 4.621.
Let’s begin by lining up the decimal points. Then we’ll follow the steps of front-end estimation.
6.819
4.621
1. Front-end estimation tells us to add the front or leftmost digits first, so we want to add the ones place.
\begin{align*}6 + 4 = 10\end{align*}.
2. Next, to refine that estimate, we’ll add the digits directly to the right of the front-end digits, or the tenths place.
\begin{align*}.8 + .6 = 1.4\end{align*}
3. Finally, we add our two estimates together.
\begin{align*}10 + 1.4 = 11.4\end{align*}.
Note: The original numbers extended to the thousandth place, but we don’t need to show this by adding unnecessary zeros to our answer. Where decimals end, it is assumed that zeros extend on infinitely.
2E. Lesson Exercises
Practice using front-end estimation to find the following sums.
- \begin{align*}3.5 + 2.34 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}12.671 + 8.123 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}15.67 + 9.345 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
Take a few minutes to check your answers with a partner.
III. Identify and Apply the Commutative and Associative Properties of Addition in Decimal Operations, Using Numerical and Variable Expressions
By now you know how to add decimals, round and add decimals, and estimate and add decimals. Now we need to examine decimal addition through the lens of the two mathematical properties which pertain to addition: the Commutative Property of Addition and the Associative Property of Addition.
In the examples that we worked with in the last section, we added three decimal addends in the order in which they appeared. What would happen if we added them in a different order? Would the sum change?
The Commutative Property of Addition states that the order of the addends does not change the sum.
Let’s test the property using simple whole numbers.
\begin{align*}4 + 5 + 9 = 18 && 5 + 4 + 9 = 18 && 9 + 5 + 4 = 18\\ 4 + 9 + 5 = 18 && 5 + 9 + 4 = 18 && 9 + 4 + 5 = 18\end{align*}
As you can see, we can add the three addends (4, 5, and 9) in any order. The Commutative Property of Addition also works for four, five, or a hundred addends, and it works for decimal addends, too.
What about the Associative Property of Addition?
The Associative Property of Addition states that the way in which addends are grouped does not change the sum. The Associative Property uses parentheses to help us with the grouping. When parentheses are inserted into a problem, according to the order of operations, you perform that operation first. What happens if you move the parentheses? This is where the Associative Property comes in.
Once again, let’s test the property using simple whole numbers.
\begin{align*}(4 + 5) + 9 = 18 && 4 + (5 + 9) = 18 && (9 + 5 + 4) = 18\end{align*}
Clearly, the different way the addends are grouped has no effect on the sum. The Associate Property of Addition works for multiple addends as well as decimal addends.
These two properties are extremely useful for solving equations. An equation is a mathematical statement that two expressions are equal. For example, we can say that \begin{align*}15 + 7 = 24 - 2\end{align*} since both sides of the equation equal 22.
A variable equation is an equation such as \begin{align*}12 + x = 14\end{align*}. It is called a variable equation because it includes an algebraic unknown, or a variable.
Example
Find the value for \begin{align*}x\end{align*} in the following equation, \begin{align*}71.321 + 42.29 = x + 71.321\end{align*}
Here is where we can use the Commutative Property of Addition. We know that one side of the equation is equal to the other side of the equation, therefore we just substitute 42.29 in for \begin{align*}x\end{align*} since the 71.321 did not change.
Our answer is \begin{align*}x\end{align*} is equal to 42.29.
We can use the Associative Property for both expressions and equations. Let’s start by looking at an expression.
Example
Find the value when \begin{align*}x\end{align*} is 2 and \begin{align*}y\end{align*} is 3. \begin{align*}(x+4)+ y\end{align*}
This is a variable expression. Notice that there are two given values for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, but that those values could be any numbers. The key here is that we could move the parentheses and the two variable expressions would have the same value. We can find the value of this expression by substituting 2 and 3 into the expression.
\begin{align*}(2+4)+ 3\end{align*}
The value of this variable expression is 3.
What happens if we move the parentheses?
\begin{align*}x+(4+y)\end{align*}
We substitute the given values in for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
\begin{align*}2+(4+3)\end{align*}
The value of this expression is 9.
Because of the Associative Property, both variable expressions have the same value regardless of the groupings.
IV. Model and Solve Real-World Problems Using Simple Equations Involving Decimal Addition
Let’s look at applying what you have learned about adding decimals and solving equations with an example.
Example
Hassan’s recipe for potato soup calls for 14.25 kilograms of potatoes. He already has 6.12 kilograms. How many more kilograms of potatoes does he need to buy?
In this problem, we are trying to find the number of kilograms of potatoes Hassan needs to buy. Let’s call that value \begin{align*}x\end{align*}. We know how many kilograms he already has, and we know the total kilograms he needs, so we can write the equation
\begin{align*}6.12 + x = 14.25\end{align*}.
Now we can solve for the value of \begin{align*}x\end{align*} using mental math. \begin{align*}x\end{align*} needs to be a value that, when added to 6.15, equals 14.25.
Let’s begin with the whole numbers. \begin{align*}\ 6 + 8 = 14\end{align*}; the whole number portion of \begin{align*}x\end{align*} is 8.
Now let’s look at the decimals. \begin{align*}.12 + ? = .25\end{align*}. The decimal parts differ in both tenths and hundredths, but if we think of them as whole numbers it is easier to see their relationship \begin{align*}12 + 13 = 25\end{align*}, so \begin{align*}.12 + .13 = .25\end{align*}.
Combining the whole number portion of \begin{align*}x\end{align*} with its decimal portion gives \begin{align*}(8 + .13)\end{align*}, we get \begin{align*}8.13\end{align*}. Remember to put the units of measurement in your answer!
The answer is 8.13 kilograms.
We can use this same strategy when working with our real-world example from the beginning of the lesson. Let’s take a look at this example once again.
Real Life Example Completed
The Relay
Here is the original problem once again. Reread the problem and underline any important information.
Last year, the girl’s middle school relay team ran the \begin{align*}4 \times 100\end{align*} meters in 53.87 and won the regional championship. This year, there are four new girls on the team and they have set a goal of running the \begin{align*}4 \times 100\end{align*} faster than the girls did the year before.
Jessica is the lead runner on the team. The second runner is Karin. The third runner is Tasha and the fourth runner, and fastest, is Uniqua. They didn’t know each prior to making the team, so it took the girls most of the season to get to know each other’s strengths and weaknesses. Their coach, Ms. Sutter, worked hard to help the girls work through their differences while maintaining a good attitude. Ms. Sutter doesn’t put up with a lot of nonsense and all of the runners show respect to her and their team mates.
Because of their hard work, the relay team made it all the way to the Regional championship.
“Just be sure to run your best race,” Ms. Sutter told them prior to the start.
All four girls shook hands and prepared to run. Here are their individual times:
Jessica = 13.00
Karin = 14.10
Tasha = 14.23
Uniqua = 12.40
Did the girls accomplish their goal of running the relay faster than the previous team?
First, we can write an equation to figure out the time of the race.
\begin{align*}j+k+t+u<53.87\end{align*}
The girls want to have a time that is less than 53.87. In this way, we are going to be combining comparing decimals and adding decimals.
Next, we fill in each girl's time into the equation.
\begin{align*}13.00 + 14.10 + 14.23 + 12.40 < 53.87\end{align*}
Now, we can line up the decimal point to add the values.
\begin{align*}& \quad 13.00\\ & \quad 14.10\\ & \quad 14.23\\ & \underline{+ \; 12.40}\\ & \quad 53.73\end{align*}
The time of the relay is 53.73.
Now we can compare this with the previous team’s time.
53.73 < 53.87
This is a true statement. The girls have beaten the previous relay team’s time. They also achieved a medal at the Regional Championship!!
Vocabulary
Here are the vocabulary words that are found in this lesson.
- Decimal
- a part of a whole represented by digits to the right of a decimal point.
- Estimate
- to find an approximate solution to a problem.
- Rounding
- a method of estimating where you rewrite a decimal or whole number according to the place value that it is closest to.
- Front-End Estimation
- a method of estimating where you only add the fronts of the numbers. You add the whole numbers if there are any, and the tenths place of the decimal digits. Then you combine the sums for the final estimate.
- Commutative Property of Addition
- states that the order in which you add numbers does not impact the sum of those numbers.
- Associative Property of Addition
- states that the grouping of addends does not impact the sum of the addends. The groupings are identified by parentheses.
- Equation
- a number sentence where the value on one side of the equals sign is the same as the value on the other side of the equals sign.
- Variable Expression
- an expression where a variable is used to represent an unknown quantity. An expression does not have an equals sign.
- Variable Equation
- an equation where a variable is used to represent an unknown quantity.
Technology Integration
James Sousa, Example of Addition of Decimals
James Sousa, Example of Addition of Three Decimals
Other Videos:
- http://www.schooltube.com/video/05733ec4ba5143739f6d/Adding-and-Subtracting-Decimals – This is a video on adding and subtracting decimals in vertical form.
Time to Practice
Directions: Find each sum.
1. 29.451 + 711.36
2. 0.956 + 0.9885
3. 13.69 + 42.23
4. 58.744 + 12.06 + 8.8
5. 67.90 + 1.2 + 3.456
Directions: Round to the nearest tenth, then find the sum.
6. 88.95 + 23.451
7. 690.026 + 831.163
8. 3.27 + 7.818
9. .99 + .909 + .048
10. 5.67 + 3.45 + 1.23
Directions: Round to the nearest hundredth, then find the sum.
11. 511.51109 + 99.0986
12. .744 + 7.005 + .7071
13. 32.368 + 303.409
14. 1.0262 + 1.0242
15. 1.309 + 3.4590
Directions: Use front-end estimation to find the following sums.
16. .48218 + .61927
17. 6.7765 + 6.421192
18. .5075412 + .859931 + .373462
19. 2.93474 + 9.72155
Directions: Find the value of \begin{align*}x\end{align*} in the following equations.
20. \begin{align*}.8603 + .292 = x + .8603\end{align*}
21. \begin{align*}(2.65 + x) + 19.35 = 22 + 2.115\end{align*}
22. \begin{align*}.306 + 1.076 = (.782 + x) + .294\end{align*}
23. \begin{align*}6.174 + 76.41 = 76.41 + x\end{align*}
Directions: Solve each problem.
24. Lamont measures the amount of water he drinks. His results for four consecutive days are as follows: 0.6 liters, 0.72 liters, 0.84 liters, 0.96 liters. If the pattern continues, how much water will Lamont drink on the fifth day?
25. Barbara is starting a jewelry making enterprise. At the supply store, she spent $19.19 on beads and $6.81 on wire and clasps. If she left the store with $24.50, how much did she start with?
26. A playground has a length of 18.36 yards and a width of 12.24 yards. What is the perimeter of the playground? (Remember: \begin{align*}P = l + l + w + w\end{align*}.)