Albert and Fernandes have two leg swimming race. Both start from opposite and of the pool. On the first leg, the boys pass each other at 18 mt from the deep end of the pool. During the II leg they pass at 10 mt from the shallow end of the pool. Both go at const speed. But one of them is faster. Each boy rests for 4 sec to see at the end of the I leg. What is the length of the pool.

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The solution is :Let the length of swimming pool be : Dlet their speed be x and y. So acc. to que the fast swimmer (let x) would start from shallow end. thuslet they first meet after time : t1x . 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 gety / x = (D + 10) / (2D-10) ----- (6)from (3) and (6)18 / (D-18) = (D+10) / (2D-10)solving we getD(D-44) = 0since D cannot be zero so, D = 44 m ans.

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The solution is :Let the length of swimming pool be : D
let their speed be x and y. So acc. to que 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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