|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 08 Dec 2006
Posts: 79
Location: United States
Concentration: Entrepreneurship, General Management
Schools: Harvard Business School (HBS) - Class of 2014
GPA: 3.64
WE: Engineering (Aerospace and Defense)
Followers: 31
Kudos [?]:
57
[0], given: 3
|
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
|
|
|
|
|
|
|
SVP
Joined: 04 May 2006
Posts: 1946
Schools: CBS, Kellogg
Followers: 10
Kudos [?]:
168
[0], given: 1
|
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
_________________
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1842
Location: New York
Followers: 20
Kudos [?]:
291
[1] , given: 5
|
1
This post received
KUDOS
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
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Manager
Joined: 27 May 2008
Posts: 205
Followers: 1
Kudos [?]:
12
[0], given: 0
|
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
|
|
|
|
|
|
Senior Manager
Joined: 30 Nov 2008
Posts: 498
Schools: Fuqua
Followers: 9
Kudos [?]:
101
[0], given: 15
|
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. 
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 232
Kudos [?]:
1301
[0], given: 346
|
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... 
_________________
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for only $0.99 (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1842
Location: New York
Followers: 20
Kudos [?]:
291
[0], given: 5
|
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!!!!!
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7
Kudos [?]:
131
[0], given: 0
|
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.
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 232
Kudos [?]:
1301
[1] , given: 346
|
1
This post received
KUDOS
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.
_________________
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for only $0.99 (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
|
|
|
|
|
|
SVP
Joined: 17 Jun 2008
Posts: 1593
Followers: 7
Kudos [?]:
131
[0], given: 0
|
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!
|
|
|
|
|
|
Manager
Joined: 28 Jul 2004
Posts: 142
Location: Melbourne
Schools: Yale SOM, Tuck, Ross, IESE, HEC, Johnson, Booth
Followers: 1
Kudos [?]:
5
[0], given: 2
|
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?
_________________
kris
|
|
|
|
|
|
CEO
Joined: 17 Nov 2007
Posts: 3596
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 232
Kudos [?]:
1301
[0], given: 346
|
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
_________________
iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit
PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for only $0.99 (read more)
Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation
The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do.
|
|
|
|
|
|
Director
Joined: 29 Aug 2005
Posts: 891
Followers: 6
Kudos [?]:
87
[0], given: 7
|
suresh, walker, thanks for explanations. +2
|
|
|
|
|
|
Manager
Joined: 13 Jan 2009
Posts: 180
Schools: Harvard Business School, Stanford
Followers: 3
Kudos [?]:
14
[0], given: 9
|
I still dont get it. I think the answer is 'C' - X=Y.
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1842
Location: New York
Followers: 20
Kudos [?]:
291
[0], given: 5
|
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?
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
|
|
|
|
|
|
Manager
Joined: 13 Jan 2009
Posts: 180
Schools: Harvard Business School, Stanford
Followers: 3
Kudos [?]:
14
[0], given: 9
|
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!
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
