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Asifpirlo
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.

Answer A
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Asifpirlo
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.

Answer: A
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Asifpirlo
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
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Can you pls explain why A is the answer?
In 5 mins A gained 250 m + x (Distance travelled by B in 5 mins)
As it does not mention that B is not moving, we cant consider x = 0
So, the answer be C
Bunuel
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.

Answer: A.
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Can you pls explain why A is the answer?
In 5 mins A gained 250 m + x (Distance travelled by B in 5 mins)
As it does not mention that B is not moving, we cant consider x = 0
So, the answer be C


B is moving, and that is already included in the 250 meters.

(1) is about the change in the gap: 5 minutes ago A was 200 meters behind, and now A is 50 meters ahead. So the gap changed by 250 meters in 5 minutes. You do not add B’s distance again, because this 250 is already A’s gain relative to B.
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