Question
A cyclist travels from town P to town Q at 12 km/h and
returns from Q to P by the same route at a speed which is 25% higher. Due to the increase in speed on the return journey, he takes 1 hour less for the entire trip than he would have taken if he had travelled at 12 km/h both ways. What is the distance between P and Q?Solution
ATQ, Let distance = d km. Time if speed is 12 km/h both ways = 2d/12 = d/6 hours. Actual speeds: P→Q: 12 km/h ⇒ time = d/12 Q→P: 25% more ⇒ 12 × 1.25 = 15 km/h ⇒ time = d/15 Actual total time = d/12 + d/15 = (5d + 4d)/60 = 9d/60 = 3d/20 Given: 3d/20 = d/6 − 1 Multiply by 60: 9d = 10d − 60 ⇒ d = 60 km.
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