We included **H****MH 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.

**Spark Your Learning**

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.

Answer:

\(\frac{3}{8}\) + \(\frac{2}{5}\) + \(\frac{1}{10}\)

\(\frac{3}{8}\) + \(\frac{4}{10}\) + \(\frac{1}{10}\)

\(\frac{3}{8}\) + \(\frac{5}{10}\)

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.

Answer:

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.

Answer:

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 ?

Answer:

\(\frac{7}{8}\) = 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?

Answer:

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. ________

Answer:

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.

\(\frac{5}{8}\) – \(\frac{3}{5}\)

Answer:

\(\frac{5}{8}\) – \(\frac{3}{5}\)

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.

\(\frac{25}{40}\) – \(\frac{24}{40}\) = \(\frac{1}{40}\)

**On Your Own**

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?

Answer:

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\(\frac{5}{12}\) – \(\frac{2}{3}\) = ____

Answer:

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.

\(\frac{7}{10}\) + \(\frac{7}{8}\) = ____

Answer:

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?

Answer:

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.

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?