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Two trains, Q and P, are traveling on straight and parallel

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Two trains, Q and P, are traveling on straight and parallel [#permalink] New post 23 May 2010, 08:24
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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.
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Re: Data sufficiency - 600 level question [#permalink] New post 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

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Re: Data sufficiency - 600 level question [#permalink] New post 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

My answer: D


sorry..i still can't get it.
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Re: Data sufficiency - 600 level question [#permalink] New post 23 May 2010, 15:14
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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.

Answer: D.

Hope it's clear.
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Re: Two trains, Q and P, are traveling on straight and parallel [#permalink] New post 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.


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

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