A uniformly gaining watch is 6 minutes slow at 4:00 pm on a Sunday and 10 2/3 minutes fast at 8:00 pm on the following Sunday. During this period, at what exact day and time was the watch correct?

Introduction / Context:
A linearly gaining watch transitions from “slow” to “fast,” implying it must be exactly correct at some intermediate instant. With uniform gain, the error varies linearly with time; we can use proportional reasoning.

Given Data / Assumptions:

• Error at t0 = Sunday 4:00 pm: −6 min (slow).
• Error at t1 = next Sunday 8:00 pm: +10 2/3 min.
• Uniform rate of change of error.

Concept / Approach:
Let ΔT be the total time from t0 to t1 and ΔE the total change in error. The time from t0 to the “zero-error” instant is proportionally (magnitude of initial slow)/(total change) * ΔT.

Step-by-Step Solution:
ΔT = 7 days + 4 h = 172 h.ΔE = (+10 2/3) − (−6) = 16 2/3 min = 50/3 min.Fraction to reach zero = 6 / (50/3) = 18/50 = 0.36.Time from t0 = 0.36 * 172 h = 61.92 h = 61 h 55 min 12 s.t0 + 61 h 55 m ≈ Wednesday 5:55 am.

Verification / Alternative check:
At ~61.92 h from start, accumulated gain relative to t0 equals 6 min, canceling the initial −6 min error → net 0.

Why Other Options Are Wrong:
1:36 am/pm and 2:36 pm come from incorrect ratios (often students divide 172 by 3 or 5). “None of these” is invalid because a precise value exists.

Common Pitfalls:
Using average of the two times instead of proportional distance based on errors; mixing true vs. indicated time is irrelevant here—only the error evolution matters.

Final Answer:
5 : 55 am (Wednesday)