# Into Math Grade 5 Module 7 Lesson 1 Answer Key Use Benchmarks and Number Sense to Estimate

We included HMH Into Math Grade 5 Answer Key PDF Module 7 Lesson 1 Use Benchmarks and Number Sense to Estimate to make students experts in learning maths.

## HMH Into Math Grade 5 Module 7 Lesson 1 Answer Key Use Benchmarks and Number Sense to Estimate

I Can use benchmarks to estimate a sum or difference of fractions with unlike denominators.

Ms. Fong mixes water, glue, and laundry detergent together to make slime. The amount of each ingredient is a fraction of a liter.

Use a visual model to estimate the total number of liters of ingredients she mixes together.

$$\frac{3}{8}$$ + $$\frac{2}{5}$$ + $$\frac{1}{10}$$
$$\frac{3}{8}$$ + $$\frac{4}{10}$$ + $$\frac{1}{10}$$
$$\frac{3}{8}$$ + $$\frac{5}{10}$$
The fractions have unlike denominators. First, find the Least Common Denominator and rewrite the fractions with the common denominator.
LCD of 8 and 10 is 40.
$$\frac{3}{8}$$ × $$\frac{5}{5}$$ + $$\frac{5}{10}$$ × $$\frac{4}{4}$$
$$\frac{15}{40}$$ + $$\frac{20}{40}$$ = $$\frac{35}{40}$$ = $$\frac{7}{8}$$

Turn and Talk Describe whether each addend is closest to 0, $$\frac{1}{2}$$, or 1.
The solution is close to 1. Because 7/8 is approximately equal to 1.

Build Understanding

1. Ms. Fong has some bottles of copper powder for science experiments. She uses $$\frac{7}{8}$$ bottle, $$\frac{2}{3}$$ bottle, and $$\frac{5}{12}$$ bottle. About how many bottles of copper powder does she use?

A. What expression represents the situation?
_______________
B. How can you use the three number lines and the benchmarks 0, $$\frac{1}{2}$$ and 1 to help you estimate the answer?

C. What expression represents the estimate?
Ms. Fong uses about ___ bottles of copper powder.

7/8 + 2/3 + 5/12 = 1.8
Ms. Fong uses about 2 bottles of copper powder.

Turn and Talk Without using a number line, how could you determine whether $$\frac{7}{8}$$ and $$\frac{5}{12}$$ are each closest to 0, $$\frac{1}{2}$$, or 1 ?
$$\frac{7}{8}$$ = 0.8
$$\frac{5}{12}$$ = 0.4
0.8 + 0.4 = 1.2
So, it is closest to 1.

Step It Out

2. Each group in science class has a 2-liter container of distilled water. Group A uses 1$$\frac{9}{10}$$ L and Group B uses 1$$\frac{3}{8}$$ L. About how much more distilled water does Group A use than Group B?
A. Represent the situation with an expression. ________
B. Determine an estimate without finding an exact answer.

• Between which two whole numbers does 1$$\frac{9}{10}$$ lie? ____
• Is the fractional part of the mixed number closest to 0, $$\frac{1}{2}$$, or 1 ? Explain.
• What whole number can you use to estimate 1$$\frac{9}{10}$$?
• Use benchmark values to estimate 1$$\frac{3}{8}$$. ____
• Write an equation to estimate the difference.

C. About how much more distilled water does Group A use than Group B?
Each group in science class has a 2-liter container of distilled water.
Group A uses 1$$\frac{9}{10}$$ L and Group B uses 1$$\frac{3}{8}$$ L.
1$$\frac{9}{10}$$ – 1$$\frac{3}{8}$$
1$$\frac{9}{10}$$ lies between 1 and 2.
The estimated whole number of 1$$\frac{9}{10}$$ is 2.
The benchmark values to estimate 1$$\frac{3}{8}$$ is 1.5
1$$\frac{9}{10}$$ – 1$$\frac{3}{8}$$
1 + $$\frac{9}{10}$$ – 1 – $$\frac{3}{8}$$
$$\frac{9}{10}$$ – $$\frac{3}{8}$$
LCD is 10 and 8 is 40.
$$\frac{36}{40}$$ – $$\frac{15}{40}$$ = $$\frac{21}{40}$$
Thus Group A use $$\frac{21}{40}$$ than Group B.

Check Understanding Math Board

Question 1.
A weather station reports that $$\frac{11}{12}$$ foot of snow fell yesterday afternoon and $$\frac{1}{3}$$ foot fell yesterday evening. Estimate the amount of snow that fell yesterday. ________
Given,
A weather station reports that $$\frac{11}{12}$$ foot of snow fell yesterday afternoon and $$\frac{1}{3}$$ foot fell yesterday evening.
$$\frac{11}{12}$$ + $$\frac{1}{3}$$
LCD is 12 and 3 is 12.
$$\frac{11}{12}$$ + $$\frac{4}{12}$$ = $$\frac{15}{12}$$ = $$\frac{5}{4}$$ = 1$$\frac{1}{4}$$
The amount of snow that fell yesterday is 1$$\frac{1}{4}$$ foot.

Use benchmark values to write an expression to represent an estimate.

Question 2.
$$\frac{9}{10}$$ + $$\frac{1}{12}$$
Answer: $$\frac{59}{60}$$
The denominator of both the fractions are not same.
So we have to find the LCD of the fractions and then add the fractions.
$$\frac{9}{10}$$ + $$\frac{1}{12}$$
LCD is 10 and 12 is 60.
$$\frac{54}{60}$$ + $$\frac{5}{60}$$ = $$\frac{59}{60}$$

Question 3.
$$\frac{5}{8}$$ – $$\frac{3}{5}$$
$$\frac{5}{8}$$ – $$\frac{3}{5}$$
The denominator of both the fractions are not same.
So we have to find the LCD of the fractions and then subtract the fractions.
LCD of 8 and 5 is 40.
$$\frac{25}{40}$$ – $$\frac{24}{40}$$ = $$\frac{1}{40}$$

Question 4.
History In the 1800s, wagon trains traveled west along the Oregon Trail. A wagon train traveled from Missouri to Wyoming in 1$$\frac{2}{3}$$ months, and from Wyoming to Utah in $$\frac{3}{5}$$ month. About how many months did it take the wagon train to travel from Missouri to Utah?
Given,
A wagon train traveled from Missouri to Wyoming in 1$$\frac{2}{3}$$ months, and from Wyoming to Utah in $$\frac{3}{5}$$ month.
1$$\frac{2}{3}$$ + $$\frac{3}{5}$$
1 + $$\frac{2}{3}$$ + $$\frac{3}{5}$$
LCS of 3 and 5 is 15.
$$\frac{2}{3}$$ + $$\frac{3}{5}$$
$$\frac{10}{15}$$ + $$\frac{9}{15}$$ = $$\frac{19}{15}$$ = 1$$\frac{4}{15}$$
1 + 1$$\frac{4}{15}$$ = 2$$\frac{4}{15}$$

Use benchmarks to estimate the sum or difference.

Question 5.
1$$\frac{5}{12}$$ – $$\frac{2}{3}$$ = ____
Rewriting our equation with parts separated
1 + $$\frac{5}{12}$$ – $$\frac{2}{3}$$
The denominators of the fractions are not same so we have to find the LCD.
LCD of 12 and 3 is 12
$$\frac{5}{12}$$ – $$\frac{8}{12}$$ = –$$\frac{3}{12}$$
1 – $$\frac{3}{12}$$ = $$\frac{3}{4}$$

Question 6.
$$\frac{7}{10}$$ + $$\frac{7}{8}$$ = ____
Rewriting our equation with parts separated
$$\frac{7}{10}$$ + $$\frac{7}{8}$$
The denominators of the fractions are not the same so we have to find the LCD.
LCD of 10 and 8 is 40
$$\frac{28}{40}$$ + $$\frac{35}{40}$$ = $$\frac{63}{40}$$
$$\frac{63}{40}$$ = 1 $$\frac{23}{40}$$

Question 7.
Use Tools Use the number lines and benchmarks to estimate 1$$\frac{1}{3}$$ – $$\frac{5}{6}$$.
mix
Estimate. Then compare the distance between each number and its benchmark. What does this tell you about the actual difference?

1$$\frac{1}{3}$$ – $$\frac{5}{6}$$
1 + $$\frac{1}{3}$$ – $$\frac{5}{6}$$
The denominators of the fractions are not the same so we have to find the LCD.
LCD of 3 and 6 is 6.
$$\frac{1}{3}$$ – $$\frac{5}{6}$$
$$\frac{2}{6}$$ – $$\frac{5}{6}$$ = –$$\frac{3}{6}$$ = –$$\frac{1}{2}$$
1 – $$\frac{1}{2}$$ = $$\frac{1}{2}$$

Question 8.
Open Ended Use mixed numbers or fractions with unlike denominators to write an addition expression and a subtraction expression, each with an estimate of 1$$\frac{1}{2}$$. Justify your answer.
2 $$\frac{1}{3}$$ – $$\frac{5}{6}$$
Rewriting our equation with parts separated
2 + $$\frac{1}{3}$$ – $$\frac{5}{6}$$
The denominators of the fractions are not the same so we have to find the LCD.
LCD of 3 and 6 is 6.
$$\frac{2}{6}$$ – $$\frac{5}{6}$$ = –$$\frac{3}{6}$$ = –$$\frac{1}{2}$$
2 – $$\frac{1}{2}$$ = 1$$\frac{1}{2}$$

I’m in a Learning Mindset!

Are the visual models effective in solving the problem?

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