A watch, which loses time uniformly, was observed to be 5 minutes fast at 8.00 p.m. on Thursday. It was noticed to be 7 minutes slow at 8.00 a.m. on the subsequent Monday. When did the watch show the correct time ? Group of answer choices 7 a.m. on Friday 10a.m. on Sunday 11 a.m. on Friday 7 a.m. Saturday
Question
A watch, which loses time uniformly, was observed to be 5 minutes fast at 8.00 p.m. on Thursday. It was noticed to be 7 minutes slow at 8.00 a.m. on the subsequent Monday. When did the watch show the correct time ?
Group of answer choices
7 a.m. on Friday
10a.m. on Sunday
11 a.m. on Friday
7 a.m. Saturday
Solution
The watch was 5 minutes fast at 8.00 p.m. on Thursday and 7 minutes slow at 8.00 a.m. on the subsequent Monday.
The total time from 8.00 p.m. on Thursday to 8.00 a.m. on Monday is 3 days and 12 hours, which is 84 hours.
In these 84 hours, the watch has lost 12 minutes (5 minutes fast to 7 minutes slow).
So, the watch loses time at a rate of 12 minutes per 84 hours, which simplifies to 1 minute every 7 hours.
The watch showed the correct time when it had lost the 5 minutes it was initially fast by.
At a rate of 1 minute every 7 hours, it would take 5 * 7 = 35 hours for the watch to lose these 5 minutes.
35 hours from 8.00 p.m. on Thursday is 7.00 a.m. on Saturday.
So, the watch showed the correct time at 7.00 a.m. on Saturday. The correct answer is 7 a.m. Saturday.
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