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…

Correct Option (4)

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

• Let $s_{1}$s1​ be the length of the route to the village and $v_{1}$v1​ be the speed. The time taken to reach the village is $T_{1} = \frac{s_{1}}{v_{1}}$T1​=v1​s1​​.
• The return route length is 1.25 times the first route, i.e., $\frac{5}{4}s_{1}$45​s1​. The return speed is twice as fast, i.e., $2v_{1}$2v1​.
• The time taken to return home, $T_{2}$T2​, is given by $T_{2} = \frac{\frac{5}{4}s_{1}}{2v_{1}} = \frac{5}{8} \left(\frac{s_{1}}{v_{1}}\right) = \frac{5}{8}T_{1}$T2​=2v1​45​s1​​=85​(v1​s1​​)=85​T1​.
• 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: $$T_{1} + 25 \text{ minutes} + T_{2} = 90 \text{ minutes}$$T1​+25 minutes+T2​=90 minutes
• Substituting $T_{2} = \frac{5}{8}T_{1}$T2​=85​T1​ into the equation: $$T_{1} + 25 + \frac{5}{8}T_{1} = 90$$T1​+25+85​T1​=90 $$\left(1 + \frac{5}{8}\right)T_{1} = 90 - 25$$(1+85​)T1​=90−25 $$\frac{13}{8}T_{1} = 65$$813​T1​=65 $$T_{1} = 65 \times \frac{8}{13} = 5 \times 8 = 40 \text{ minutes}$$T1​=65×138​=5×8=40 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.