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# m03 #36

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m03 #36 [#permalink]  26 Mar 2011, 20:37
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!

A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above

Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.
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Re: m03 #36 [#permalink]  26 Mar 2011, 22:06
Hello mate !
Read the stem as you read a CR. And read it critically - you may not have to do any of that math.

A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?

See the italics. First italic implies she had to face resistance of the river one way. The second time she swam in still water, she did not face any resistance. So the text is alluding that Y is less than X.

gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!

No the concept of average speed tells me that the even when she traveled the same distance in both cases, her average speed was low the first time and it was high the second time reducing the time marginally. Hope that helps !
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Re: m03 #36 [#permalink]  26 Mar 2011, 23:36
Thanks a lot Dude!

No the concept of average speed tells me that the even when she traveled the same distance in both cases, her average speed was low the first time and it was high the second time reducing the time marginally. ! Hope that helps

But it just states that the speed is the same in the first and second time. It doesn't mention any variation. It also states that the distance is the same. I can't see any reasonable explanation for variation in time using the premises stated in this question.

The fact that still water has nothing to do with this question is definitely true. It didn't mention if the speed is related to ground or water. I assume that it is ground.

distance/speed = time

Sorry, I didn't get it your explanation.

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Re: m03 #36 [#permalink]  27 Mar 2011, 00:28
gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!

A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above

Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.

Let the speed of the swimmer in still water be $$v_{still}$$
Let the speed of the stream be $$v_{stream}$$
Let the one way distance be $$D$$

Time going upstream
$$T_{upstream} = \frac{D}{v_{still}-v_{stream}}$$
$$T_{downstream} = \frac{D}{v_{still}+v_{stream}}$$

$$Total Time = X = T_{upstream}+T_{downstream} = \frac{D}{v_{still}-v_{stream}}+\frac{D}{v_{still}+v_{stream}}$$
$$X = \frac{D*v_{still}+D*v_{stream}+D*v_{still}-D*v_{stream}}{(v_{still})^2-(v_{stream})^2}$$
$$X = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$

Likewise;
$$Y = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
But here;
$$v_{stream}=0$$
When $$v_{stream}=0$$, the denominator becomes maximum making the result to be minimum.

Thus, we can say that $$Y$$ will be minimum time taken by the swimmer at $$v_{stream}=0$$.

Hence;
$$X>Y$$

Ans: "A"
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Re: m03 #36 [#permalink]  27 Mar 2011, 00:49
fluke wrote:
gsagula wrote:
Why the answer is A? If something travels at the same speed and at the same distance twice is it not the same number of hours? Thanks!

A swimmer makes a round trip up and down the river which takes her X hours. If the next day she swims the same distance with the same speed in still water, which takes her Y hours, which of the following statements is true?
• X>Y
• X<Y
• X=Y
• X=1/2*Y
• none of the above

Pick numbers and then check them against the options. Take 12 km as the distance traveled up/down the river, and assume the swimmer's speed to be 4 km/h; the current being 2 km/h, which means 6 km/h down the river and 2 km/h up the river. Going upriver takes 2 hours, return journey takes 6, thus a total of 8 hours. In still water, 24 km requires 6 hours. Thus and x=8 and y=6.
1. or True.
2. or False.
3. or False.
4. or False.
5. None of the above. False because A is true.
The correct answer is A.

Let the speed of the swimmer in still water be $$v_{still}$$
Let the speed of the stream be $$v_{stream}$$
Let the one way distance be $$D$$

Time going upstream
$$T_{upstream} = \frac{D}{v_{still}-v_{stream}}$$
$$T_{downstream} = \frac{D}{v_{still}+v_{stream}}$$

$$Total Time = X = T_{upstream}+T_{downstream} = \frac{D}{v_{still}-v_{stream}}+\frac{D}{v_{still}+v_{stream}}$$
$$X = \frac{D*v_{still}+D*v_{stream}+D*v_{still}-D*v_{stream}}{(v_{still})^2-(v_{stream})^2}$$
$$X = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$

Likewise;
$$Y = \frac{2D*v_{still}}{(v_{still})^2-(v_{stream})^2}$$
But here;
$$v_{stream}=0$$
When $$v_{stream}=0$$, the denominator becomes maximum making the result to be minimum.

Thus, we can say that $$Y$$ will be minimum time taken by the swimmer at $$v_{stream}=0$$.

Hence;
$$X>Y$$

Ans: "A"

Thanks Dude!

What did the question mean was that in the second event she had constant speed? In this case it was inaccurately stated, because I can assume that she had the same speed of the day before.

What the question states is that speed is the same.

We don't have two speeds:
v_{still} v_{stream}

In other words, the question states that both events had the same distance and the same speed.
In this case it doesn't matter where she is swimming.
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Re: m03 #36 [#permalink]  27 Mar 2011, 00:58
gsagula wrote:

Thanks Dude!

What the question states is that speed is the same.

We don't have two speeds:
v_{still} v_{stream}

In other words, the question states that both events had the same distance and the same speed.

In this case it doesn't matter where she is swimming.

Speed of the swimmer is the same but the speed of the stream vary.

$$v_{still}$$ is the speed of the swimmer, which I agree, is same in both instances.
$$v_{stream}$$ is the speed of the stream(flowing water) is different in both instances (in the first instance it has some speed more than 0 and in second, it is still(not flowing) and is 0)
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Re: m03 #36 [#permalink]  27 Mar 2011, 01:22
fluke wrote:
gsagula wrote:

Thanks Dude!

What the question states is that speed is the same.

We don't have two speeds:
v_{still} v_{stream}

In other words, the question states that both events had the same distance and the same speed.

In this case it doesn't matter where she is swimming.

Speed of the swimmer is the same but the speed of the stream vary.

$$v_{still}$$ is the speed of the swimmer, which I agree, is same in both instances.
$$v_{stream}$$ is the speed of the stream(flowing water) is different in both instances (in the first instance it has some speed more than 0 and in second, it is still(not flowing) and is 0)

Ow... Thanks again!

So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.
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Re: m03 #36 [#permalink]  27 Mar 2011, 01:26
Thank you very much fluke and gmat1220!
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Re: m03 #36 [#permalink]  27 Mar 2011, 07:14
gsagula wrote:
Ow... Thanks again!

So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.

Sorry, this inference is wrong. Lets says you travel from SF to NY with the wind and make 150mph and travel back to SF from NY at 100mph against the wind. The average speed is not (150 + 100)/2 = 125mph. It will be less than 125mph. Similarly, if the swimmer travels 50mph with the river and back 30mph against the river. Her average speed is less than (50+30)/2 i.e. 40mph. Try this when your cruising in your car next time.
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Re: m03 #36 [#permalink]  27 Mar 2011, 08:38
gmat1220 wrote:
gsagula wrote:
Ow... Thanks again!

So, I believe that this question is flaw because it's not speed, It's relative speed. The question should specify if the speed is in relationship to ground or water (search for Relative Velocity). One more thing, if she has resistance to go up, she will have this energy in her favor to go down, and in this case the average will remain exactly the same.

Sorry, this inference is wrong. Lets says you travel from SF to NY with the wind and make 150mph and travel back to SF from NY at 100mph against the wind. The average speed is not (150 + 100)/2 = 125mph. It will be less than 125mph. Similarly, if the swimmer travels 50mph with the river and back 30mph against the river. Her average speed is less than (50+30)/2 i.e. 40mph. Try this when your cruising in your car next time.

Right, it is Total Distance/Total Time, however the question states that "speed" is the same, and it kills the argument.
Re: m03 #36   [#permalink] 27 Mar 2011, 08:38
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# m03 #36

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