A clock is set right at 5 AM. The clock loses 16 minutes in 24 hours.What will be the true time when the clock indicates 10 p.m. on 4th day?

A clock is set right at 5 AM. It loses 16 minutes in 24 hours.

In 24 hours of correct time, the faulty clock runs only:
$24 - \frac{16}{60}$ hours $= 24 - \frac{4}{15} = \frac{360-4}{15} = \frac{356}{15}$ hours

So, speed ratio (Faulty : Correct) is:
$\dfrac{\text{Faulty time}}{\text{Correct time}} = \dfrac{24 - \frac{16}{60}}{24} = \dfrac{356/15}{24} = \dfrac{356}{360} = \dfrac{89}{90}$

That is, in the time when correct clock shows 90 hours, faulty clock shows 89 hours.
So,
$89$ hours of faulty clock $= 90$ hours of correct clock.

Now, find how many hours the faulty clock indicates from 5 AM (Day 1) to 10 PM (Day 4):
From 5 AM Day 1 to 5 AM Day 4 $= 3$ days $= 72$ hours
From 5 AM Day 4 to 10 PM Day 4 $= 17$ hours
So total indicated (faulty) time:
$72 + 17 = 89$ hours

Since $89$ faulty hours correspond to $90$ correct hours, the true elapsed time is:
$90$ hours $= 3$ days $+ 18$ hours

Start time was 5 AM (Day 1). After 3 days, it will be 5 AM (Day 4).
Add 18 hours to 5 AM (Day 4):
$5\ \text{AM} + 18\ \text{hours} = 11\ \text{PM}$

Therefore, when the faulty clock indicates 10 PM on the 4th day,
the true time is $11\ \text{PM}$.