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
Cars J and K are making the trip from City A to City B. Car
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Let the speed of Car J is x mph, thus, the speed of Car K = 80% of x = 0.8x mph;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
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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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: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
rt = 0.8r(t + 15)
rt = 0.8rt + 12r
t = 0.8t + 12
0.2t = 12
t = 60
Answer: C
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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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: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
rt = 0.8r(t + 15)
rt = 0.8rt + 12r
t = 0.8t + 12
0.2t = 12
t = 60
Answer: C
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