gmattesttaker2 wrote:
David and Stacey are riding bicycles on a flat road at a constant rate. If Stacey is now three miles ahead of David, in how many minutes will Stacey be just two miles ahead of David?
1. Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
2.45 minutes ago Stacey was 4.5 miles ahead of David.
David is currently 3 miles behind Stacey.
The question stem asks how much time must pass for David to be only 2 miles behind Stacey.
In other words, we must determine how long it takes for David to CATCH UP by 1 mile.
Statement 1: Stacey is traveling at rate of 10 mph and David is traveling at a rate of 12 mph.
The CATCH-UP rate is equal to the DIFFERENCE between the two rates:
12-10 = 2 miles per hour.
Since David travels 12 miles per hour, while Stacey travels only 10 miles per hour, every hour David CATCHES UP by 2 miles.
Time for David to catch up by 1 mile = (distance to catch-up)/(catch-up rate) = 1/2 hour.
SUFFICIENT.
Statement 2: 45 minutes ago Stacey was 4.5 miles ahead of David.
Since 3/4 of an hour ago David was 4.5 miles behind Stacey, but now he is only 3 miles behind Stacey, it took David 3/4 of hour to CATCH UP by 3/2 miles.
Thus, the catch-up rate = (distance caught up)/(time) = (3/2)/(3/4) = (3/2)(4/3) = 2 miles per hour.
Since Statement 2 implies the same catch-up rate as Statement 1, SUFFICIENT.
The correct answer is
D.
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