Albert and Fernandes have two leg swimming race. Both start from opposite ends of the pool. On the first leg, the boys pass each other at 18 m from the deep end of the pool. During the second leg they pass at 10 m from the shallow end of the pool. Both go at constant speed but one of them is faster. Each boy rests for 4 seconds at the end of the first leg. What is the length of the pool?

• 44m
  A covers D-18 m when F covers 18
  A covers 2D-10 when Fcovers D+10
  equating both we get D^2-44D=0
  D=0,44
  omitting 0 we get 44
• 10 years agoHelpfull: Yes(1) No(0)
• The solution is :Let the length of swimming pool be : D
  let their speed be x and y. So according to question the fast swimmer (let x) would start
  from shallow end.
  Thus
  Let they first meet after time: t1
  x×t1=D–18
  (1)
  y×t1=18
  (2)
  (2) / (1)we get
  yx=18(D–18)
  --- (3)
  Let t2 be the time after which they meet 2nd time (the 4 sec delay is cancelled as both
  wait for 4 sec)
  So
  x×t2=2D–10
  ---- (4)
  (as x travelled one length complete to deep end + length from deep end to 10 m before
  shallow end)
  4y×t2=D+10
  ----- (5)
  (as y travelled one length complete to shallow end + 10 m from shallow end)
  (5) / (4)we get
  yx=(D+10)(2D–10)
  ----- (6)
  from (3) and (6)
  18(D–18)=(D+10)(2D–10)
  solving we get
  D x (D – 44) = 0
  Since D cannot be zero
  So D = 44 m answer.
• 6 years agoHelpfull: Yes(1) No(0)
• Does these questions helps in actual aptitude round of infosys?
• 6 years agoHelpfull: Yes(1) No(0)