Comparing journey times to determine clock speed differences
Question
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
Options
10 minutes slow
5 minutes slow
10 minutes fast
5 minutes fast
Explanation
Establish the true time boundaries using the reliable home clock: Departure = 14:30. Return = 16:00. Total time away from home = 90 minutes.
Subtract the resting time to find the total driving time: Total driving time = 90 - 25 = 65 minutes.
Let the forward journey distance be D and speed be v. Forward time t_1 = D/v. Return journey distance is 1.25D and speed is 2v. Return time t_2 = 1.25D/2v = 0.625 \left(D/v\right) = 0.625 t_1.
Set up the total driving time equation: t_1 + 0.625 t_1 = 65 1.625 t_1 = 65 \implies t_1 = 65/1.625 = 40 minutes.
Calculate the true arrival time at the village: True arrival time = Departure (14:30) + 40 mins = 15:10. The village clock displayed 15:15 upon arrival. Therefore, the village clock is 5 minutes fast.
Answer: (d).
Question details
Year
2022
Paper
CSAT
Question
Q46
Section
Numerical Ability
Sub-topic
Time, Speed & Distance
Type
Factual single
Difficulty
Hard
Source hint
Man driving to a village and back, clock synchronization
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