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?

Answer 1

hope it will help you..

Answer 2

Answer:

The watch will show correct time on Saturday at 9 P.M

Step-by-step explanation:

Total hours from Thursday 8 P.M. to Monday 8 A.M. = 84 hours

Now, first the watch is 5 minutes fast then the watch is 7 minutes slow

⇒ The variation is +5 and -7 minutes

So, total time difference = 5 - (-7) = 5 + 7 = 12 minutes

Now, 12 minutes variation in 84 hours

1 minutes variation in 7 hours

Currently the watch is 7 minutes slow, so we eed to make the watch 7 minutes faster

So, 7 minutes variation is 7 × 7 = 49 hours

Hence, The watch will show correct time after 49 hours from Thursday 8 P.M.

Therefore, The watch will show correct time on Saturday at 9 P.M