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# Train T leaves town A for town B and travels at a constant rate of spe

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Math Expert
Joined: 02 Sep 2009
Posts: 44284
Train T leaves town A for town B and travels at a constant rate of spe [#permalink]

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17 Oct 2017, 08:14
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68% (00:40) correct 32% (00:45) wrong based on 91 sessions

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Train T leaves town A for town B and travels at a constant rate of speed. At the same time, train S leaves town B for town A and also travels at a constant rate of speed. Town C is between A and B. Which train is traveling faster? Towns A, C, and B lie on a straight line.

(1) Train S arrives at town C before train T.
(2) C is closer to A than to B.
[Reveal] Spoiler: OA

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Train T leaves town A for town B and travels at a constant rate of spe [#permalink]

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17 Oct 2017, 12:09
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KUDOS
Bunuel wrote:
Train T leaves town A for town B and travels at a constant rate of speed. At the same time, train S leaves town B for town A and also travels at a constant rate of speed. Town C is between A and B. Which train is traveling faster? Towns A, C, and B lie on a straight line.

(1) Train S arrives at town C before train T.
(2) C is closer to A than to B.

Let the speed of train T be $$u$$, time taken to reach C be $$t_1$$ and distance between A & C be $$x$$

Let the speed of train S be $$v$$, time taken to reach C be $$t_2$$ and distance between B & C be $$y$$

we need to find which speed between $$u$$ & $$v$$ is greatest

Statement 1: implies $$t_2<t_1$$ $$=> \frac{y}{v}<\frac{x}{u}$$

or $$u<\frac{x}{y}v$$. Now we need to know the ratio of $$\frac{x}{y}$$ to determine whether $$u$$ or $$v$$ is greater. But the value of $$x$$ & $$y$$ is not given. Hence Insufficient

Statement 2: implies that $$x<y => ut_1<vt_2$$. but nothing given about $$u$$, $$v$$ or $$t_1$$ & $$t_2$$. Hence Insufficient

Combining 1 & 2 we get $$x<y => \frac{x}{y}<1$$, multiply both sides by $$v$$ to get $$\frac{x}{y}v<v$$

from statement 1 we get $$u<v$$. Hence sufficient

Option C
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Re: Train T leaves town A for town B and travels at a constant rate of spe [#permalink]

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23 Oct 2017, 10:35
C, I had to do this one visually. So I drew a line with points A and B at each end. I drew T over point A and drew S over point B. I drew a point C halfway in between A and B, but I reminded myself that C could be anywhere between A and B.

St 1, says train S arrives at C before train T. Insufficient, but if we need the location of point C in relation to A and B, then that would suffice.
St 2, says point C is closer to point A.
-Insufficient by itself, but nevertheless, I crossed out where I had drawn C and moved it closer to A.

1+2, Since C is closer to point A and train S which is travelling a farther distance beats train T, then it must be the faster train.
_________________

I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.

Re: Train T leaves town A for town B and travels at a constant rate of spe   [#permalink] 23 Oct 2017, 10:35
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