Question

Keitaro walks at a pace of 3 miles per hour and runs at a pace of 6 miles per hour. Each month, he wants to complete at least 36 miles but not more than 90 miles. The system of inequalities represents the number of hours he can walk, w, and the number of hours he can run, r, to reach his goal. 3w + 6r ≥ 36 3w + 6r ≤ 90 Which combination of hours can Keitaro walk and run in a month to reach his goal? 2 hours walking; 12 hours running 4 hours walking; 3 hours running 9 hours walking; 12 hours running 12 hours walking; 10 hours running

Answer 1

Answer:

Option 1 is correct. i.e  2 hours walking; 12 hours running

Step-by-step explanation:

We are given the equations

3w + 6r ≥ 36

3w + 6r ≤ 90

We will check which of the option satisfy the above equations.

1)  2 hours walking; 12 hours running

w = 2 and r =12

3w + 6r ≥ 36

3(2) + 6(12) ≥ 36

6+72 ≥ 36

78 ≥ 36

3w + 6r ≤ 90

3(2) + 6(12) ≤ 90

6+72 ≤ 90

78 ≤ 90

Both equations are satisfied. Option 1 is correct.

2) 4 hours walking; 3 hours running

w = 4 and r =3

3w + 6r ≥ 36

3(4) + 6(3) ≥ 36

12+18 ≥ 36

30 ≥ 36 (this equation doesn't hold as 30 < 36 and not < or equal to 36)

3w + 6r ≤ 90

3(4) + 6(3) ≤ 90

12+18 ≤ 90

30 ≤ 90

So, Option 2 is incorrect.

3) 9 hours running 12 hours walking

w = 9 and r =12

3w + 6r ≥ 36

3(9) + 6(12) ≥ 36

27+72 ≥ 36

99 ≥ 36

3w + 6r ≤ 90

3(9) + 6(12) ≤ 90

27+72 ≤ 90

99 ≤ 90 (this equation doesn't hold because 99 is greater than 90 and not less than 90)

Option 3 is incorrect.

4) 12 hours walking; 10 hours running

w = 12 and r =120

3w + 6r ≥ 36

3(12) + 6(10) ≥ 36

36+60 ≥ 36

96 ≥ 36

3w + 6r ≤ 90

3(12) + 6(10) ≤ 90

36+60 ≤ 90

99 ≤ 90 (this equation doesn't hold because 96 is greater than 90 and not less than 90)

Option 4 is incorrect.