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Cars J and K are making the trip from City A to City B. Car

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Cars J and K are making the trip from City A to City B. Car

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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

OA C

Source: Princeton Review

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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

OA C

Source: Princeton Review
We can let r = the rate of Car J, and so 0.8r = the rate of Car K. We also let t = the time of Car J, in minutes, and (t + 15) = the time of Car K, in minutes, when Car J catches up to Car K. We can create the equation:

rt = 0.8r(t + 15)

rt = 0.8rt + 12r

t = 0.8t + 12

0.2t = 12

t = 60

Answer: C

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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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