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?

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Himanshu

  • Dec 5th, 2005
 

ans.

length of the pool is 44

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Himanshu

  • Dec 7th, 2005
 

hey i solved ur prob there z a ques. above  which z very similar to this

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sharad vasisht

  • Dec 7th, 2005
 

Thnx buddy. actually i solved on the same day and you posted it.thnx for the ans.

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gaurav4067

  • Dec 8th, 2010
 

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.

 

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