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# Station Q is to the East of Station T. At 12 noon, a train s

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Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  30 Sep 2013, 10:45
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Question Stats:

56% (02:17) correct 44% (01:12) wrong based on 102 sessions
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
[Reveal] Spoiler: OA

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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  15 Oct 2013, 04:39
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SaraLotfy wrote:
I don't know how to solve these distance/rate question in the two "meeting" at some point in the middle. Some questions ask at what time would they meet and others ask about the distance at which they would meet.

Can someone please clarify the equation I should use. I know I'm supposed to subtract rates, but can someone please give me a quick run down over the procedures.

Thanks

Remember 2 things :

I.Two objects moving in the same direction,subtract their speeds, call it $$V_{sd}$$

II.Two objects moving in opposite direction, add their speeds, call it $$V_{od}$$

The distance between them be fixed as D. Thus, as we know that Distance = Speed*Time, for the first case, we have Time taken to meet = $$\frac{D}{V_{sd}}$$

Similarly, for the second case, we have Time = $$\frac{D}{V_{od}}$$

Now, if you want the distance at which they meet, just multiply the time calculated above, fix the reference from which object do you want to calculate the distance, and multiply this time with the speed of that object.

Also, you can refer to this post : cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1264794

Hope this helps,
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  30 Sep 2013, 11:11
1
KUDOS
honchos 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

haha .. there is no mention of distance between Q and T .. neither in question nor in any of the statements .. distance could be 10000 miles or 10 miles .. not possible to get the answer ..

Its E ..

funny one ..
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  15 Oct 2013, 03:30
I don't know how to solve these distance/rate question in the two "meeting" at some point in the middle. Some questions ask at what time would they meet and others ask about the distance at which they would meet.

Can someone please clarify the equation I should use. I know I'm supposed to subtract rates, but can someone please give me a quick run down over the procedures.

Thanks
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Joined: 24 Apr 2013
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Kudos [?]: 8 [0], given: 23

Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  15 Oct 2013, 23:48
mau5 wrote:
SaraLotfy wrote:
I don't know how to solve these distance/rate question in the two "meeting" at some point in the middle. Some questions ask at what time would they meet and others ask about the distance at which they would meet.

Can someone please clarify the equation I should use. I know I'm supposed to subtract rates, but can someone please give me a quick run down over the procedures.

Thanks

Remember 2 things :

I.Two objects moving in the same direction,subtract their speeds, call it $$V_{sd}$$

II.Two objects moving in opposite direction, add their speeds, call it $$V_{od}$$

The distance between them be fixed as D. Thus, as we know that Distance = Speed*Time, for the first case, we have Time taken to meet = $$\frac{D}{V_{sd}}$$

Similarly, for the second case, we have Time = $$\frac{D}{V_{od}}$$

Now, if you want the distance at which they meet, just multiply the time calculated above, fix the reference from which object do you want to calculate the distance, and multiply this time with the speed of that object.

Also, you can refer to this post : cars-p-and-q-started-simultaneously-from-opposite-ends-of-a-159355.html#p1264794

Hope this helps,

Thank you, this is exactly what I needed
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Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]  31 Dec 2013, 07:10
honchos 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

Oh I see, they got me on this one.

E for sure

Cheers!
J
Re: Station Q is to the East of Station T. At 12 noon, a train s   [#permalink] 31 Dec 2013, 07:10
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