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# Swimming Race Rate Problem

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Swimming Race Rate Problem [#permalink]

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13 Mar 2010, 11:11
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Hardy and Andy start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lanes at uniform speeds, but Hardy is faster than Andy. They 1st pass at a point 18.5m from the deep end and having completed one length each 1 is allowed to rest on the edge for exactly 45 sec. After setting off on the return length, the swimmers pass for the 2nd time just 10.5m from the shallow end. How long is the pool?

A. 55.5 m
B. 45 m
C. 66 m
D. 49 m
E. 54 m
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Re: Swimming Race Rate Problem [#permalink]

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13 Mar 2010, 11:42
Really tough question!!. Couldn't solve it yet..
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Re: Swimming Race Rate Problem [#permalink]

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13 Mar 2010, 22:47
Hussain15 wrote:
Hardy and Andy start a two-length swimming race at the same moment but from opposite ends of the pool. They swim in lanes at uniform speeds, but Hardy is faster than Andy. They 1st pass at a point 18.5m from the deep end and having completed one length each 1 is allowed to rest on the edge for exactly 45 sec. After setting off on the return length, the swimmers pass for the 2nd time just 10.5m from the shallow end. How long is the pool?

A. 55.5 m
B. 45 m
C. 66 m
D. 49 m
E. 54 m

Whats the source of problem?
Retired Moderator
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Re: Swimming Race Rate Problem [#permalink]

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14 Mar 2010, 22:22
I got this question from the past thread of this forum.

Kindly answer the question with explanation.
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Re: Swimming Race Rate Problem [#permalink]

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15 Mar 2010, 08:12
Let Length of Pool be L, speed of Hardy be H and that of Andy be A. Given H>A

They 1st pass at a point 18.5m from the deep end

(L - 18.5)/H = 18.5/A --- eqn1

1/A = (L - 18.5)/18.5 H ------eqn2

They cross for 2nd time at a point 10.5m from the shallow end

(L - 10.5)/ H = 10.5/A ----eqn3

now total time spent by both from 1st end to 2nd end ,plus wait period and crossing for second time will be equal i.e.

L/H + 45 + (L - 10.5)/ H = L/A +45 + 10.5/A

on solving this we get
(2L - 10.5)/H = (L+10.5)/A----eqn 4

substituting the value of 1/A from eqn2 in eqn 4 we get

(2L - 10.5)/H = (L+10.5) * (L - 18.5)/18.5 H, on solving this we get L = 45
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Re: Swimming Race Rate Problem [#permalink]

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15 Mar 2010, 08:21
kp1811 wrote:
(L - 18.5)/H = 18.5/A --- eqn1

Awesome solution man...just one question. How can you say that Hardy started from the shallow end?
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Re: Swimming Race Rate Problem [#permalink]

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15 Mar 2010, 09:07
sidhu4u wrote:
kp1811 wrote:
(L - 18.5)/H = 18.5/A --- eqn1

Awesome solution man...just one question. How can you say that Hardy started from the shallow end?

even if we consider the opposite it doesn't affect the final answer as we substitute for A or H and H or A cancels out.

Moreover from answer options we know the midpoint is 22.5 (considering the lowest value of 45 among the answer choices) and since H>A we can intuitively assume that Hardy started from shallow end.

solving it the other way i.e. Hardy is at Deep end initially
Let Length of Pool be L, speed of Hardy be H and that of Andy be A. Given H>A

They 1st pass at a point 18.5m from the deep end

(L - 18.5)/A = 18.5/H --- eqn1

1/A = 18.5/(L - 18.5) H ------eqn2

They cross for 2nd time at a point 10.5m from the shallow end

(L - 10.5)/ A = 10.5/H ----eqn3

now total time spent by both from 1st end to 2nd end ,plus wait period and crossing for second time will be equal i.e.

L/A + 45 + (L - 10.5)/ A = L/H +45 + 10.5/H

on solving this we get
(2L - 10.5)/A = (L+10.5)/H----eqn 4

substituting the value of 1/A from eqn2 in eqn 4 we get

(L + 10.5)/H = (2L -10.5) * 18.5/(L - 18.5) H, on solving this still we get L = 45
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Re: Swimming Race Rate Problem [#permalink]

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15 Mar 2010, 11:50
Thanks...understood now
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Re: Swimming Race Rate Problem   [#permalink] 15 Mar 2010, 11:50
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# Swimming Race Rate Problem

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