A man started from home at 14:30 hours and drove to village, arriving there when the village clock indicated 15:15 hours. After staying for 25 minutes, he drove back by a different route of length 1·25 times the first route at a rate twice as fast reaching home at 16:00 hours. As compared to the clock at home, the village clock is

(a) 10 minutes slow
(b) 5 minutes slow
(c) 10 minutes fast
(d) 5 minutes fast ✓ UPSC's answer

Why the answer is (d)

• Let the time taken for the first leg (Home to Village) be $t$ hours. The distance is $d$.

• The return leg distance is $1.25d$ and the speed is $2v$ (where $v = d/t$).

• Time for return leg = Distance / Speed = $1.25d / (2d/t) = 0.625t$.

• Total time elapsed from home clock: Departure 14:30 to Return 16:00 is 1 hour 30 minutes (90 minutes).

• This 90 minutes comprises: Time to Village ($t$) + Stay (25 min) + Time back ($0.625t$).

• Equation: $t + 25 + 0.625t = 90 \Rightarrow 1.625t = 65 \Rightarrow t = 40$ minutes.

• The man arrived at the village 40 minutes after 14:30, so the home clock time was 15:10.

• The village clock showed 15:15, which is 5 minutes ahead of the home clock time (15:10).

Why the other options are wrong

(a) 10 minutes slow: The calculation shows the village clock is 5 minutes fast, not 10 minutes slow.
(b) 5 minutes slow: The village clock is 5 minutes fast, not slow, as it reads 15:15 when the actual time is 15:10.
(c) 10 minutes fast: The difference is 5 minutes, not 10 minutes, based on the computed travel time of 40 minutes.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.