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Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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13 Mar 2010, 11:11

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45% (02:06) correct
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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?

Re: Hardy and Andy start a two-length swimming race at the same moment but [#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?

Re: Hardy and Andy start a two-length swimming race at the same moment but [#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

Re: Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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18 Aug 2016, 04:31

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Re: Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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18 Aug 2016, 06:39

kp1811 wrote:

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

HI, your explanation is indeed excellent. But I don't get the point how the quoted parts is equal. According the point L/H=L/A. But these two can't be equal as their relative speed is not same. Correct my understanding, please.

Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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18 Aug 2016, 20:30

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let x=length of pool at first meeting, combined distance=x at second meeting, combined distance=3x if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x andy's total distance to second meeting=x+10.5 m x+10.5=55.5 m x=45 m

Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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19 Aug 2016, 12:48

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NaeemHasan wrote:

How have you got 3x. I don't get it. explain please.

Posted from my mobile device

Hi Naeem,

x=length of pool Andy's distance to second meeting= 18.5+(x-18.5)+10.5=x+10.5 m Hardy's distance to second meeting= (x-18.5)+18.5+(x-10.5)=2x-10.5 m combined distance to second meeting= (x+10.5)+(2x-10.5)=3x m

Re: Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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23 Aug 2016, 01:09

gracie wrote:

let x=length of pool at first meeting, combined distance=x at second meeting, combined distance=3x if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x andy's total distance to second meeting=x+10.5 m x+10.5=55.5 m x=45 m

Hi Gracie

Although I do understand (at least i think i do) this part: if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x Could you please explain that mathematically? Is it because the distance covered by each party till the point of meeting is in proportion to their speed?

Hardy and Andy start a two-length swimming race at the same moment but [#permalink]

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27 Aug 2016, 10:21

saumverm wrote:

gracie wrote:

let x=length of pool at first meeting, combined distance=x at second meeting, combined distance=3x if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x andy's total distance to second meeting=x+10.5 m x+10.5=55.5 m x=45 m

Hi Gracie

Although I do understand (at least i think i do) this part: if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x Could you please explain that mathematically? Is it because the distance covered by each party till the point of meeting is in proportion to their speed?

Thanks Saumya

Hi Saumya,

yes, because time of each to meeting is equal, distance of each is proportional to their speeds combined distance to first meeting is x meters andy swims 18.5 of those x meters thus, the ratio of andy's distance to combined distance is 18.5/x meters assuming consistent speeds and multiplying by 3, the same ratio will apply to second meeting: 55.5/3x meters

I hope this helps. gracie

gmatclubot

Hardy and Andy start a two-length swimming race at the same moment but
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27 Aug 2016, 10:21

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