A gang of criminals hijacked a train heading due south. At exactly the same time, a police car located \(50\) miles

[BTGmoderatorLU]

Source: Manhattan Prep

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 OA is C

[Brent@GMATPrepNow]

Source: Manhattan Prep

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 OA is C

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: C

Cheers,
Brent