A man started from home to 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

Correct Option (4)

To determine the discrepancy of the village clock, the actual travel times must be calculated based on the home clock.

• Let $s_1$ be the length of the route to the village and $v_1$ be the speed. The time taken to reach the village is $T_1=\frac{s_1}{v_1}$.
• The return route length is 1.25 times the first route, i.e., $\frac{5}{4}{s}_1$. The return speed is twice as fast, i.e., $2v_1$.
• The time taken to return home, $T_2$, is given by $T_2=\frac{\frac{5}{4}{s}_1}{2v_1}=\frac{5}{8}\left(\frac{s_1}{v_1}\right)=\frac{5}{8}{T}_1$.
• The man started from home at 14:30 and returned home at 16:00. The total duration of the trip, as per the home clock, is 1 hour 30 minutes, which is 90 minutes.
• The total time spent can be expressed as the sum of travel time to the village, the stay duration, and travel time back home: $$T_1+25\text{ minutes}+T_2=90\text{ minutes}$$
• Substituting $T_2=\frac{5}{8}{T}_1$ into the equation: $$T_1+25+\frac{5}{8}{T}_1=90$$ $$(1+\frac{5}{8})T_1=90-25$$ $$\frac{13}{8}{T}_1=65$$ $$T_1=65\times \frac{8}{13}=5\times 8=40\text{ minutes}$$
• The actual time of arrival at the village, according to the home clock, would be 14:30 + 40 minutes = 15:10.
• The village clock indicated 15:15 upon arrival. Comparing this with the actual arrival time (15:10), the village clock is 15:15 - 15:10 = 5 minutes fast.

Incorrect Options:

Options 1, 2, and 3 are incorrect as the calculations demonstrate that the village clock is 5 minutes fast, not slow or fast by a different margin.