Question 84

A man makes four trips of equal distances. His speed on first trip was 60 km/hr and in each subsequent trip his speed was half of the previous trip. What is the average speed (in km/hr) of the man in these four trips?


Let distance travelled in each trip = $$d$$ km

=> Total distance travelled = $$4d$$ km

Time taken in each trip = distance/speed

= $$\frac{d}{60}+\frac{d}{30}+\frac{d}{15}+\frac{d}{7.5}$$

= $$\frac{d}{7.5}(\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+1)$$

= $$\frac{d}{7.5}(\frac{1+2+4+8}{8})$$

= $$\frac{d}{7.5}\times\frac{15}{8}=\frac{d}{4}$$

$$\therefore$$ Average speed = total distance / total time

= $$4d\div(\frac{d}{4})$$

= $$4d\times \frac{4}{d}=16$$

=> Ans - (A)

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