sandy 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
When elements compete, SUBTRACT THEIR RATES.
The rate for the police = 80 mph, while the rate for the gang = 50 mph.
Thus, every hour the police travel 80 miles, while the gang travels only 50 miles.
The difference between their rates = 80-50 = 30 mph.
Implication of this rate difference:
Every hour the police travel 30 more miles than the gang.
As a result, every hour the police CATCH-UP by 30 miles.
Since the police must catch up by 50 miles, and their catch-up rate is 30 mph, we get:
\(Catch-up-time = \frac{catch-up-distance}{catch-up-rate} = \frac{50}{30} = \frac{5}{3}\) hours = 1 hour and 40 minutes.
Alternate approach:
MAP OUT the distances in 30-minute increments until the police have traveled the same total distance as the gang.
Since the gang's rate = 50 mph, the gang travels 25 miles every 30 minutes.
Since the police's rate = 80 mph, the police travel 40 miles every 30 minutes.
Start --> gang = 50 miles, police = 0 miles
30 minutes later --> gang = 50+25 = 75 miles, police = 0+40 = 40 miles
1 hour later --> gang = 75+25 = 100 miles, police = 40+40 = 80 miles
1.5 hours later --> gang = 100+25 = 125 miles, police = 80+40 = 120 miles
2 hours later --> gang = 125+25 = 150 miles, police = 120+40 = 160 miles
Since the police are 5 miles behind after 1.5 hours (120 miles for the police versus 125 miles for the gang) but 10 miles ahead after 2 hours (160 miles for the police versus 150 miles for the gang), the time for the police to catch up must be between 1.5 hours and 2 hours.
Only C works.
_________________
GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY at gmail
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY at gmail