Two cyclists are racing up a mountain at different constant : GMAT Data Sufficiency (DS)
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# Two cyclists are racing up a mountain at different constant

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Two cyclists are racing up a mountain at different constant [#permalink]

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21 Aug 2013, 17:08
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Two cyclists are racing up a mountain at different constant rates. Cyclist A is now 50 meters ahead of cyclist B. How many minutes from now will cyclist A be 150 meters ahead of cyclist B?

(1) 5 minutes ago, cyclist A was 200 meters behind cyclist B.
(2) Cyclist A is moving 25% faster than cyclist B.
[Reveal] Spoiler: OA

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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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21 Aug 2013, 20:37
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In 5 mins he could cover 250 meters. Hence he is 50 m/mins faster. So it will take 2 mins
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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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21 Aug 2013, 21:13
Asifpirlo wrote:
Two cyclists are racing up a mountain at different constant rates. Cyclist A is now 50 meters ahead of cyclist B. How many minutes from now will cyclist A be 150 meters ahead of cyclist B?

(1) 5 minutes ago, cyclist A was 200 meters behind cyclist B.
(2) Cyclist A is moving 25% faster than cyclist B.

The question basically asks the relative speed of A wrt B. This means that we don't need to find the individual speed of A & B

Statement 1
In 5 minutes A cycled 250 m more than the distance traveled by B.
In other words, Relative Speed of A is = 250/5 = 50m/min
Time to cover extra 100 m will be 2 Min
Thus Sufficient.

Statement 2
In 5 minutes A cycled 250 m more than the distance traveled by B.
Speed of A is 25% more than B's.
This option will give different answers depending on the speed of B.
Thus Insufficient.

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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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22 Aug 2013, 00:51
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Two cyclists are racing up a mountain at different constant rates. Cyclist A is now 50 meters ahead of cyclist B. How many minutes from now will cyclist A be 150 meters ahead of cyclist B?

(1) 5 minutes ago, cyclist A was 200 meters behind cyclist B. In 5 minutes A gained 200+50=250 meters --> 50 meters in 1 minute. To gain additional 100 meters he'll need 2 more minutes. Sufficient.

(2) Cyclist A is moving 25% faster than cyclist B. If both cyclist are very slow A'll need more minutes than if both cyclist are very fast. Not sufficient.

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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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10 May 2015, 03:17
Hello from the GMAT Club BumpBot!

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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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18 Aug 2015, 09:26
I marked C but i do realize my mistake now . still sorry guys not able to understand your explanation 700+ problms distance/rate kills me

i was able to form the answer in some other way ;

let x be the distance travlled by B in those 5 min in which x got ahead .
Speed of A = A , Speed of B=B

distance travlled by A will 200m + 50m + x(distance by B ) = A * 5min ------ 1

distance by B = X = B * 5 min ------2

Subtract 2 from 1

250 + X - X = 5(A-B)

250/(A-B)=5min

so 250 distance will be covered in 5 min
so 150 can be done in 3 min
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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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29 Aug 2016, 23:17
Hello from the GMAT Club BumpBot!

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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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08 Nov 2016, 16:16
Since both are moving at the same time wouldn't the actual speed be the sum of speed of A and the speed of B making the answer C

As it might take A 2 min but then B would have moved too
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Re: Two cyclists are racing up a mountain at different constant [#permalink]

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11 Nov 2016, 07:14
Asifpirlo wrote:
Two cyclists are racing up a mountain at different constant rates. Cyclist A is now 50 meters ahead of cyclist B. How many minutes from now will cyclist A be 150 meters ahead of cyclist B?

(1) 5 minutes ago, cyclist A was 200 meters behind cyclist B.
(2) Cyclist A is moving 25% faster than cyclist B.

We are given that two cyclists are racing up a mountain at different constant rates and that cyclist A is now 50 meters ahead of cyclist B. We need to determine in how many minutes cyclist A be 150 meters ahead of cyclist B.

Statement One Alone:

5 minutes ago, cyclist A was 200 meters behind cyclist B.

Since we know that 5 minutes ago cyclist A was 200 meters behind cyclist B and that cyclist A is now 50 meters ahead of cyclist B, we can determine the “catch-up rate” of cyclist A. Since rate = distance/time, the catch up rate of cyclist A is: 250/5 = 50 meters per minute.

We now can determine how long it takes cyclist A to travel 150 meters ahead of cyclist B. Since cyclist A is now 50 meters ahead of cyclist B, we are determining how long it will take cyclist A to travel 100 meters farther than cyclist B.

Since time = distance/rate, it will take cyclist A 100/50 = 2 minutes to be 150 meters ahead of cyclist B. Statement one alone is sufficient to answer question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

Cyclist A is moving 25% faster than cyclist B.

Although we know that cyclist A is traveling 25% faster that cyclist B, we still cannot determine the “catch-up rate” of cyclist A and thus cannot determine in how many minutes cyclist A will be 150 meters ahead of cyclist. Statement two is not sufficient.

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Two cyclists are racing up a mountain at different constant [#permalink]

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29 Nov 2016, 05:08
Asifpirlo wrote:
Two cyclists are racing up a mountain at different constant rates. Cyclist A is now 50 meters ahead of cyclist B. How many minutes from now will cyclist A be 150 meters ahead of cyclist B?

(1) 5 minutes ago, cyclist A was 200 meters behind cyclist B.
(2) Cyclist A is moving 25% faster than cyclist B.

We need either their individual speeds or their relative velocities in miles/minute

from 1

Speed of A - Speed of B = 250/5 = 50m/m thus with this relative velocity A will be ahead by an extra 100 m in 2 minutes... suff

from2

Speed of A = 1.25 speed of B thus relative velocity is 0.25 Speed of A ... no idea about actual speed of A ... insuff

A
Two cyclists are racing up a mountain at different constant   [#permalink] 29 Nov 2016, 05:08
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