Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

56% (02:16) correct
44% (01:13) wrong based on 98 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?

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

1

This post received 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 ..

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.

Re: Station Q is to the East of Station T. At 12 noon, a train s [#permalink]
15 Oct 2013, 04:39

2

This post received KUDOS

Expert's post

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.

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.

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

gmatclubot

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

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...