Latest Infosys Aptitude Question SOLUTION: How much does a watch lose per day, if its hands coincide every 64 minutes? A. 32 8 min. 11 B. 36 5 min. 11 C. 90 min. D. 96 min.

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(#M40003022) INFOSYS Numerical Ability question Keep an EYE

How much does a watch lose per day, if its hands coincide every 64 minutes?

A.
32 8 min.
11
B.
36 5 min.
11
C.
90 min.
D.
96 min.

Asked In Infosys jai Narayan Mishra (12 years ago)

Unsolved Read Solution (6)

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Read Solution (6) :

pranali dhete  (12 years ago)

55 min. spaces are covered in 60 min.

60 min. spaces are covered in 60/55x 60min.= 65 5/11min.

Loss in 64 min. = 65 5/11 - 64 = 16/11min.

Loss in 24 hrs = ((16/11)x(1/64) x 24 x 60)min. =32 8/11min.

Is this solution Helpfull? Yes (7) | No (4)  

pankaj  (12 years ago)

correction in last solution:
total delay per coincide=16/11
for 64 min,delay is=16/11
for 1 min,delay is=(16/11)*(1/64)
in 1 day delay is=(1/44)*(24*60)=32.73 min

Is this solution Helpfull? Yes (4) | No (4)  

Srinivas Reddy  (12 years ago)

In 1 hour both the hands cover 55 min space.
=> 60/55 = 12/11 min space covered in 1min of the actual time.

for the hand to coincide hands have to cover 60 min space
=> 12/11 * 60 = 720/11 = 65.5/11 min in actual clock.

But the clock coincides every 64 min.
=>65.5/11 - 64 = 1.5/11 = 16/11 min loss in 64min.
=>16/11 * 1/64 = 1/44 min loss in 1min.

Loss In 24 Hrs => 1/44 * (24 *60) = 360/11 = 32.8/11 min the clock looses in a day.

Is this solution Helpfull? Yes (3) | No (1)  

vipin  (12 years ago)

in an hour hnads concide but given in 64 min it is to be concide so 4 min late in 24 hrs it will be (24*4)=96 min late........

Is this solution Helpfull? Yes (2) | No (8)  

alka rathore  (12 years ago)

normal watch's hands coincides after every 55 minures
if the given watch coincides every 64 minutes then its losing 64-55 minutes every hour or in total 24 hours it will lose 24*9 minutes every day

Is this solution Helpfull? Yes (1) | No (7)  

pankaj  (12 years ago)

after 1 hr. diff. b/w two hands is of 5 minute.
for 1 hour hand will cover 5 min.
for 1 min
hour hand will cover 5/60=1/12min.
so time after which they coincide:let it be x min. + 1hr.
time taken by minute hand=x min.
time taken by hour hand=5+x/12
5+x/12=x
x=60/11
total time after which two hands coincide=60+60/11=720/11minutes
in 1 day these will coincide=24*60*11/720=22times
total delay=(720/11-64)*22=32

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