If A and B with the speed of 40 km/hr and 60 km/hr respectively, starts journey heading towards each other and found that they are 25 Km away from each other 15 minutes before they meet.What is the initial distance between them ?

• the ans i.e distance between them will be 50km
  The distance between them before 15 mins is 25 km
  Now let the distance A has already travelled be x and the distance B has already travelled be y
  In 15 mins A travels 10km(40*1/4)
  In 15 mins B travels 15km(60*1/4)
  So (x+10)=40*t
  and (y+15)=60*t
  Solving (x+10)/(y+15)=2/3
  Put x=10 and y=15 and it matches the eqn...
  So total distance=10+15+25=50km
  The previous eqn can have many other soln but this is the closest...
• 12 years agoHelpfull: Yes(5) No(2)
• It can be any distance more than 25 kms.
  
  For all cases, time taken to travel last 25 kms will be 15 minutes as relative speed of A and B is 100 km/hr.
• 12 years agoHelpfull: Yes(2) No(1)
• Data is incomplete for getting unique answer.
• 12 years agoHelpfull: Yes(1) No(2)
• @ Anurag,
  
  How have u decided the values of x and y as 10 and 15 only ?
  
  The eqn (x+10)/(y+15)=2/3
  
  can be satisfied by any value of x and y such that y = 1.5x.
  
  In each case, you will get different value of total distance between A and B.
  Take x=20, y=30, total distance will be 20+30+25 = 75 kms
  Take x=30, y=45, total distance will be 30+40+25 = 100 kms.
  
  so as stated by Garima and SS, no unique answer is possible.
• 12 years agoHelpfull: Yes(0) No(0)
• @anurag
  if (x+10)/(y+15) =40/60
  i can also put x=30
  y=45
  then total =30+45=75
  so ur ans is not sufficient i think
• 12 years agoHelpfull: Yes(0) No(0)
• I have stated that the eqn can have many solns. in my last line....
  (10,15),(20,30),(2,3),(4,6)all satisfy the eqn...
  the option has to be selected from the choices:)
  
• 12 years agoHelpfull: Yes(0) No(0)