Motorboat A leaves shore P as B leaves Q; they move across the lake at a constant speed. They meet first time 600 yards from P. Each returns from the opposite shore without halting, and they meet 200 yards from Q. How long is the lake?

• Let, A's speed x & B's speed y & distance d
  600/x=(d-600)/y
  d/x+200/x=d/y+(d-200)/y
  Solving those eqns. d = 1600yards
• 10 years agoHelpfull: Yes(2) No(0)
• According to first condition i.e . they meet first time 600 yards from P
  let p is the speed of P and q is the speed of Q and d is the width of river,
  600/p = (d-600)/q ----(i)
  solving first we get, p/q = 600/(d-600) ----(a)
  According to Second condition i.e . they meet 200 yards from Q,
  (d-600)/p + 200/p = (d-200)/q + 600/q ------(ii)
  solving second we get p/q = (d-400)/(d+400) -----(b)
  
  now solving (a) and (b) we get
  => 600/(d-600) = (d-400)/(d+400)
  => d = 1600 yards
• 6 years agoHelpfull: Yes(1) No(0)