Question 119

A bus left 60 minutes later than the scheduled time but in order to reach its destination 48 km away in time, it had to increase the speed by 4 km/hr from the usual speed. What is usual speed (in km/hr) of the bus?

Solution

Given $$D = 48$$. Then $$S \times T = 48$$ ....(1)

$$(T - 1) (S + 4) = 48$$

$$TS + 4T - S - 4 = 48$$

Substitute (1) in the above equation

$$48 + 4(\frac{48}{S}) - S - 4 = 48$$

$$S^{2} + 4S - 192 = 0$$

$$S = 12, -16$$(which cannot be a solution)

Hence, option B is the correct answer.

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