Discussion
A boy increases his speed to 9/5 times of his original speed. By this he reaches his school 30 minutes before the usual time. How much time does he takes usually?
Answer: D
Speed and time are inversely proportional over a fixed distance, so multiplying the speed by \( \tfrac{9}{5} \) multiplies the time by \( \tfrac{5}{9} \).
\[ \text{time saved} = t - \frac{5}{9}t = \frac{4}{9}t \]
\[ \frac{4}{9}t = 30 \quad\Longrightarrow\quad t = \frac{270}{4} = 67.5 \text{ min} \]
Checking: at the higher speed the trip takes 37.5 minutes, exactly 30 less.
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