BTGmoderatorDC wrote:Cars J and K are making the trip from City A to City B. Car J departs from City A 15 minutes after Car K does, and both cars travel along the same route. If Car K travels at a constant speed that is 80% the constant speed of Car J, then how many minutes will elapse before Car J catches up to Car K?
A) 20
B) 45
C) 60
D) 75
E) 1230
Since J catches up to K, J and K travel the same distance.
Let t = J's time.
Since K leaves 15 minutes earlier than J, K's time = t+15.
Rate and time have a RECIPROCAL RELATIONSHIP.
Since K's rate is 80% of J's rate, the ratio of K's rate to J's rate = 8/10 = 4/5.
Thus, the ratio of K's time to J's time = 5/4:
(t+15)/t = 5/4
4t + 60 = 5t
60 = t
The correct answer is
C.
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