A watch goes fast by 15 minutes compared to the right time everyday. If it is corrected and set to the standard time at 12O' clock at noon, which of the following will be the time shown by it at 4 : 00 a.m. in the morning?

### Concept & Proportionality The watch gains time uniformly over a 24-hour cycle. We need to calculate the proportional gain for the exact number of hours elapsed between the starting standard time and the target actual time. ### Step-by-Step Solution 1. **Determine the total elapsed hours:** From 12:00 noon to 12:00 midnight is 12 hours. From 12:00 midnight to 4:00 a.m. is 4 hours. Total time elapsed = $12 + 4 = 16 \text{ hours}$. 2. **Calculate the proportional time gained:** The watch gains 15 minutes in 24 hours. Gain in 1 hour = $\frac{15}{24} \text{ minutes} = \frac{5}{8} \text{ minutes}$. Gain in 16 hours = $16 \times \left(\frac{5}{8}\right) \text{ minutes}$. $$16 \times \frac{5}{8} = 2 \times 5 = 10 \text{ minutes}.$$ 3. **Determine the shown time:** The actual time is 4:00 a.m. Because the watch is fast, it will show a time ahead of the actual time. Time shown = 4:00 a.m. + 10 minutes = 4:10 a.m. ### Exam Strategy & Shortcut Set up a quick ratio: $\frac{\text{Gain}}{\text{Actual Hours}}$. $\frac{15}{24} = \frac{x}{16}$. $x = \frac{15 \times 16}{24} = 10$ minutes. Add to 4:00 to get 4:10. ### Common Pitfall A standard trap is applying the full 15-minute gain regardless of the elapsed time, leading to option (c) 4:15 a.m. Always scale the gain to the specific time window. ### Final Answer Therefore, the correct answer is **4 : 10 a.m.**