It is currently 14 Dec 2018, 01:43
My Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

A gang of criminals hijacked a train heading due south. At e

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1659 [1] , given: 396

CAT Tests
A gang of criminals hijacked a train heading due south. At e [#permalink] New post 18 Jun 2018, 14:56
1
This post received
KUDOS
Expert's post
00:00

Question Stats:

66% (01:38) correct 33% (02:13) wrong based on 18 sessions
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

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Intern
Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 15 [0], given: 16

Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 19 Jun 2018, 22:57
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


Sandy please, can you provide the explanation for this?
3 KUDOS received
Manager
Manager
Joined: 29 Nov 2017
Posts: 190
Location: United States
GRE 1: Q142 V146
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 77 [3] , given: 99

Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 21 Jun 2018, 09:35
3
This post received
KUDOS
The distance is 50 mile.

Train speed is 80.

police speed is 50

Remember - whenever there is a chase - the speed are substracted to find the rate at which the gap is shrinking- here

80-50 = 30

We need to find Time=

Distance
---------
Speed

50/30*60(hour)=
1.40 hour

annswer is C
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1659 [0], given: 396

CAT Tests
Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 10 Jul 2018, 05:29
Expert's post
Explanation

In this “chase” problem, 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 the D = RT formula to find the time:

50 = 30t
\(\frac{5}{3} = t\)

The time it takes to catch up is \(\frac{5}{3}\) hours, or 1 hour and 40 minutes.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Intern
Joined: 09 Jul 2018
Posts: 10
Followers: 1

Kudos [?]: 10 [0], given: 0

Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 10 Jul 2018, 12:11
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.

[Reveal] Spoiler:
C


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

Intern
Intern
Joined: 04 Jun 2018
Posts: 8
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 10 Jul 2018, 19:08
To simplify:
The range between a police car and a train: 50 miles
The difference in 2 vehicles speed: 30 miles/hour
=> In 1 hour, the range reduces 30 miles. Then only 20 miles are left after 1 hour, obviously it does not need to take a full 1 hour to reduce 20 miles range. Only need: (20/30)* 60mins = 40 mins
The answer is: 1hour + 40 mins
Manager
Manager
Joined: 09 Nov 2018
Posts: 75
Followers: 0

Kudos [?]: 5 [0], given: 1

CAT Tests
Re: A gang of criminals hijacked a train heading due south. At e [#permalink] New post 18 Nov 2018, 21:06
Can anybody explain with picture/ diagram ?
Re: A gang of criminals hijacked a train heading due south. At e   [#permalink] 18 Nov 2018, 21:06
Display posts from previous: Sort by

A gang of criminals hijacked a train heading due south. At e

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.