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Station Q is to the East of Station T. At 12 noon, a train starts from

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Station Q is to the East of Station T. At 12 noon, a train starts from  [#permalink]

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New post 17 Jun 2018, 05:34
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Question Stats:

78% (01:19) correct 22% (01:18) wrong based on 107 sessions

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Station Q is to the East of Station T. At 12 noon, a train starts from Station Q and travels at a constant speed of x mph towards Station T. At 12 noon of the same day, another train starts from Station T and travels at a constant speed of y mph towards Station Q. At what time will the trains meet?

(1) y = 4x/3
(2) x = 100 mph

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Station Q is to the East of Station T. At 12 noon, a train starts from  [#permalink]

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New post Updated on: 17 Jun 2018, 08:52
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Let the distance travaled by 1st train be A and distance travaled by 2st train be B.
Since trains started at the same time, they wil spend the same time to meet each other.

Time=A/x=B/y. Clearly we have 1 equation and 4 variables. To solve the problem we have to have extra 3 equations. Let's check:

(1) y = 4x/3. Then A/x=B/4x/3, A=3/4B. Not sufficient

(2) x = 100 mph. Then A/100=B/y. Not sufficient

(1)+(2) A=3/4B and A/100=B/y. 3/4A/100=B/y, y=400/3. Not sufficient.

Answer E.

P.S. Notice that both statements give us only 2 of 3 necessary equations. If you are lack in time, you can solve the problem based on reviewing how number of variables and equations match without doing calculations. It would save a good amount of time.

Originally posted by Hero8888 on 17 Jun 2018, 06:00.
Last edited by Hero8888 on 17 Jun 2018, 08:52, edited 2 times in total.
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Station Q is to the East of Station T. At 12 noon, a train starts from  [#permalink]

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New post 17 Jun 2018, 08:41
let total distance between the two stations = D miles

let the the time after they meet = t hours

distance covered by train from station Q in t hours = t * x miles

distance covered by train from station T in t hours = t * y miles

tx + ty = D

t = D/(x+y)

therefore we need values of x , y and total distance between the two stations

statement 1 y = 4x/3 not sufficient

statement 2 x = 100mph no info about Distance and y. therefore not sufficient

combined no info about total distance, therefore not sufficient
Answer E
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Re: Station Q is to the East of Station T. At 12 noon, a train starts from  [#permalink]

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New post 21 Jun 2018, 23:24
Bunuel wrote:
Station Q is to the East of Station T. At 12 noon, a train starts from Station Q and travels at a constant speed of x mph towards Station T. At 12 noon of the same day, another train starts from Station T and travels at a constant speed of y mph towards Station Q. At what time will the trains meet?

(1) y = 4x/3
(2) x = 100 mph


Given two trains travelling in opposite direction at speeds of x mph & y mph each.

Hence their relative speed = (x + y) mph

Let D be the distance between the two stations & T be the time taken for the trains to meet each other.

We get, D = (x + y) * T, hence T = D/(x + y)

Statement 1:

y = 4x/3

Hence T = D/(x +4x/3)

RHS has two unknowns. Hence Statement 1 is Not Sufficient.


Statement 2:
x = 100 mph

Hence T = D/(100 + y)

RHS has two unknowns. Hence Statement 2 is Not Sufficient.

Combining both statements we get,

x = 100 & y = 4x/3 = 400/3

Hence T = D/(100 + 400/3)

D is still not known.

Hence combining is Not Sufficient.

Answer E.


Thanks,
GyM
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Re: Station Q is to the East of Station T. At 12 noon, a train starts from  [#permalink]

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New post 19 Dec 2018, 04:13
. Since the distance between the two stations is not mentioned. It could not be solved by taking
only speeds in consideration. Hence, E.
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Re: Station Q is to the East of Station T. At 12 noon, a train starts from &nbs [#permalink] 19 Dec 2018, 04:13
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