Find all School-related info fast with the new School-Specific MBA Forum

It is currently 16 Sep 2014, 21:43

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A swimmer makes a round trip up and down the river. It takes

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
A swimmer makes a round trip up and down the river. It takes [#permalink] New post 11 Oct 2007, 18:49
A swimmer makes a round trip up and down the river. It takes him X hours. Next day he swims the same distance in the still water. It takes him Y hours. What can we say about X and Y?

A) X>Y
B) Y>X
C) Y^2=X^2
D) abs(Y)= abs (X)
E) none of the above
Manager
Manager
avatar
Joined: 03 Oct 2007
Posts: 61
Followers: 4

Kudos [?]: 13 [0], given: 0

 [#permalink] New post 11 Oct 2007, 19:32
what's the source of this question?
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Oct 2007, 19:35
Hyperstorm wrote:
what's the source of this question?


Challanges =)

NOTE: Just as I mentioned before: after #8 they are better. I took them in reverse order. So I have problems with ambiguity only now as I get closer to one. After #8 they are fine.
Manager
Manager
User avatar
Joined: 07 Sep 2007
Posts: 121
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: distance, speed [#permalink] New post 11 Oct 2007, 19:51
IrinaOK wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. Next day he swims the same distance in the still water. It takes him Y hours. What can we say about X and Y?

A) X>Y
B) Y>X
C) Y^2=X^2
D) abs(Y)= abs (X)
E) none of the above


A. Easy way to imagine it is if the drag from the water makes the swimmer take 1000 hours to get to the top, and <1 second to get back to the bottom. We realize we can achieve this condition no matter how long the distance is or how fast/slow the swimmer is.
Manager
Manager
User avatar
Joined: 07 Sep 2007
Posts: 121
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: distance, speed [#permalink] New post 11 Oct 2007, 19:58
Or another way to think about it: for the same distance, he is swimming a S-x speed for a longer time (A), and a faster speed (S-x) speed for a shorter time (B), where x is the drag from the water.

if we know A > B

then we also know Ax > Bx

where Ax is the speed decrease
and Bx is the speed increase
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
Re: distance, speed [#permalink] New post 11 Oct 2007, 20:13
Agree A it is.

here is my explanation:

S- speed of swimmer
C- current
D- distance up= distance down the river

X=D/(S-C) +D/(S+C) = 2SD/S^2-C^2
Y=2D/S

compare X to Y

X/Y = S^2/ (S^2-C^2) given that S always bigger than C
(logically speed of swimmer must be more than current`s, it can not be equal as well)

X/Y always more than 1, thus X is always more than Y.
Manager
Manager
avatar
Status: Post MBA, working in the area of Development Finance
Joined: 09 Oct 2006
Posts: 170
Location: Africa
Followers: 1

Kudos [?]: 2 [0], given: 1

Re: distance, speed [#permalink] New post 11 Oct 2007, 21:40
From the eqn. X/Y = S^2/ (S^2-C^2)
X will be greater than Y if S^2>(S^2-C^2) which implies that 0>-C^2 or C^2>0. This just implies that conclusion (X>Y) holds as long as the river has a current!

Regards.

IrinaOK wrote:
Agree A it is.

here is my explanation:

S- speed of swimmer
C- current
D- distance up= distance down the river

X=D/(S-C) +D/(S+C) = 2SD/S^2-C^2
Y=2D/S

compare X to Y

X/Y = S^2/ (S^2-C^2) given that S always bigger than C
(logically speed of swimmer must be more than current`s, it can not be equal as well)

X/Y always more than 1, thus X is always more than Y.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jul 2004
Posts: 5097
Location: Singapore
Followers: 19

Kudos [?]: 143 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Oct 2007, 21:42
Assume:
1) River is flowing in only direction at 2m/s
2) River is 100m long
3) Swimmer swims constantly at 10m/s on both days

So:
Time taken when river is not still = x hours over 200m
Total time = 100/12 + 100/8 = 125/6 hours = x

Time taken when river is still = y hours over 200m
Total time = 100/10 + 100/10 = 20 hours = y

So x>y --> Ans = A
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
Re: distance, speed [#permalink] New post 11 Oct 2007, 21:42
Artemov wrote:
From the eqn. X/Y = S^2/ (S^2-C^2)
X will be greater than Y if S^2>(S^2-C^2) which implies that 0>-C^2 or C^2>0. This just implies that conclusion (X>Y) holds as long as the river has a current!

Regards.

IrinaOK wrote:
Agree A it is.

here is my explanation:

S- speed of swimmer
C- current
D- distance up= distance down the river

X=D/(S-C) +D/(S+C) = 2SD/S^2-C^2
Y=2D/S

compare X to Y

X/Y = S^2/ (S^2-C^2) given that S always bigger than C
(logically speed of swimmer must be more than current`s, it can not be equal as well)

X/Y always more than 1, thus X is always more than Y.


Agree,

up and down river implies the C is more than zero.... :)
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Oct 2007, 21:43
ywilfred wrote:
Assume:
1) River is flowing in only direction at 2m/s
2) River is 100m long
3) Swimmer swims constantly at 10m/s on both days

So:
Time taken when river is not still = x hours over 200m
Total time = 100/12 + 100/8 = 125/6 hours = x

Time taken when river is still = y hours over 200m
Total time = 100/10 + 100/10 = 20 hours = y

So x>y --> Ans = A



Always, HAPPY to see you!
CEO
CEO
User avatar
Joined: 29 Mar 2007
Posts: 2593
Followers: 16

Kudos [?]: 189 [0], given: 0

GMAT Tests User
Re: distance, speed [#permalink] New post 11 Oct 2007, 22:12
IrinaOK wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. Next day he swims the same distance in the still water. It takes him Y hours. What can we say about X and Y?

A) X>Y
B) Y>X
C) Y^2=X^2
D) abs(Y)= abs (X)
E) none of the above


Seems like we are pretty close on doing the same challenges.

I said A b/c all we have to imagine here is that the time for day X (x hours) would be greater b/c the swimmer has to fight through more intense water.
Intern
Intern
avatar
Joined: 03 Sep 2007
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: distance, speed [#permalink] New post 12 Oct 2007, 01:11
IrinaOK wrote:
Artemov wrote:
From the eqn. X/Y = S^2/ (S^2-C^2)
X will be greater than Y if S^2>(S^2-C^2) which implies that 0>-C^2 or C^2>0. This just implies that conclusion (X>Y) holds as long as the river has a current!

Regards.

IrinaOK wrote:
Agree A it is.

here is my explanation:

S- speed of swimmer
C- current
D- distance up= distance down the river

X=D/(S-C) +D/(S+C) = 2SD/S^2-C^2
Y=2D/S

compare X to Y

X/Y = S^2/ (S^2-C^2) given that S always bigger than C
(logically speed of swimmer must be more than current`s, it can not be equal as well)

X/Y always more than 1, thus X is always more than Y.


Agree,

up and down river implies the C is more than zero.... :)



Not sure I understand why the answer is not E. I agree it's implicit that the rate of the swimmer is greater than that of the current, but there are a host of other assumptions which cannot be inferred from the question stem. First, how can we be sure that the swimmer's own speed (independent from the current) is constant on both days?

Second -- and admittedly, I may be overreaching here -- how can we know that the river's water flow rate stays constant? By way of illustration, I am quoting the following from http://www.fwee.org/hpar.html, The Foundation for Water and Energy Education Website: "The Guadalupe is an ever-changing river. Easily the most popular river to float in Texas, its character changes as the flow fluctuates with the release rate of water from Canyon dam turbines."


-GeoMATrace
Intern
Intern
User avatar
Joined: 09 Oct 2007
Posts: 17
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: distance, speed [#permalink] New post 12 Oct 2007, 01:57
IrinaOK wrote:
A swimmer makes a round trip up and down the river. It takes him X hours. Next day he swims the same distance in the still water. It takes him Y hours. What can we say about X and Y?

A) X>Y
B) Y>X
C) Y^2=X^2
D) abs(Y)= abs (X)
E) none of the above


It is A.

Let
V - speed of the swimmer.
Vr - speed of the river.
M - half of the distance the swimmer made.
Total time for the river (X) will be:
M/(V-Vr)+ M/(V+Vr)=X
Total time for the still water will be:
M/V+M/V = Y

X = (M(V+Vr)+M(V-Vr))/(V-Vr)*(V+Vr)
after simplification:
X = 2MV/(V^2-Vr^2). Y = 2M/V

Or

X = 2V divided by (V^2-Vr^2)/V

We get X = 2V/smth and Y = 2V/smth.

We should just compare those somethings.:)
(V^2-Vr^2)/V and V.

Mulitply both by V.
V^2 - Vr^2 < V^2. It is A. :)
Director
Director
avatar
Joined: 22 Aug 2007
Posts: 573
Followers: 1

Kudos [?]: 12 [0], given: 0

GMAT Tests User
Re: distance, speed [#permalink] New post 12 Oct 2007, 07:22
geomatrace wrote:
IrinaOK wrote:
Artemov wrote:
From the eqn. X/Y = S^2/ (S^2-C^2)
X will be greater than Y if S^2>(S^2-C^2) which implies that 0>-C^2 or C^2>0. This just implies that conclusion (X>Y) holds as long as the river has a current!

Regards.

IrinaOK wrote:
Agree A it is.

here is my explanation:

S- speed of swimmer
C- current
D- distance up= distance down the river

X=D/(S-C) +D/(S+C) = 2SD/S^2-C^2
Y=2D/S

compare X to Y

X/Y = S^2/ (S^2-C^2) given that S always bigger than C
(logically speed of swimmer must be more than current`s, it can not be equal as well)

X/Y always more than 1, thus X is always more than Y.


Agree,

up and down river implies the C is more than zero.... :)



Not sure I understand why the answer is not E. I agree it's implicit that the rate of the swimmer is greater than that of the current, but there are a host of other assumptions which cannot be inferred from the question stem. First, how can we be sure that the swimmer's own speed (independent from the current) is constant on both days?

Second -- and admittedly, I may be overreaching here -- how can we know that the river's water flow rate stays constant? By way of illustration, I am quoting the following from http://www.fwee.org/hpar.html, The Foundation for Water and Energy Education Website: "The Guadalupe is an ever-changing river. Easily the most popular river to float in Texas, its character changes as the flow fluctuates with the release rate of water from Canyon dam turbines."


-GeoMATrace


GeoMATrace,

Thank you for detailed explanations and interesting info,

here are few more notes:

1. That the river`s rate is less than that of swimmer`s is clearly stated in the fact that swimmer could still swim up the river (if the river`s rate is more, swimming up would be impossible).
2. Whether river`s rate changes next day or not does not matter. Stem says that next day the swimmer swims in the still water.
3. Agree, that the swimmer`s rate could change over day (swimmer felt tired, or something boosted his/her overall energy level), then the right asnwer is E. Whether X is more or less than Y will depend on the difference between the rates of swimmer and current, and the difference between the swimmer`s rates on the first and second day.
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1816
Followers: 8

Kudos [?]: 89 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2007, 07:27
I'm going with (E) as well... So many assumptions and the existence of this option make me choosing it :).... It's simply too much unrealistic :)
Senior Manager
Senior Manager
User avatar
Joined: 27 Aug 2007
Posts: 255
Followers: 1

Kudos [?]: 8 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2007, 07:48
Fig wrote:
I'm going with (E) as well... So many assumptions and the existence of this option make me choosing it :).... It's simply too much unrealistic :)


I have just plugged in some numbers and figured out that it is X>Y always.
I suggest you to do the same

Ans: A
Senior Manager
Senior Manager
avatar
Joined: 27 Jul 2006
Posts: 299
Followers: 1

Kudos [?]: 7 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Oct 2007, 07:58
A is right. I struggled with a question of this sort for some time. However, it takes more time to swilm against a current than it saves you swimming with it.

Kind of the same idea as appreciation. If every day a stock goes up 10 percent and then down 10 percent, eventually you will have a really really bad investment.
  [#permalink] 12 Oct 2007, 07:58
    Similar topics Author Replies Last post
Similar
Topics:
2 A swimmer makes a round trip up and down the river. It takes bigfernhead 11 06 Nov 2008, 15:03
3 Experts publish their posts in the topic M03 Q 36 icandy 20 12 Oct 2008, 14:56
1 Experts publish their posts in the topic A swimmer makes a round trip up and down the river. It takes seongbae 8 17 May 2008, 17:41
Experts publish their posts in the topic A swimmer makes a round trip up and down the river. It takes JDMBA 5 11 Nov 2007, 12:08
A swimmer makes a Round trip UP and DOWN the river. It takes Ozmba 2 09 Nov 2007, 08:37
Display posts from previous: Sort by

A swimmer makes a round trip up and down the river. It takes

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.