Fast clock — gains 10 minutes/day: A clock is correct at 9:00 p.m. It gains 10 minutes in 24 hours. What is the true time when this fast clock indicates 2:00 p.m. on the following day?

Introduction / Context:Here the clock runs fast, so its indicated interval exceeds the true interval. We must convert the shown time back to true time.

Given Data / Assumptions:

• Clock correct at 9:00 p.m. (start).
• Gain = 10 minutes per 24 true hours.
• Shown (indicated) time later: 2:00 p.m. next day.

Concept / Approach:In 24 true hours, indicated = 24 h + 10 min = 24 + 1/6 = 145/6 h. Thus indicated/true = (145/6)/24 = 145/144. Therefore true elapsed = indicated × (144/145).

Step-by-Step Solution:1) Indicated elapsed from 9:00 p.m. to next-day 2:00 p.m. = 17 h.2) True elapsed = 17 × (144/145) = 2448/145 h = 16 h + (128/145) h.3) Convert fractional hour: (128/145) × 60 ≈ 52.97 min ≈ 53 min.4) True time ≈ 9:00 p.m. + 16 h 53 m = next day about 1:53 p.m.

Verification / Alternative check:Check proportion: a rate of +10 min/day ≈ +0.694% fast. Over ~17 h, that is ~7.07 minutes of fastness; 2:00 p.m. minus ~7 min ≈ 1:53 p.m., consistent.

Why Other Options Are Wrong:“48 past …” choices assume ~+48 min/day or misuse the ratio; 2:48 or 1:48 do not match the precise ratio 145/144.

Common Pitfalls:Subtracting 10 minutes instead of scaling by the correct factor; mixing “true” and “indicated” intervals.

Final Answer:About 1 : 53 pm