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Two trains, Q and P, are traveling on straight and parallel [#permalink]
 23 May 2010, 09:24
Question Stats:
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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.
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Re: Data sufficiency - 600 level question [#permalink]
23 May 2010, 09:59
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]
23 May 2010, 15:52
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]
23 May 2010, 16:14
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: Data sufficiency - 600 level question
[#permalink]
23 May 2010, 16:14
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