Quantitative Aptitude
CLOCK MCQs
Let us denote the rate at which the first clock gains time per hour as x and the rate at which the second clock gains time per hour as y.
Hence, x = 2 minutes per hour and y = x/2 = 1 minute per hour.
From 10 a.m. on Friday to 2 p.m. on Monday, the duration is 4 days, i.e. 96 hours.
Therefore, the total time gained by the first clock in 96 hours = x × 96 = 192 minutes.
Similarly, the total time gained by the second clock in 96 hours = y × 96 = 96 minutes.
Therefore, the total time difference between the two clocks after 96 hours = 192 – 96 = 96 minutes.
Now, at 10 a.m. on Friday, the two clocks read the same time.
Therefore, at 2 p.m. on Monday, the time shown by the second clock will be 2 p.m. plus 96 minutes, i.e. 3.16 p.m.
Hence, the correct answer is Option C - 3.16 p.m.
If you think the solution is wrong then please provide your own solution below in the comments section .
To solve this problem, we need to calculate the time gained by the clock and add it to the time shown on the clock to get the actual time. Let's first calculate the time gained by the clock in 54 hours (3 days and 15 hours).
Time gained by the clock in 1 hour = 30 secondsTime gained by the clock in 54 hours = (30/3600) x 54 = 27/20 minutes
So, the clock is ahead of the actual time by 27/20 minutes at 3:39 p.m. on the 3rd day afternoon.
Now, let's find out the actual time by subtracting the time gained by the clock from the time shown on the clock.
Time shown on the clock = 3:39 p.m.Time gained by the clock = 27/20 minutes
Subtracting 27/20 minutes from 3:39 p.m., we get:
Actual time = 3:39 p.m. - 27/20 minutesActual time = 2:54 p.m.
Therefore, the actual time when the clock indicates 3.39 p.m. on the 3rd day afternoon is 2:54 p.m. Option A (3 p.m.) is not correct, as the actual time is before 3 p.m.
Hence, the correct answer is option A (2:54 p.m.).If you think the solution is wrong then please provide your own solution below in the comments section .
To solve the problem, we can use the formula:
Time gained or lost = (Time elapsed) x (Gain or loss per unit time)
Let's denote the time the watch was correct by X, and the gain per hour by G. We know that:
At 12 a.m. on Monday, the watch was 6 minutes slow, which means it showed 11:54 p.m. The time elapsed from Monday midnight to Wednesday 9 p.m. is 57 hours.
At 12 a.m. on Friday, the watch was 6 minutes 48 seconds fast, which means it showed 12:06:48 a.m. The time elapsed from Friday midnight to Wednesday 9 p.m. is 84 hours.
Using the formula, we can set up two equations:
X - 11:54 = 57G (1)12:06:48 - X = 84G (2)
To solve for X, we can add the two equations and simplify:
12:06:48 - 11:54 = 141G12:06:48 - 11:54 = 12 x 3600 + 6 x 60 + 48 - 11 x 3600 - 54 x 60= 732G = 732/141G = 5.19 seconds per hour
Now we can substitute G into equation (1) and solve for X:
X - 11:54 = 57 x 5.19X = 9:00 p.m.
Therefore, the watch was correct on Wednesday at 9 p.m., and the answer is option D.
Note: We can also solve the problem using a proportion. Let T be the correct time, then:
From Monday midnight to Wednesday 9 p.m., the watch gained 6 minutes or 360 seconds, which is 6/57 of the total time elapsed.
From Friday midnight to Wednesday 9 p.m., the watch gained 6 minutes 48 seconds or 408 seconds, which is 408/84 of the total time elapsed.
Therefore, we have:
(360 seconds) / (57 hours) = (408 seconds) / (84 hours) = (T - 11:54) / (57 hours)
Solving for T, we get T = 9:00 p.m.If you think the solution is wrong then please provide your own solution below in the comments section .