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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

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

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

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.

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...

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

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.

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.

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.

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?

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

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 }