Walking at 5/7 of his usual speed, a boy reaches school 6 minutes late. What is his usual time to reach the school (in minutes)?

Introduction / Context:For a fixed distance, time is inversely proportional to speed. If speed is scaled by a factor, time is scaled by the reciprocal. Use the late time (excess) to back out the usual time.

Given Data / Assumptions:

• New speed = (5/7) * usual speed.
• Delay = 6 minutes.

Concept / Approach:Let T be usual time. New time = T / (5/7) = (7/5)T. Lateness = (7/5)T − T = (2/5)T = 6 ⇒ T = 15 min.

Step-by-Step Solution:

Verification / Alternative check:New time = (7/5)*15 = 21 min, which is 6 min more than the usual 15 min.

Why Other Options Are Wrong:10/12/18 min do not satisfy the proportionality equation.

Common Pitfalls:Adding/subtracting speeds instead of using inverse proportionality for time.

Final Answer:15 min