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:

Cars P and Q started simultaneously from opposite ends of a [#permalink]
07 Sep 2013, 23:41

5

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

50% (02:39) correct
50% (01:43) wrong based on 169 sessions

Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P. 2) Up to location X, the average speed of car Q was \(1 \frac{1}{3}\)times that of car P.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
07 Sep 2013, 23:49

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

fozzzy wrote:

Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P. 2) Up to location X, the average speed of car Q was 1 1/3 times that of car P.

Please provide detailed explanations...

The time taken for the 2 cars to meet at point X = \(\frac{300}{p+q}\) , where p and q are the respective speeds. Thus, the distance travelled by car P = \(p*\frac{300}{p+q}\)

F.S 1 states that q = p+15. Thus, substituting this above, we get \(p*\frac{300}{2p+15}\). Clearly depends on the value of p. Insufficient.

F.S 2 states that \(q =p* \frac{4}{3}\), and this yields = \(p*\frac{300}{p+q}\) = \(\frac{900}{7}\). Sufficient.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
08 Sep 2013, 00:01

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

fozzzy wrote:

The time taken for the 2 cars to meet at point X = \(\frac{300}{p+q}\) , where p and q are the respective speeds.

Can you explain how you formed that equation so quick?

This is nothing but applied concept of relative velocity.

For two objects moving in the opposite direction, always add their speeds. For objects moving in the same direction, subtract their speeds. For example, just as this example, as they are moving in opposite directions, treat their relative velocity to be a sum total of their individual speeds, and this is their relative speed. Thus, the time taken, as we know is \(\frac{Distance}{Speed}\), thus, \(t = \frac{300}{p+q}\).

To understand this using algebra, Imagine this : Let the time taken for both to meet be t.

Now, in time t, car P covers say distance x. Thus,\(x = p*t\). Again, car Q must have covered a distance of (300-x). Thus, \(300-x = q*t\) Replace the value of x from the first equation, we get \(300-p*t = q*t\) Thus,\(t = \frac{300}{p+q}\)

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
12 Sep 2013, 07:51

fozzzy wrote:

Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P. 2) Up to location X, the average speed of car Q was \(1 \frac{1}{3}\)times that of car P.

Please provide detailed explanations...

Time taken for P to meet Q at a common point is same for both P, and Q.

Time for P, = Distance / speed = 'x'(assume) / P(assume) Time for Q, = 300-x / (P+20) from A

Equate both as they are equal. so x / P = 300-x / (p+20) cannot solve..

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
14 Oct 2013, 15:56

fozzzy wrote:

Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P. 2) Up to location X, the average speed of car Q was \(1 \frac{1}{3}\)times that of car P.

Please provide detailed explanations...

@Bunuel: Please move this question to Distance/Speed DS forum as it is not Work/Rate DS question. Thanks.

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
31 Dec 2013, 07:13

1

This post was BOOKMARKED

fozzzy wrote:

Cars P and Q started simultaneously from opposite ends of a straight 300-mile express way and travelled towards each other, without stopping, until they passed at location X. To the nearest mile, how many miles of the express way had car P travelled when the two cars passed each other?

1) Up to location X, the average speed of car Q was 15 miles per hour faster than that of car P. 2) Up to location X, the average speed of car Q was \(1 \frac{1}{3}\)times that of car P.

Please provide detailed explanations...

Same-o, I've seen couple of questions such as this one

When one has a constant distance, all we need to know what the problem is asking is the relative rate, that is the RATIO between both rates. Not the difference

Re: Cars P and Q started simultaneously from opposite ends of a [#permalink]
14 Jan 2015, 00:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

Booth allows you flexibility to communicate in whatever way you see fit. That means you can write yet another boring admissions essay or get creative and submit a poem...