Discussion
Walking 7/6 of his usual rate, a boy reaches his school 4 min early. Find his usual time to reach the school?
Answer: D
Speed and time are inversely proportional over a fixed distance, so walking at \( \tfrac{7}{6} \) of his rate takes \( \tfrac{6}{7} \) of the time.
\[ t - \frac{6}{7}t = \frac{1}{7}t = 4 \text{ min} \]
\[ t = 28 \text{ minutes} \]
Checking: the faster walk takes 24 minutes, exactly 4 less.
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