There is a circular track of length 440m. There are 3 persons A, B and C standing at different points on the track ready to start the race.A an B are standing diametrically opposite to each other while C is exactly midway between A an B, such that A, B and C are standing in clockwise order. The race started at 10 am. The speeds of A,B and C are respectively 5m/s,10m/s and 8m/s. at what time would B and C meet for the second time,if all the three of them run in the clockwise direction.
option
a) 10:06:25 am
b) 10:04:35 am
c) 10:00:50 am
d) none of these.

• answer is (B)
  speed of B is 10m/s, speed of C is 8m/s.
  speed of B with respect to C in clockwise direction is 2m/s.
  B is 110m behind C in clockwise direction.
  Hence to pass C twice, B will travel 110+440=550m clockwise with the relative speed of 2m/s.
  time taken by B to do so is (550/2)=275sec or 4min35sec.
  
• 11 years agoHelpfull: Yes(39) No(4)
• answer is B
  distance between b and c is 110m, relative speed of b over c= 2m/s
  1st meet after t1=110/2=55sec
  2nd meet after LCM (Time taken by C in one round, same for B)
  LCM(55,44)=220sec
  total time = 220+55=275 sec (4 minute 35 sec)
• 11 years agoHelpfull: Yes(6) No(2)
• when v r considering relative vel. how the person with smaller vel vl move 4ward??the solution vl b---The relative speed between Harb and Inz: 10 - 8 = 2 m/s and they are 330m apart
  Therefore, 330/2 = 165 sec; 2 min 45 sec; when they meet the 1st time it will be 10:02:45
  When they meet the 2nd time: 440/2 = 220 sec; 3 min 40 sec
  10:02:45
  + 03:40
  --------------
  10:06:25 when they meet the 2nd time
• 10 years agoHelpfull: Yes(4) No(4)
• The relative speed between Harb and Inz: 10 - 8 = 2 m/s and they are 330m apart
  Therefore, 330/2 = 165 sec; 2 min 45 sec; when they meet the 1st time it will be 10:02:45
  When they meet the 2nd time: 440/2 = 220 sec; 3 min 40 sec
  10:02:45
  + 03:40
  --------------
  10:06:25 when they meet the 2nd time
• 10 years agoHelpfull: Yes(3) No(5)
• ans a) 10:06:25 am...explanation...
  relative speed of A and B 10-8 = 2 m/s
  for the second meeting of A and B distance to be covered is 330+440=770 m
  time to cover this distance 770/2 =385 sec = 77/12 min= 6 min 25 sec...
• 11 years agoHelpfull: Yes(2) No(10)
• oops ...there should be B and C in place of A and B in above solution...
• 11 years agoHelpfull: Yes(1) No(6)
• ans is A but
  
  first meet = 330/2= 165s
  second meet after completing one round bcoz clock waise movement and speed is difer. so after 1 round difference b/w them is 440-88. now cover this distance by a is in the tyme 352/2= 176. 44s taken by B to i round = 176+44=220.
  total tym= 220+165= 375/60= 6:25s.
• 10 years agoHelpfull: Yes(1) No(3)
• after meeting for the first time B and C will be at the same point and start from there but since B has higher speed and c has lower so C will always follow B but can not cross him.
• 11 years agoHelpfull: Yes(0) No(1)
• i think it's depends on the standing position of A and B if clockwise sequence is BCA then option b if ACB then option a
• 11 years agoHelpfull: Yes(0) No(1)
• i think correct ans is D:
  distance between b and c = 110
  relative speed b/w b,c = 10-8 =2
  first time they meet after 110/2= 55 sec
  and they again(2 time) meet {LCM of 10 ,8 is 40} ;
  so b and c always meet after 40sec +55
  total time 95 sec.mean they meet 10:01:35
• 11 years agoHelpfull: Yes(0) No(3)
• can any one make me understand that in all the solution how value of 'z' and 'w' became 9..:(.. its quite confusing
  
• 8 years agoHelpfull: Yes(0) No(0)
• answer is d) none of these
  B and C meet for the first time at 10:03:36 am
  for the second time they meet at 10:07:18 am
  calculate yourself
• 5 years agoHelpfull: Yes(0) No(0)