NIMCET Clock Previous Year Questions (PYQs) – Page 1 of 2

Watch gains 10s in 5 min
$\text{Gain rate} = \dfrac{10}{5} = 2 \text{ s per minute}$
$\text{Every } 60 \text{ s real time} \Rightarrow \text{watch shows } 62 \text{ s}$

Watch time elapsed:
$9:00 \text{ am to } 7:20 \text{ pm} = 10 \text{ hr } 20 \text{ min} = 620 \text{ min}$

Real time formula:
$\text{Real time} = \text{Watch time} \times \dfrac{60}{62}$

$= 620 \times \dfrac{60}{62}$

$= \dfrac{37200}{62}$

$= 600 \text{ minutes}$

$= 10 \text{ hours}$

Correct Time:
$9:00 \text{ am} + 10 \text{ hours} = \boxed{7:00 \text{ pm}}$

In 12 hours, the hour and minute hands overlap 11 times. 

Garvita’s time is correct, so current time = 9:35 AM.
Manya’s watch is 5 minutes fast, so 9:25 AM on Manya’s watch means actual time = 9:20 AM.
Buses come every 20 minutes, so next bus after 9:20 AM is at 9:40 AM (actual time).
Waiting time = 9:40 AM − 9:35 AM = 5 minutes.

Answer: 5 minutes

Clock reading from $8$ a.m. to next day $1$ p.m.: $ 29\ \text{hours (clock time)} $ Clock gains $10$ minutes in $24$ hours ⇒ it runs $\dfrac{145}{144}$ times faster. Real time: $ 29 \times \dfrac{144}{145} = 28.8\ \text{hours} $ $ 0.8\text{ hrs} = 48\text{ min} $ Add $28$ hours $48$ minutes to $8$ a.m.: $ 8\text{ a.m.} + 28\text{ hrs} = 12\text{ noon next day} $ Then $+48$ minutes = $12:48$ p.m.

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 $11\ \text{PM}$.

$\theta=30H-5.5M$