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

Question

1 month ago

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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

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-02-27T19:28:43+03:00

Answer:

Option 1 is valid, which entails 2 hours of walking and 12 hours of running.

Step-by-step explanation:

The equations provided are:

3w + 6r ≥ 36

3w + 6r ≤ 90

We'll assess which options comply with these 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 valid.

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 does not hold since 30 < 36)

3w + 6r ≤ 90

3(4) + 6(3) ≤ 90

12+18 ≤ 90

30 ≤ 90

Thus, Option 2 is invalid.

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 does not hold since 99 > 90)

Option 3 is invalid.

4) 12 hours walking; 10 hours running

w = 12 and r = 10

3w + 6r ≥ 36

3(12) + 6(10) ≥ 36

36+60 ≥ 36

96 ≥ 36

3w + 6r ≤ 90

3(12) + 6(10) ≤ 90

36 + 60 ≤ 90

96 ≤ 90 (this does not hold since 96 > 90)

So, Option 4 is invalid.