Uniformly gaining watch corrected time: A watch is 2 minutes slow at noon on Sunday and 4 minutes 48 seconds fast at 2 pm the following Sunday (uniform gain). When was it correct?

Introduction / Context:
A uniform-gain (or uniform-loss) watch changes its error at a constant rate. If we know the error at two instants and the elapsed real time between them, we can compute the rate of change of error and then find when the error would be exactly zero (i.e., watch shows correct time).

Given Data / Assumptions:

• Error at start (Sun 12:00): −2 min.
• Error at end (next Sun 2:00 pm): +4 min 48 s = +4.8 min.
• Elapsed real time: 7 days + 2 hours = 170 hours.

Concept / Approach:
Change in error = (+4.8) − (−2) = +6.8 min over 170 h ⇒ rate = 6.8/170 = 0.04 min/h. To go from −2 min to 0, the watch must gain 2 minutes at this rate.

Step-by-Step Solution:

Verification / Alternative check:
From Tue 2 pm to next Sun 2 pm is 120 h; error accrued = 120 × 0.04 = 4.8 min (fits final reading).

Why Other Options Are Wrong:
Other listed times do not lie exactly 50 h from the initial reference.

Common Pitfalls:
Converting 48 seconds as 0.48 min instead of 0.8 min; forgetting the sign of the initial error (slow).

Final Answer:
2 pm on Tuesday