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Two trains, Q and P, are traveling on straight and parallel [#permalink]
23 May 2010, 08:24

00:00

A

B

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D

E

Difficulty:

15% (low)

Question Stats:

79% (01:47) correct
21% (01:05) wrong based on 75 sessions

Two trains, Q and P, are traveling on straight and parallel tracks. Q and P are traveling at different constant rates. If train Q is now 3 miles ahead of train P, in how many minutes will train Q be 6 miles ahead of train P ?

(1) Train Q is traveling at a speed of 150 mph and train P is traveling at a speed of 120 mph.

(2) Four minutes ago, train Q was 1 mile ahead of train P.

Re: Data sufficiency - 600 level question [#permalink]
23 May 2010, 08:59

shekar123 wrote:

Could anyone help me with this question.

Two trains, Q and P, are traveling on straight and parallel tracks. Q and P are traveling at different constant rates. If train Q is now 3 miles ahead of train P, in how many minutes will train Q be 6 miles ahead of train P ?

(1) Train Q is traveling at a speed of 150 mph and train P is traveling at a speed of 120 mph. (2) Four minutes ago, train Q was 1 mile ahead of train P.

It really helps if the stop watch is included in these questions .

Anyways, here is my take on this:

speed of P = v1 speed of Q = v2 Assuming they are traveling in the same direction combined speed of P and Q = v2-v1 Distance in between them = 3 miles Time required by train Q to be 6 miles ahead = t Distance needed by Q = 3 miles

1) v2 = 150 ad v1 = 120 t = d/v = 3/(150-120) = 3/30 hrs = 1/10 *(60) = 6 mins Sufficient

2) So 4 mins ago train Q was 1 mile ahead, and now it is 3 miles ahead, this means it traveled 2 miles in 4 mins

Thus, combined speed = d/4 = 2/4 v2 - v1 = 2/4 miles per minute (note it is not mph)

from this we can find t = 3/(v2 - v1) = 3*4/2 = 6 mins Sufficient

My answer: D

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Re: Data sufficiency - 600 level question [#permalink]
23 May 2010, 14:52

dimitri92 wrote:

shekar123 wrote:

Could anyone help me with this question.

Two trains, Q and P, are traveling on straight and parallel tracks. Q and P are traveling at different constant rates. If train Q is now 3 miles ahead of train P, in how many minutes will train Q be 6 miles ahead of train P ?

(1) Train Q is traveling at a speed of 150 mph and train P is traveling at a speed of 120 mph. (2) Four minutes ago, train Q was 1 mile ahead of train P.

It really helps if the stop watch is included in these questions .

Anyways, here is my take on this:

speed of P = v1 speed of Q = v2 Assuming they are traveling in the same direction combined speed of P and Q = v2-v1 Distance in between them = 3 miles Time required by train Q to be 6 miles ahead = t Distance needed by Q = 3 miles

1) v2 = 150 ad v1 = 120 t = d/v = 3/(150-120) = 3/30 hrs = 1/10 *(60) = 6 mins Sufficient

2) So 4 mins ago train Q was 1 mile ahead, and now it is 3 miles ahead, this means it traveled 2 miles in 4 mins

Thus, combined speed = d/4 = 2/4 v2 - v1 = 2/4 miles per minute (note it is not mph)

from this we can find t = 3/(v2 - v1) = 3*4/2 = 6 mins Sufficient

Re: Data sufficiency - 600 level question [#permalink]
23 May 2010, 15:14

1

This post received KUDOS

Expert's post

shekar123 wrote:

Could anyone help me with this question.

Two trains, Q and P, are traveling on straight and parallel tracks. Q and P are traveling at different constant rates. If train Q is now 3 miles ahead of train P, in how many minutes will train Q be 6 miles ahead of train P ?

(1) Train Q is traveling at a speed of 150 mph and train P is traveling at a speed of 120 mph.

(2) Four minutes ago, train Q was 1 mile ahead of train P.

Given distance between T and Q now is 3 miles (T---Q). Question in how many minutes distance will become 6 miles (T------Q), or in how many minutes Q will gain 3 more miles over T.

(1) Q is gaining 150-120=30 miles per hour over T. 3 miles will be gained in 1/10th of an hour, or in 6 minutes. Sufficient.

(2) As distance between T and Q now is 3 miles, therefore in 4 minutes Q gained over T 3-1=2 miles. 2 miles per 4 minute --> 3 miles will be gained in 1.5*4=6 minutes. Sufficient.

Re: Two trains, Q and P, are traveling on straight and parallel [#permalink]
16 Oct 2013, 06:46

(1) Train Q is traveling at a speed of 150 mph and train P is traveling at a speed of 120 mph.

Distance between 2 objects is known ; and statement 1 gives speeds of both the objects; Now this information should be enough to get all the parameters, as speed=distance/time is just one equation with 3 variables.

(2) Four minutes ago, train Q was 1 mile ahead of train P.

According to statement 2, Q takes 4 minutes to create a gap of 2 miles, so wrt to P, we can calculate its speed (relative speed). Hence this is sufficient.

There is no need to solve this numerically.

-- Kudos are appreciated.

gmatclubot

Re: Two trains, Q and P, are traveling on straight and parallel
[#permalink]
16 Oct 2013, 06:46

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