a car A starts from a point P towards another point Q.Another car B starts (also from p) 1 h after the first car and overtakes it after covering 30% of the distance PQ .after that .the car contniue and on reaching Q, Car B reverses and meet car A at 23(1/3) of the distance QP.find the time taken by car B to cover the distance PQ(in hours)

• (Pardon my last sloution)
  This question is from cat 2012 ppr. And the mistake in this question is in the % sign which was omitted. It is 23(1/3)% of PQ.
  
  PQ=x
  When B overtakes A , distances travelled by both are same=3x/10
  
  When they meet for the second time, further distance travelled by B = 7x/10 + 23(1/3)x/100 = 28x/30
  
  Further Distance travelled by A = 7x/10 - 7x/30=14x/30
  
  Ratio of speeds, A:B= 14x/30:28x/30=1:2
  Ratio of time, A:B= 2:1
  A starts 1 hr ahead of B, implies
  1hr+1hr : 1hr
  B covered a distance of 0.3x in 1hr
  Therefore, it will cover x distance in 10/3hrs
  
• 9 years agoHelpfull: Yes(9) No(0)
• The second part of the question ie. When B reverses its direction of motion and starts its journey towards P from Q, B must have met A at a distance 1/3 from Q. (23(1/3) is wrong as it implies that the car has crossed the path PQ. Also it cannot be 2(1/3) as it is the first meeting point of A and B itself.)
  
  P_______.________.________Q
  mn
  
  m is the first meeting point at 1/3distance of PQ. n is the second meeting point after B reverses its direction.
  
  In the first journey, time of A=ta
  Time of B= ta-1
  D=s*t
  Da/Db=Sa*ta/Sb*(ta-1)
  Da=Db=x
  
  » Sa/Sb=(ta-1)/tb
  
  » in the second half of B's journey after the first meeting point..
  Distance travelled by B= 2x+x=3x
  Distance travelled by A in the same time= x
  Ratio of dist= x/3x=1/3
  
  »Ratio of dist = ratio of speeds
  1/3=(ta-1)/ta
  ta=3/2
  Time of B to travel x distance=3/2 -1=1/2hrs
  Time of B to travel PQ(3x)= 3/2hrs
  
• 9 years agoHelpfull: Yes(2) No(1)
• Let the distance between the point P and Q= a
  When B overtakes A for the first time, both of them cover 3a/10.
  When B meets A after that, it (B) covers (7a/10+7a/30) or 28a/30,
  while A covers (23a/30 – 9a/30) or 14a/30.
  Therefore, B is twice as fast as A.
  A starts 1 H after B, it catches up with in 1 H. Therefore, covers 0.3a in 1 H. or x in 10/3H.
• 7 years agoHelpfull: Yes(1) No(0)