Two cars travel the same distance starting at 10:00 am and 11:00 am,…

Concept: When two objects cover the same distance, speed and time are inversely related — Speed = Distance ÷ Time, so Speed₂ ÷ Speed₁ = Time₁ ÷ Time₂. If one travel time has a fixed lower bound and the two times always differ by a constant gap, the ratio of times (and hence the percentage difference in speed) is largest when both times are pushed down to their smallest permitted values — a fixed gap changes a small number by a bigger fraction than it changes a large number.

1. Let Car 1's travel time be n hours (n ≥ 6, since it starts at 10:00 am and travels at least 6 hours). Car 2 starts an hour later, at 11:00 am, and both reach the destination at the same clock time, so Car 2's travel time is n − 1 hours (≥ 5).

2. Since both cars cover the same distance, Speed of Car 2 ÷ Speed of Car 1 = n ÷ (n − 1). This ratio decreases as n grows (for example 6/5 = 1.2 but 7/6 ≈ 1.167), so it is maximum at the smallest allowed n.

3. The smallest allowed value is n = 6 hours for Car 1, which makes Car 2's time 5 hours.

4. Take the common distance as the LCM of 6 and 5, i.e. 30 km, so both speeds come out as whole numbers.

5. Car 1's speed = 30 ÷ 6 = 5 km/h. Car 2's speed = 30 ÷ 5 = 6 km/h.

6. Percentage by which Car 2's speed exceeds Car 1's speed = [(6 − 5) ÷ 5] × 100 = (1/5) × 100 = 20%.

Cross-check: at the next possible pair of times, n = 7 and n − 1 = 6, the speeds would be 30/7 ≈ 4.29 km/h and 30/6 = 5 km/h — an increase of only about 16.7%, which is indeed lower than 20%. This confirms the maximum increase occurs at the smallest permitted travel times.

So the highest possible percentage by which the second car's speed can exceed the first car's speed is 20%.