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A gang of criminals hijacked a train heading due south.

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stonecold
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A gang of criminals hijacked a train heading due south. [#permalink] New post   03 May 2017, 03:07
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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
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C
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   03 May 2017, 03:53
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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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A gang of criminals hijacked a train heading due south. [#permalink] New post   Updated on: 01 Apr 2020, 05:31
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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:
Show Spoiler
C


Cheers,
Brent

Originally posted by BrentGMATPrepNow on 03 May 2017, 07:49.
Last edited by BrentGMATPrepNow on 01 Apr 2020, 05:31, edited 1 time in total.
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   06 May 2017, 17:51
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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


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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A gang of criminals hijacked a train heading due south. [#permalink] New post   19 Dec 2020, 10:08
Top Contributor
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


\(Distance = Speed*Time\)

The train is going to the south and the police car was at the north. So, the Police car was behind the train and chasing the tran.

The Poice care 50 miles behind the train with a parallel roadway. While police will cover this 50 miles the train will go ahead and that will \(50T\). So, the police will have to cover the \(50+50T\) distance.

The police will also take T hours to go \(50+50T\) distance.

Thus,

\(80T\)=\(50+50T\)

\(30T=50\)

\(T=\frac{50}{30}*60=100\) minutes or 1 hour 40 minutes

The answer is \(C\)
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   09 Nov 2022, 14:42
BrentGMATPrepNow wrote:
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:
Show Spoiler
C


Cheers,
Brent



Brilliant explanation BrentGMATPrepNow. If solving algebraic way, is the below word equation correct? Thanks Brent
train's distance = Police's car distance
50T= 50+80T
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   09 Nov 2022, 15:07
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Kimberly77 wrote:

Brilliant explanation BrentGMATPrepNow. If solving algebraic way, is the below word equation correct? Thanks Brent
train's distance = Police's car distance
50T= 50+80T


No, we can't say that those distances are the same, since the police car starting 50 miles north of the criminals.
This means the police car travelled for 50 miles more than the criminals travel.
So, we can write: (police's travel distance) - (criminal's travel distance) = 50 miles

Or we can write: (police's travel TIME) = (criminal's travel TIME)
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   09 Nov 2022, 15:30
BrentGMATPrepNow wrote:
Kimberly77 wrote:

Brilliant explanation BrentGMATPrepNow. If solving algebraic way, is the below word equation correct? Thanks Brent
train's distance = Police's car distance
50T= 50+80T


No, we can't say that those distances are the same, since the police car starting 50 miles north of the criminals.
This means the police car travelled for 50 miles more than the criminals travel.
So, we can write: (police's travel distance) - (criminal's travel distance) = 50 miles

Or we can write: (police's travel TIME) = (criminal's travel TIME)


Make sense now and thanks BrentGMATPrepNow for the clarificatin :thumbsup: :please:
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Re: A gang of criminals hijacked a train heading due south. [#permalink] New post   10 Dec 2023, 04:47
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Re: A gang of criminals hijacked a train heading due south. [#permalink]
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