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?

• The solution is :Let the length of swimming pool be : D
  let their speed be x and y. So acc. to ques 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 y / x =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)
  
  y . t2 =D + 10 ----- (5)
  (as y travelled one length complete to shallow end + 10 m from shallow end)
  
  (5) / (4)we get
  
  y / x =(D + 10) / (2D-10) ----- (6)
  
  from (3) and (6)
  
  18 / (D-18)= (D+10) / (2D-10)
  
  solving we get
  
  D(D-44) 0
  
  since D cannot be zero
  
  so D 44 m ans.
• 10 years agoHelpfull: Yes(14) No(0)
• @vignesh why u did d+10 n d-10 in 2 case
• 10 years agoHelpfull: Yes(1) No(1)
• as d-10 and d+10
  bocz in the question it is written that they pass at 10 m
  similar as we done earlier in 18m when both of them pass at 18 m
• 10 years agoHelpfull: Yes(1) No(1)
• 44 is the solution
• 10 years agoHelpfull: Yes(1) No(2)
• suppose the length of swimming pool is X and speed of albert is A and speed of fernandes is B,
  at the first time meet at t1 the eqtn is A*t1=X-18----(1) and B*t1=18----(2)
  at next time meet at t2 is A*t2=X+8---(3) and B*t2=X-8----(4)
  by dividing eqstn 1 by 2 we get A/B=(X-18)/18----(5)
  from eqstn 3 and 4 we get A/B=(X+8)/(X-8)----(6)
  by solving these two eqstn we find the length of swimming pool
  
• 10 years agoHelpfull: Yes(0) No(2)
• not understand this....
• 10 years agoHelpfull: Yes(0) No(0)
• The solved tag on this question is false. The solution by Vignesh Kumar falsely assumes that the 4-second duration cancels out since it is not stated anywhere in the question that the faster one waits till the slower one reaches the end of the pool. If the faster one does not wait then he would have finished his 4-second wait sometime before the slower one and then the slower one will have a disadvantage.
• 7 years agoHelpfull: Yes(0) No(0)