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09 Jun 2021, 07:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
44% (01:40) correct
56%
(02:07)
wrong
based on 32
sessions
History
Date
Time
Result
Not Attempted Yet
The two hands of an incorrect clock coincide after every 65 min. How much time does the clock gain or lose in one day?
(A) 9.069 min
(B) 10.069 min
(C) 11.069 min
(D) 12.069 min
(E) 13.069 min
(A) 9.069 min
(B) 10.069 min
(C) 11.069 min
(D) 12.069 min
(E) 13.069 min
09 Jun 2021, 07:54
Kudos
Bookmarks
It should be B. If any one can try this with an explanation.
09 Jun 2021, 09:51
Kudos
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Bunuel
Regardless of how fast or slow the clock is, the rule is always the minute hand travels 12 rotations for each rotation the hour hand travels. In other words, the minute hand is 12x as fast as the hour hand.
We can set the speeds as \(x\) and \(12x\). Coinciding means after 65 min they meet again, with the minute hand being faster. Thus we need the hands to be exactly 360 degrees apart after 65 min. We can let 360 be a distance value and the corresponding unit for x would be degrees per minute.
We have: \(65x + 360 = 12x*65\), \(360 = 65*11x\) and \(x = \frac{360}{65*11}\), thus is how many degrees the hand hour travels every minute. Therefore it takes \(\frac{360}{x} = 65*11 = 650+65 = 715\) minutes for the hand hour to travel one rotation, on the other hand for a normal clock it would take 60*12 = 720 minutes to travel one rotation, so we know every 12 hours this clock is faster by 5 minutes.
Finally this means for 24 hours this clock is faster by exactly 10 minutes.
Ans: B
