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

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by BTGmoderatorDC » Wed Oct 17, 2018 5:07 pm

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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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by Jay@ManhattanReview » Wed Oct 17, 2018 11:19 pm
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
Let the speed of Car J is x mph, thus, the speed of Car K = 80% of x = 0.8x mph;

Distance covered by Car K in 15 minutes = (0.8x)/4 = 0.2x miles

Since both the cars are moving on the same track and in the same direction, the relative speed of Car J = Speed of Car J - Speed of Car K = x - 0.8x = 0.2x mph

Thus, Car J had to cover the distance of 0.2x miles with the relative speed of 0.2x mph.

Time taken = Distance / Relative speed = (0.2x) / (0.2x) = 1 hours = 60 minutes

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Thu Oct 25, 2018 5:08 pm
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
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 and 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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by [email protected] » Thu Oct 25, 2018 6:57 pm
Hi All,

We're told that 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, both cars travel along the same route and Car K travels at a constant speed that is 80% the constant speed of Car J. We're asked for the number of MINUTES that will elapse before Car J catches up to Car K. This question can be solved rather easily by TESTing VALUES.

To start, we need two values - for the two speeds - in which one value is 80% of the other. You might immediately think about 80 and 100, but you could save some 'math time' if you use 8 and 10, or even 4 and 5. Since the question is based in MINUTES, we should set the speed in miles/minute.

Car J = 5 miles/minute
Car K = 4 miles/minute

Since Car K has a 15-minute "head start", it travels (15)(4) = 60 miles before Car J gets moving.
Each minute after that, Car J will travel 5 miles and Car K will travel 4 miles, so Car J will "catch up" 1 mile/minute.
With a 60 mile lead, Car J will need 60/1 = 60 minutes to catch Car K.

Final Answer: C

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by Scott@TargetTestPrep » Tue Oct 30, 2018 10:23 am
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
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, and create the equation:

rt = 0.8r(t + 15)

rt = 0.8rt + 12r

t = 0.8t + 12

0.2t = 12

t = 60

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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