Slow clock — loses 16 minutes/day: A clock is set right at 6:00 a.m. It loses 16 minutes in every 24 hours. If it shows 11:00 p.m. on the 4th day, what is the true time?

Introduction / Context:
We convert a shown time from a slow clock into true time using a rate ratio.

Given Data / Assumptions:

• Clock correct at 6:00 a.m. (Day 1).
• Loss = 16 minutes per 24 true hours.
• Shown time: 11:00 p.m. on the 4th day.

Concept / Approach:
In 24 true hours, the clock shows 24 − 16/60 = 23 44/60 = 23 11/15 hours. Thus indicated/true = (23 11/15)/24 = (356/15)/24 = 89/90. Hence true elapsed = indicated × (90/89).

Step-by-Step Solution:
1) Indicated elapsed from Day 1, 6:00 a.m. to Day 4, 11:00 p.m.: Day1 6→Day4 6 = 72 h; plus 6→11 p.m. = 17 h; total indicated = 89 h.2) True elapsed = 89 × (90/89) = 90 h.3) True time = Day 1, 6:00 a.m. + 90 h = Day 5, 12:00 a.m. (i.e., midnight as Day 4 ends).

Verification / Alternative check:
Loses 16 min/day ⇒ in 3 days 18 h (90 h) it would have lost 60 minutes, so it would read one hour behind: 11 p.m. shown while true is 12 o’clock.

Why Other Options Are Wrong:
11 p.m. ignores the loss; 2 p.m./9 p.m. are unrelated.

Common Pitfalls:
Using 16/24 hours directly; forgetting to compute indicated elapsed first.

Final Answer:
12 o' clock