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A gang of criminals hijacked a train heading due south.

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A gang of criminals hijacked a train heading due south. [#permalink]

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A gang of criminals hijacked a train heading due south. At exactly the same time, a police car located 50 miles north of the train started driving south toward the train on an adjacent roadway parallel to the train track. If the train traveled at a constant rate of 50 miles per hour, and the police car traveled at a constant rate of 80 miles per hour, how long after the hijacking did the police car catch up with the train?

A. 1 hour
B. 1 hour and 20 minutes
C. 1 hour and 40 minutes
D. 2 hours
E. 2 hours and 20 minutes
[Reveal] Spoiler: OA

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Re: A gang of criminals hijacked a train heading due south. [#permalink]

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stonecold wrote:
A gang of criminals hijacked a train heading due south. At exactly the same time, a police car located 50 miles north of the train started driving south toward the train on an adjacent roadway parallel to the train track. If the train traveled at a constant rate of 50 miles per hour, and the police car traveled at a constant rate of 80 miles per hour, how long after the hijacking did the police car catch up with the train?

A. 1 hour
B. 1 hour and 20 minutes
C. 1 hour and 40 minutes
D. 2 hours
E. 2 hours and 20 minutes


The two vehicles are moving in the same direction, with one chasing the other.
To determine how long it will take the rear vehicle to catch up, subtract the rates to find out how quickly the rear vehicle is gaining on the one in front.
The police car gains on the train at a rate of 80 - 50 = 30 miles per hour. Since the police car needs to close a gap of 50 miles, plug into D = RT to find the time:
50 = 30t
5/3 = t
The time it takes to catch up is 5/3 hours, or 1 hour and 40 minutes

Hence option C is correct
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Re: A gang of criminals hijacked a train heading due south. [#permalink]

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stonecold wrote:
A gang of criminals hijacked a train heading due south. At exactly the same time, a police car located 50 miles north of the train started driving south toward the train on an adjacent roadway parallel to the train track. If the train traveled at a constant rate of 50 miles per hour, and the police car traveled at a constant rate of 80 miles per hour, how long after the hijacking did the police car catch up with the train?

A. 1 hour
B. 1 hour and 20 minutes
C. 1 hour and 40 minutes
D. 2 hours
E. 2 hours and 20 minutes


This is a shrinking gap question.

Train's speed = 50 miles per hour
Police card's speed = 80 miles per hour
80 miles per hour - 50 miles per hour = 30 miles per hour
So, the gap between the train and the police car DECREASES at a rate of 30 miles per hour

Original gap (aka distance) = 50 miles
Time = distance/rate
So, time to close gap = 50/30 hours
= 5/3 hours
= 1 2/3 hours
= 1 hour and 40 minutes

Answer:
[Reveal] Spoiler:
C


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Re: A gang of criminals hijacked a train heading due south. [#permalink]

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New post 06 May 2017, 16:51
stonecold wrote:
A gang of criminals hijacked a train heading due south. At exactly the same time, a police car located 50 miles north of the train started driving south toward the train on an adjacent roadway parallel to the train track. If the train traveled at a constant rate of 50 miles per hour, and the police car traveled at a constant rate of 80 miles per hour, how long after the hijacking did the police car catch up with the train?

A. 1 hour
B. 1 hour and 20 minutes
C. 1 hour and 40 minutes
D. 2 hours
E. 2 hours and 20 minutes


Since the police car travels 80 - 50 = 30 mph faster than the criminals, the distance between the police and the criminals decreases 30 miles every hour. Since they were 50 miles apart and time = distance/rate, the police car will catch up to the train in 50/30 = 5/3 hours, or 1 and ⅔ hours. Since ⅔ of an hour is 40 minutes, the police car will catch up to the train 1 hour and 40 minutes.

Answer: C
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Re: A gang of criminals hijacked a train heading due south.   [#permalink] 06 May 2017, 16:51
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