Question

On the birthday of Sam, Param gave him an alarm clock, now Sam sets alarm at 5 am, and Sam notices that:
- The clock loses 960 seconds in last one day.
- Effectively, the clock indicates 10:00 pm on the 4th day.
Find out the original time of the clock.

• A. 9 PM
• B. 10 PM
• C. 11 PM
• D. 8 PM

Hint:

Note how the indicated time of 89 hours perfectly cancels out with the denominator of the conversion ratio ($90/89$), yielding exactly 90 hours. These numerical simplifications are typical in competitive exams.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a "faulty clock" problem where we must calculate the ratio of actual time passed to the incorrect time shown by the slow clock.
Detailed Explanation:
First, convert the time lost into minutes: \[ 960 \text{ seconds} = \frac{960}{60} = 16 \text{ minutes per day} \] Calculate the time elapsed on the incorrect clock:
- Start: 5:00 AM on Day 1.
- End: 10:00 PM on Day 4.
- Total days passed from Day 1, 5:00 AM to Day 4, 5:00 AM = 3 full days = 72 hours.
- From Day 4, 5:00 AM to Day 4, 10:00 PM = 17 hours.
- Total indicated time on the faulty clock ($T_i$) = $72 + 17 = 89$ hours.
Find the ratio of actual (correct) time ($T_c$) to incorrect time ($T_i$):
In 24 hours of correct time, the faulty clock runs for: \[ 24 \text{ hours} - 16 \text{ minutes} = 23 \text{ hours } 44 \text{ minutes} = 23 + \frac{44}{60} = \frac{356}{15} \text{ hours} \] So the ratio is: \[ \frac{T_c}{T_i} = \frac{24}{\frac{356}{15}} = \frac{360}{356} = \frac{90}{89} \] Using this ratio, calculate the actual time passed ($T_c$) when the faulty clock shows 89 hours: \[ T_c = 89 \times \frac{90}{89} = 90 \text{ hours} \] Since the correct time elapsed is exactly 90 hours, which is 1 hour more than the 89 hours indicated: \[ \text{Correct Time} = \text{Indicated Time } (10:00 \text{ PM}) + 1 \text{ hour} = 11:00 \text{ PM} \]

Step 2: Final Answer:
The correct time of the clock is 11 PM, matching Option (C).