A and B stand at distinct points of a circular race track of length 135 m. They cycle at a speed of a m/s and b m/s. They meet for the 1st time 9sec. After they start the race and for the second time 24 sec from the time they start the race. Now if b had started in the opposite direction to the one he had originally started. They would have met for the first time after the 18sec. If B is quicker then A , then find b?

• b=6m/s
  a=3m/s
  a+b=135÷15
  So a+b=9
  Nd thus they will totally cover 9×9=81m nd on reverse they will be 135-81=54m apart so when b changes direction
  
  54/18=a-b(relative speed concept)
  a-b=3
  
  Frm here a+b=9 and a-b=3
  We get b=6
  a=3 as said in ques b is quicker.
• 8 years agoHelpfull: Yes(8) No(7)
• It is given they stand in distinct position in race track and after 9 secs they meet each other. So we can take this point (say P) as a relative starting point for the next meet i.e. after 24 secs from the original starting position.
  So from P to next meet up point it took 24-9=15 sec.
  A/C to formula Distance/(relative Speed)= time taken for 1st meet.
  so 135/(a+b)=15 secs => (a+b)=9 m/sec........................(1)
  Now we have relative speed and getting back to starting of the question: both A and B were at distinct position and met after 9 secs.
  so distance covered = Relative speed x time taken= 9 m/sec x 9 secs=81 m. ( they are 81 m apart at their original position)
  Now if B changes the direction then they will run away from each other from their original starting position
  so the distance to be covered now is 135-81= 54 m
  D/s=t => 54/(a-b)=18 => (a-b)=3 m/sec.........................(2)
  so we got a+b=9 and a-b=3
  by solving we get a=3 and b=6 (as b is quicker).
  
• 8 years agoHelpfull: Yes(6) No(0)
• not understood
• 8 years agoHelpfull: Yes(1) No(3)
• Not understand
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