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A swimmer makes a round trip up and down the river. It takes [#permalink]
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

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.

Re: distance, speed [#permalink]
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.

Re: distance, speed [#permalink]
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

Re: distance, speed [#permalink]
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.

Re: distance, speed [#permalink]
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....

Re: distance, speed [#permalink]
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.

Re: distance, speed [#permalink]
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."

Re: distance, speed [#permalink]
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.

Re: distance, speed [#permalink]
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.

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.