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Hardy and Andy start a two-length swimming race at the same moment but

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

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New post 13 Mar 2010, 10:11
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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: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 19 Sep 2016, 21:40
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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


Let length of the pool be L.

Deep

A —————————————><———————————----———————————— H
.................(18.5m)


H ——————————————————————————-———><————————- A
..............................................................................................................(10.5)


In same time, Andy covers 18.5 and Hardy covers L - 18.5.
In same time, Andy covers L - 18.5 + 10.5 while Hardy covers 18.5 + L - 10.5
The ratio of these distances covered would be the same since the speeds of both people are constant.

18.5/(L - 18.5) = (L - 8)/(L + 8)
Plug in the options to find that L = 45 works.
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Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post Updated on: 03 Sep 2016, 11:15
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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 of x, then he will swim 3*18.5=55.5 m of 3x
andy's total distance to second meeting=x+10.5 m
x+10.5=55.5 m
x=45 m

Originally posted by gracie on 18 Aug 2016, 19:30.
Last edited by gracie on 03 Sep 2016, 11:15, edited 1 time in total.
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 13 Mar 2010, 10:42
Really tough question!!. Couldn't solve it yet..
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 13 Mar 2010, 21:47
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?
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 14 Mar 2010, 21:22
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I got this question from the past thread of this forum.

Kindly answer the question with explanation.
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 15 Mar 2010, 07:12
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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: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 15 Mar 2010, 07:21
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: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 15 Mar 2010, 08:07
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: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 15 Mar 2010, 10:50
Thanks...understood now
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 18 Aug 2016, 05:29
kp1811 wrote:
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


How can we equate times from eqn3 and further when Hardy swims faster than Andy? Doesn't Hardy takes less time than Andy?
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 18 Aug 2016, 05:39
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.
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 18 Aug 2016, 22:12
How have you got 3x. I don't get it. explain please.

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

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New post 19 Aug 2016, 11:48
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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

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

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New post 19 Aug 2016, 20:28
Clear now. thanks a ton.

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

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New post 23 Aug 2016, 00:09
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?

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

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New post 27 Aug 2016, 09:21
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.
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 02 Sep 2016, 23:44
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


Hey, i had a doubt in your solution :
I get that the combined distance travelled by both H and A is 3x

But what do you mean by if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x ?

Is it that Andy covers 18.5 m in time units and hence he will cover 18.5*3 in 3x time units ? (provided speed remains constant)
But then again, 3x is the distance

So you must be inferring that Andy covers 18.5 m in a total length of x and hence he must cover 18.5*3 in a total length of 3x. Please confirm on this. Thanks
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Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 03 Sep 2016, 11:04
aniketm.87@gmail.com 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


Hey, i had a doubt in your solution :
I get that the combined distance travelled by both H and A is 3x

But what do you mean by if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x ?

Is it that Andy covers 18.5 m in time units and hence he will cover 18.5*3 in 3x time units ? (provided speed remains constant)
But then again, 3x is the distance

So you must be inferring that Andy covers 18.5 m in a total length of x and hence he must cover 18.5*3 in a total length of 3x. Please confirm on this. Thanks


Hi aniketm,
See my response to Saumya above.
I hope this helps.
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Re: Hardy and Andy start a two-length swimming race at the same moment but  [#permalink]

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New post 19 Sep 2016, 22:52
Took 14 mins to solve but managed to get it right.
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Re: Hardy and Andy start a two-length swimming race at the same moment but &nbs [#permalink] 19 Sep 2016, 22:52

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