| A watch which gain uniformly, is 5 min slow at 7 O'clock in the morning on Sunday and it is 5 min 48 s fast at 7 pm on following Sunday. Which day was it correct? |
A) Friday
B) Saturday
C) Tuesday
D) Wednesday
Correct Answer: D
Solution :
| Time from 7 am on Sunday to 7 pm on following Sunday = 7 days 12 h = 180h |
| \[\therefore \]Watch gains \[\left( 5+5\frac{4}{5} \right)\min \]or\[\frac{54}{5}\min \] in \[180\,\,\text{h}.\] |
| Now,\[\frac{54}{5}\min \] are gained in \[180\,\,h.\] |
| \[\therefore \]5 min are gained in \[=\left( 180\times \frac{5}{54}\times 5 \right)\text{h}\] |
| \[=83\,\,\text{h}\,\,20\min \] |
| = 3 days 11 h and 20 min |
| \[\because \]Watch is correct after 3 days 11 h and 20 min after 7 am of Sunday. |
| It will be correct at 20 min past 6 pm on Wednesday. |
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