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.
Answer: A
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