A watch gains uniformly. It is 2 minutes slow at noon on Monday and 4 minutes 48 seconds fast at 2 p.m. the following Monday. When was it correct?

Problem restatement
The watch's error changes uniformly from −2 min at Monday 12:00 to +4 min 48 s at the next Monday 14:00. Find the instant when the error was zero (i.e., watch showed correct time).

Given data

• Start error (E₀) = −2 min.
• End error (E₁) = +4 min 48 s = +4.8 min.
• Elapsed real time between readings: 7 days + 2 hours = 170 hours.

Concept/Approach
With uniform gain, error increases linearly. The fraction of the interval to reach zero error is |E₀| ÷ (E₁ − E₀).

Step-by-Step calculation
Total error change = 4.8 − (−2) = 6.8 minutesFraction to get to zero from start = 2 ÷ 6.8 = 5 ÷ 17Time from Monday 12:00 to correction = 170 h × (5 ÷ 17) = 50 hours50 hours after Monday 12:00 = Wednesday 14:00

Verification/Alternative
Remaining time to the second reading: 170 − 50 = 120 h. Error accrued during first 50 h: 2 min (to reach zero). During next 120 h: gain = (6.8 − 2) = 4.8 min, matching the final fastness.

Common pitfalls
Converting 48 s incorrectly (must be 0.8 minute) or forgetting the 2 extra hours in the 7-day span.

Final Answer
Wednesday 2 p.m.