Boat A leaves shore P and Boat B leaves shore Q.(P and Q are opposite shores of a river.) A and B travel at constant speed. But the speeds are not same. Both boats meet at 600m from P for the first time. In their return journeys (i.e. after touching the shores), they meet again at 200m from Q. Find distance between P and Q.

• If x is the distance, then
  
  600/a = (x-600)/b or 600/(x-600)= a/b.... (1)
  also
  (x+200)/a= (2x-200)/b or (x+200)/(2x-200) = a/b ...(b)
  
  from equating LHS of (1) and (2),
  we get
  x=0,1600
  
  but x can not be zero.
  
  so distance = 1600 mtrs
• 12 years agoHelpfull: Yes(40) No(0)
• P.......................................…
  ...... 600 .... X ..........................
  ............................... Y ... 200 ..
  
  when they meet first, they have covered the full length of PQ (say, d)
  when they meet the 2nd time, they've covered 2d,
  so each boat has travelled twice the previous distance
  considering, say, boat A
  (d-600) + 200 = 2*600
  d= 1600m
  ---------------
• 12 years agoHelpfull: Yes(5) No(4)
• @ Santosh,
  
  Pls check.
  
  Total Distance travelled by two boats is 3d and not 2d as mentioned in ur solution.
  But u hv managed to give an answer which is accepted by many persons and may be correct because Engineers can bring answer if they know the answer in any way, .. :P
• 12 years agoHelpfull: Yes(4) No(2)
• let the speed of boats be s1 and s2 respectively and total distance be x.....
  then it is given that they first meet at 600m from p(offcourse taking same time to reach that point)......so we can write
  (x-600)/s1=600/s2..........(1)
  similarly after touching the opposite shore we can again write the equations as they take equal time
  (600+x-200)/s1=(x-400)/s2...........(2)
  now dividing 1 by 2....
  (x-600)/(x+400)=600/(x-400)
  solving this we get x=1600
• 12 years agoHelpfull: Yes(1) No(1)
• ans:800m
  just add 600m+200m
• 12 years agoHelpfull: Yes(0) No(20)
• plz any body solve dis qusn..its urgent..thank u
• 12 years agoHelpfull: Yes(0) No(0)