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Question:

Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is

Explanation:

The speeds of both the cars in the three legs of the journey are equal. Therefore the second car will reach point B as many minutes after the first car as it started after the first car.

For the first 50 km of the journey, the speed of the cars is 100 kmph. Therefore, the first car reaches a distance of 20 km

from point A after = $\frac{20}{100} = 0.2$ hours

= 0.2 × 60 = 12 minutes.

When the first car reaches point B, the second car is 12 minutes away from point B. The speed of the car in the last 50 km distance is 25 kmph. Therefore the distance of the second car from point

$B = 25 \times \frac{12}{60} = 5 k m$

Hence, 5.

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