A , B and C are standing on a circular track of 4300m such that A and C are diametrically opposite and B is in between A and C .There clockwise order are ABC , B and C runs clockwise while A runs anti-clockwise .If the speed of A B and C are 5m/s , 8m/s and 10m/s respectively . They all start running at 10:00 AM. calculate the distance covered by B when A and C meet for the first time after 10:05AM .
I dont remember the options , anyway options cant actually help in such type of questios!!

• Relative speed of A & C will be.. 15 m/s.. hence for first they will meet at
  2150/15= 143.33 sec (around 2min 23sec).. so time will be 10:02:23..this case will not be considered as it is before 10:05.. Again they will meet at
  4300/15=286.66 sec(around 4min 46sec).. time will be 10:07:10.. this case will be considered
  Hence in 7min10 sec i.e 430
  In 430sec B will cover 430*8 metres=3440 metres
  
• 8 years agoHelpfull: Yes(80) No(0)
• I think ans is 3440......because in 5 mnt A goes 1500 m and C goes 3000 m....so after 5 mnt distance between A and C is 1950.....after 5 mnt B covers distance 2400 mtr...
  relative speed of A and C is 15 m/s....... so time needed to cover the distance 1950/15=130 sec......
  in 130 sec B will cover distance 130*8=1040 mtr....so total distance covered by B will be 3440 mtr........
• 8 years agoHelpfull: Yes(6) No(0)
• I think answer will be approx 3545m.
  
  In the first 5 min from 10:00 to 10:05 B has covered 2400m.
  After that A &C took (2 min & 23 sec = 143 sec) to meet. by this time B has covered (143*8 = 1144m).
  so total distance covered by B is (2400+1144) = 3544.
  
  I am not sure this option was there in the question or not.
• 8 years agoHelpfull: Yes(2) No(5)
• speed of A & C=15m/s
  so,2150/15=143.33s
  so time=10:02:23
  discarded
  next they will meet at 4300/15=286.66s=10:07:10 i.e 430 sec
  so B will cover 430*8=3440m=3.440km
• 8 years agoHelpfull: Yes(2) No(0)
• A and C will meet second time at 10:10 Am
  hence B covered 80 mts.
• 8 years agoHelpfull: Yes(1) No(7)
• since A & C both traveling towards each other so
  
  relative speed =15m/s
  
  when they meet 1st time its between A and C so s=4300/2=> time taken =distance/speed=430/3 < 60*5
  
  => when they meet 2nd time distance =4300 time taken=860/3< 300
  for third time distance=6350 time=6350/15=430 second>300
  distance traveled = 8*430=3440
  
• 8 years agoHelpfull: Yes(1) No(0)
• How 7.10 Will come... when 4.46 Sec come out after calculation...
• 8 years agoHelpfull: Yes(1) No(1)
• plzz give complete solution
  
• 8 years agoHelpfull: Yes(0) No(0)
• sorry for the previous solution. Answer will be 3440.
  
  After 2 min 10 sec (130 sec) A&C will meet. In the mean while B will cover (8*130 = 1040m)
  total distance by B is (2400+1040) =3440m.
• 8 years agoHelpfull: Yes(0) No(0)
• since A & C both traveling towards each other so
  
  relative speed =15m/s
  
  when they meet 1st time its between A and C so s=4300/2=> time taken =distance/speed=430/3 < 60*5
  
  => when they meet 2nd time distance =4300 time taken=860/3
• 8 years agoHelpfull: Yes(0) No(1)
• 860 is the answer.
  
• 8 years agoHelpfull: Yes(0) No(2)
• in all the solution all of you just consider only a and c and wha about b
  first take relative speed of b ans that is 2 and thid add to a that gives 7m/s.
  now devide this to distance gives first time where they meet. this time when multiply any speed give meeting point
  
• 8 years agoHelpfull: Yes(0) No(0)