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

Introduction / Context:This is the same structure as other fast-watch problems: convert indicated time to true time using the multiplicative rate factor.

Given Data / Assumptions:

• Gain: 5 s per 3 min → rate = 37/36.
• Set right at 7:00 am; indicated later is 4:15 pm.

Concept / Approach:True elapsed = indicated elapsed × 36/37.

Step-by-Step Solution:Indicated elapsed = 9 h 15 m = 9.25 h.True elapsed = 9.25 × 36/37 = 9 h.True time = 4:00 pm.

Verification / Alternative check:Over 9 true hours the watch gains 15 minutes (5/3 min per hour × 9 h) → shows 9 h 15 m indicated = 4:15 pm.

Why Other Options Are Wrong:All other minute values mismatch the proportionality.

Common Pitfalls:Treating the gain as an additive offset instead of a rate.

Final Answer:4 pm