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swimming!!!!

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Manager
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swimming!!!! [#permalink]  18 Jan 2009, 18:30
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A swimmer makes a round trip up and down the river. It takes him X hours. The next day he swims the same distance with the same speed in the still water. It takes him Y hours. What can we say ABOUT X and Y ?
1. X>Y
2.X<Y
3.X=Y
4. X= 1/2 Y
5. NONE OF THE ABOVE
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Re: swimming!!!! [#permalink]  19 Jan 2009, 01:49
Abhishek.pitti wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. The next day he swims the same distance with the same speed in the still water. It takes him Y hours. What can we say ABOUT X and Y ?
1. X>Y
2.X<Y
3.X=Y
4. X= 1/2 Y
5. NONE OF THE ABOVE

I assumed that "the same distance" is different from " a round trip", so a is the choice
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Re: swimming!!!! [#permalink]  19 Jan 2009, 02:38
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Abhishek.pitti wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. The next day he swims the same distance with the same speed in the still water. It takes him Y hours. What can we say ABOUT X and Y ?
1. X>Y
2.X<Y
3.X=Y
4. X= 1/2 Y
5. NONE OF THE ABOVE

Time = Distance travelled/speed
assume v= speead in the still water
s = stream speed
D = Distance
X = time taken for upstream +time taken for downstream
= D/(v-s) + D/(v+s)
= D(2v)/(v^2-s^2)
= 2D/(v-(s^2)/v)
Y = 2D/v

from the above clearly X>Y
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Re: swimming!!!! [#permalink]  19 Jan 2009, 05:45
Distance = D
speed in upstream S-a, speed in downstream S+a
speed in still water = S

for Still water, for both up and down,
So 2D = $S/Y$
Y = $S/(2D)$ ----- a)

For upstream and downstream,
So D = S-a/t1(upstream )
D = S+a/t2 (downstream)

Dt1 = S-a --- 1
Dt2 = S+a ----2

Adding 1) and 2) t1+t2 = X
DX = 2S ===> X=2S/D ---- b)

from a) and b)

x=4D
S0 X>Y
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Re: swimming!!!! [#permalink]  19 Jan 2009, 07:27
x2suresh wrote:
Abhishek.pitti wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. The next day he swims the same distance with the same speed in the still water. It takes him Y hours. What can we say ABOUT X and Y ?
1. X>Y
2.X<Y
3.X=Y
4. X= 1/2 Y
5. NONE OF THE ABOVE

Time = Distance travelled/speed
assume v= speead in the still water
s = stream speed
D = Distance
X = time taken for upstream +time taken for downstream
= D/(v-s) + D/(v+s)
= D(2v)/(v^2-s^2)
= 2D/(v-(s^2)/v)
Y = 2D/v

from the above clearly X>Y

I missed the conversion below and was wondering how to attack the problem.

= 2D/(v-(s^2)/v)
Y = 2D/v

Thank you.
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Re: swimming!!!! [#permalink]  19 Jan 2009, 13:17
Expert's post
Another approach

Let's imagine we slowly change speed of stream from 0 to speed of a swimmer: the time change from Y to Infinity (the swimmer cannot swim against a strong stream). As speed of stream is greater than 0 - therefore, X>Y

Sometimes another approach could help when we suddenly forget necessary formulas or have no time to write out something...
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Re: swimming!!!! [#permalink]  19 Jan 2009, 16:14
walker wrote:
Another approach

Let's imagine we slowly change speed of stream from 0 to speed of a swimmer: the time change from Y to Infinity (the swimmer cannot swim against a strong stream). As speed of stream is greater than 0 - therefore, X>Y

Sometimes another approach could help when we suddenly forget necessary formulas or have no time to write out something...

you are machine... great approach!!!!!
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Re: swimming!!!! [#permalink]  21 Jan 2009, 22:57
walker wrote:
Another approach

Let's imagine we slowly change speed of stream from 0 to speed of a swimmer: the time change from Y to Infinity (the swimmer cannot swim against a strong stream). As speed of stream is greater than 0 - therefore, X>Y

Sometimes another approach could help when we suddenly forget necessary formulas or have no time to write out something...

Excellent explanation! But, this can be understood only for swimming against the stream. In round trip, initial understanding comes as the time lost against the stream will equal the time gained along the stream and hence total time will be the same.
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Re: swimming!!!! [#permalink]  22 Jan 2009, 03:10
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Expert's post
In my approach I take two edge points:

1) speed of stream = speed of swimmer ---> Y=X
2) speed of stream = speed of swimmer ---> Y=infinity >> X

Now, I assume that Y changes gradually from X to infinity between these points. I agree with you that it is an assumption and there is no any "special" point in the interval. At the same time I see the assumption pretty natural as at point 2 downstream time decreases only by two times (relative speed of a swimmer doubles) but upstream time increases by infinite times (relative speed of a swimmer is zero). In other words, downstream time decreases much slowly than upstream time increases.
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Re: swimming!!!! [#permalink]  22 Jan 2009, 04:18
walker wrote:
In my approach I take two edge points:

1) speed of stream = speed of swimmer ---> Y=X
2) speed of stream = speed of swimmer ---> Y=infinity >> X

Now, I assume that Y changes gradually from X to infinity between these points. I agree with you that it is an assumption and there is no any "special" point in the interval. At the same time I see the assumption pretty natural as at point 2 downstream time decreases only by two times (relative speed of a swimmer doubles) but upstream time increases by infinite times (relative speed of a swimmer is zero). In other words, downstream time decreases much slowly than upstream time increases.

crystal clear. Thanks walker!
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Re: swimming!!!! [#permalink]  23 Jan 2009, 15:34
scthakur wrote:
walker wrote:
In my approach I take two edge points:

1) speed of stream = speed of swimmer ---> Y=X
2) speed of stream = speed of swimmer ---> Y=infinity >> X

Now, I assume that Y changes gradually from X to infinity between these points. I agree with you that it is an assumption and there is no any "special" point in the interval. At the same time I see the assumption pretty natural as at point 2 downstream time decreases only by two times (relative speed of a swimmer doubles) but upstream time increases by infinite times (relative speed of a swimmer is zero). In other words, downstream time decreases much slowly than upstream time increases.

crystal clear. Thanks walker!

Thakur, can you put Walker's explanation in some other words? I don't get it. I read it 3-4 times but just don't get it. Just wondering if this can be presented in alternate way?
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Re: swimming!!!! [#permalink]  23 Jan 2009, 22:48
Expert's post
Perhaps, it is because a misprint. It should be:

1) speed of stream = 0 ---> Y=X
2) speed of stream = speed of swimmer ---> Y=infinity >> X
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Re: swimming!!!! [#permalink]  24 Jan 2009, 03:54
suresh, walker, thanks for explanations. +2
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Re: swimming!!!! [#permalink]  26 Jan 2009, 03:05
I still dont get it. I think the answer is 'C' - X=Y.
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Re: swimming!!!! [#permalink]  26 Jan 2009, 07:25
Ibodullo wrote:
I still dont get it. I think the answer is 'C' - X=Y.

Time = Distance travelled/speed
assume v= speead in the still water
s = stream speed
D = Distance
X = time taken for upstream +time taken for downstream
= D/(v-s) + D/(v+s)
= D(2v)/(v^2-s^2)
= 2D/(v-(s^2)/v)
= 2D/(v-k)

{ say s^2/v = k, which is always postive unless stream speed =0 (in this case k=0)}

Y = 2D/v

from the above clearly X>Y (because v-k is always less than v }

Did you get it now?
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Re: swimming!!!! [#permalink]  26 Jan 2009, 20:34
x2suresh wrote:
Ibodullo wrote:
I still dont get it. I think the answer is 'C' - X=Y.

Time = Distance travelled/speed
assume v= speead in the still water
s = stream speed
D = Distance
X = time taken for upstream +time taken for downstream
= D/(v-s) + D/(v+s)
= D(2v)/(v^2-s^2)
= 2D/(v-(s^2)/v)
= 2D/(v-k)

{ say s^2/v = k, which is always postive unless stream speed =0 (in this case k=0)}

Y = 2D/v

from the above clearly X>Y (because v-k is always less than v }

Did you get it now?

Understood and agreed. Thanks a lot Suresh!
Re: swimming!!!!   [#permalink] 26 Jan 2009, 20:34
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