Clock – Fast watch (gains 5 s in 3 min): A watch gains 5 seconds in every 3 minutes and is set right at 7:00 am. Later that afternoon, when the watch shows 4:15 pm, what is the true time?

Introduction / Context:
A fast watch advances quicker than real time. We relate indicated time to true time via a rate factor built from its gain rate.

Given Data / Assumptions:

• Gain = 5 s per 3 min.
• Set correct at 7:00 am.
• Indicated time later: 4:15 pm (same day).

Concept / Approach:
In 3 real minutes (180 s), the watch gains 5 s → indicated/true rate = 185/180 = 37/36. Thus true elapsed = indicated × 36/37.

Step-by-Step Solution:
1) Indicated elapsed from 7:00 to 4:15 = 9 h 15 m = 9.25 h.2) True elapsed = 9.25 × 36/37 = 9 hours.3) True time = 7:00 am + 9 h = 4:00 pm.

Verification / Alternative check:
Check proportional gain: Over 9 h true, watch gains 9 × (5/3) = 15 minutes, so it would show 9 h 15 m indicated, matching 4:15 pm.

Why Other Options Are Wrong:
Other minute values do not match the computed proportional relationship.

Common Pitfalls:
Subtracting a fixed “5 seconds” rather than using the multiplicative rate.

Final Answer:
4 pm