Clock – Uniform gain/lose period: A watch is 5 minutes slow at 7:00 am on Sunday and 5 minutes 48 seconds fast at 7:00 pm the following Sunday. On which day was it exactly correct?

Introduction / Context:
When a watch gains uniformly, knowing two offsets at two times lets us find the instant when the error was zero. Here the watch goes from being 5 minutes slow to 5 minutes 48 seconds fast over one week plus 12 hours.

Given Data / Assumptions:

• T1: Sunday 7:00 am, error = −5:00 (slow).
• T2: Next Sunday 7:00 pm, error = +5:48 (fast).
• Uniform gain across the interval.

Concept / Approach:
Total gain = 5:48 − (−5:00) = 10:48 = 648 s. Total real time = 7 days 12 hours = 180 hours = 648,000 s. Rate of gain = 648 / 648,000 = 0.001 s/s = 3.6 s per hour. Time from T1 to zero error = 300 seconds / 3.6 s per hour = 83 1/3 hours.

Step-by-Step Solution:
1) Compute gain rate: 10 m 48 s over 180 h → 3.6 s/h.2) Time to correct from −5 m (−300 s): 300 / 3.6 = 83 1/3 h = 83 h 20 m.3) From Sunday 7:00 am + 83 h 20 m → Wednesday 6:20 pm.

Verification / Alternative check:
Check mid-interval symmetry: the correction time should lie roughly midweek; Wednesday evening fits.

Why Other Options Are Wrong:
They correspond to different weekdays; only Wednesday aligns with 83 h 20 m after the initial Sunday morning.

Common Pitfalls:
Confusing “fast” with “slow” sign conventions; mixing minutes and seconds during rate computation.

Final Answer:
Wednesday