All the solutions provided in **McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 9 Reasonable Estimate **are as per the latest syllabus guidelines.

## McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 9 Reasonable Estimate

**Solve**

**Estimate the sum or difference. Use front-end estimation.**

Question 1.

32\(\frac{2}{9}\) + 12\(\frac{1}{7}\) _______________

Answer:

30 + 10 = 40

Question 2.

45\(\frac{1}{2}\) – 33\(\frac{1}{5}\) _______________

Answer:

50 – 30 = 20.

Explanation:

The difference between \(\frac{91}{2}\) – \(\frac{166}{5}\). The front end estimated is 50-30 which is 20.

Question 3.

77\(\frac{5}{9}\) – 21\(\frac{2}{5}\) _______________

Answer:

80 – 20 = 50.

Explanation:

The difference between \(\frac{698}{9}\) – \(\frac{107}{5}\). The front end estimated is 80-20 which is 50.

Question 4.

34\(\frac{7}{8}\) + 22\(\frac{2}{9}\) _______________

Answer:

30 + 20 = 50.

Explanation:

The addition of \(\frac{279}{8}\) + \(\frac{200}{9}\). The front end estimated is 30+20 which is 50.

Question 5.

Answer:

1 + 1 = 2.

Explanation:

The addition of \(\frac{5}{6}\) + \(\frac{6}{7}\). The front end estimated is 1+1 which is 2.

Question 6.

Answer:

1 – 0 = 1.

Explanation:

The difference between \(\frac{5}{8}\) – \(\frac{3}{7}\). The front end estimated is 1-0 which is 1.

Question 7.

Answer:

1 – 0 = 1.

Explanation:

The difference between \(\frac{2}{3}\) – \(\frac{1}{4}\). The front end estimated is 1-0 which is 1.

Question 8.

Answer:

0 + 1 = 1.

Explanation:

The addition of \(\frac{1}{9}\) + \(\frac{7}{11}\). The front end estimated is 0+1 which is 1.