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.