A student decreases his speed to 75% of his original speed (a 25% reduction)…

Concept

For a fixed distance, time taken is inversely proportional to speed: if speed becomes a fraction k of the original speed, the time taken becomes 1/k of the original time (distance = speed x time stays constant).

Application

1. Let the usual speed be S and the usual time be T minutes, so the distance covered is D = S x T.

2. A 25% decrease in speed means the new speed is 75% of S, i.e. new speed = (3/4)S.

3. Since distance stays the same, new time = D / new speed = (S x T) / ((3/4)S) = (4/3)T.

4. Extra time taken = new time - usual time = (4/3)T - T = (1/3)T.

5. This extra time equals the given delay: (1/3)T = 8 minutes.

6. Solving, T = 8 x 3 = 24 minutes.

Cross-check

If the usual time is 24 minutes, the new (slower) time is (4/3) x 24 = 32 minutes, and 32 - 24 = 8 minutes - exactly the delay given in the question, confirming the usual time is 24 minutes.