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.

• 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.
  
• 13 years agoHelpfull: Yes(7) No(4)
• 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
• 13 years agoHelpfull: Yes(4) No(4)
• 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.
• 13 years agoHelpfull: Yes(3) No(1)
• 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........
• 13 years agoHelpfull: Yes(2) No(8)
• 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
• 13 years agoHelpfull: Yes(1) No(7)
• 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
• 13 years agoHelpfull: Yes(1) No(4)