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

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

by AAPL » Fri Nov 02, 2018 5:00 am
Princeton Review

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.
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by Brent@GMATPrepNow » Fri Nov 02, 2018 5:18 am
AAPL wrote:Princeton Review

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
C.
Car K travels at a constant speed that is 80% the constant speed of Car J
Let's assign some nice values to the speeds.
Let's say Car J travels 40 miles per hour
So, Car K travels 32 miles per hour

Car J departs from City A 15 minutes after Car K does
Let's calculate the distance that Car K travels in those 15 minutes
Distance = (speed)(time)
So, distance = (32 mph)(1/4 hours) = 8 miles

Once Car J starts driving, it is 8 miles behind Car K
So, we want to SHRINK the 8-mile gap to zero.

Since Car K's speed is 8 mph greater than Car J's speed, we know the gap is shrinking at a rate of 8 mph

Time = distance/rate
So, the time to shrink the 8-mile gap = 8/8
= 1 hour
= 60 minutes

Answer: C

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Brent
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by [email protected] » Fri Nov 02, 2018 3:36 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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Rich
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by Scott@TargetTestPrep » Sat Nov 03, 2018 4:54 pm
AAPL wrote:Princeton Review

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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