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 500,000 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:

Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

04 Mar 2012, 13:14

6

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

65% (01:48) correct
35% (02:02) wrong based on 317 sessions

HideShow timer Statistics

Hayden began walking from F to G, a distance of 40 miles, at the same time Ava began walking from G to F on the same road. If Haydens walking speed was x miles per hour and Avas was y miles per hour, how many miles away from F were they, in terms of x and y, when they met?

Hayden's speed = x miles/hr Ava speed = y miles/hr

Relative speed = (x+y) miles/hr as they are travelling in opposite direction (F to G and G to F)

Total time taken = \(\frac{40}{x+y}\) ---------------------------[D = Speed/Time]

So Hayden will take = \(\frac{40x}{x+y}\)-------------------------[I have guessed this and this is the right answer]. But not 100% why though? Can someone please help?

Total time(T) taken is the time for Hayden and Eva to meet. Now you know that Hayden must tavel for time T before they meet. Thus distance Hayden will cover is = Time T * Hayden Speed.
_________________

The question is not can you rise up to iconic! The real question is will you ?

Hayden began walking from F to G, a distance of 40 miles, at the same time Ava began walking from G to F on the same road. If Haydens walking speed was x miles per hour and Avas was y miles per hour, how many miles away from F were they, in terms of x and y, when they met? (A) \(\frac{40(x-y)}{x+y}\) (B) \(\frac{40(x-y)}{x+y}\) (C) \(\frac{(x-y)}{x+y}\) (D) \(\frac{40y}{x+y}\) (E) \(\frac{40x}{x+y}\)

Combined rate of Hayden and Ava is \(x+y\) miles per hour, hence they will meet in \(time=\frac{distance}{rate}=\frac{40}{x+y}\) hours. In \(\frac{40}{x+y}\) hours Hayden will cover \(distance=rate*time=x*\frac{40}{x+y}\) miles.

Thanks Bunuel. But why we are only calculating it for Hayden? That's where I am getting confused buddy.

Hayden began walking from F and we are asked how many miles away from F were Hayden and Ava when they met. In 40/(x+y) hours Hayden, at x miles per hour, will cover 40/(x+y)*x miles from F.
_________________

1. Time taken for the two to meet which is also the time walked by Hayden = 40 /x+y hrs 2. Speed of Hayden = x miles/hr 3. Distance traveled by Hayden = 40x / (x+y)
_________________

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

09 Sep 2013, 03:06

Hayden began walking from F to G, a distance of 40 miles, at the same time Ava began walking from G to F on the same road. If Haydens walking speed was x miles per hour and Avas was y miles per hour, how many miles away from F were they, in terms of x and y, when they met?

combined speed = x+y miles/hour

time in which it took them to meet = 40/x+y

So, we know that it takes them 40/x+y hours to meet up. The question states that Hayden traveled from F to G and Ava, from G to F. It asks us how far from F (Hayden's starting point) did they meet at. Therefore, to solve for distance (d=r*t) we solve by multiplying Hayden's rate (x) by the time it took them to meet (40/x+y) to get 40x/x+y

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

22 Apr 2014, 07:10

Bunuel wrote:

enigma123 wrote:

Thanks Bunuel. But why we are only calculating it for Hayden? That's where I am getting confused buddy.

Hayden began walking from F and we are asked how many miles away from F were Hayden and Ava when they met. In 40/(x+y) hours Hayden, at x miles per hour, will cover 40/(x+y)*x miles from F.

Still much is left in explanation :

Ans you will get will be same if you multiply Ava rate with time when they meet :

As Distance travel by ava when they meet : y* 40/(x+y)

Now question has asked distance from F i.e. you have to subtract distance travel by ava from 40.

therefore - 40 - (40y/x+y) = on solving you will get 40x/x+y i.e E

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

14 Jun 2015, 09:25

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.
_________________

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

09 Nov 2016, 09:30

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.
_________________

Hayden starts from Point F, Ava from Pt G and they meet at Pt M. Since the question is "how far from F do they meet", we are required to figure out FM i.e. the distance that Hayden covers [this should answer Enigma123's question]. Since they start walking towards each other at the same time and continue walking until they meet, they take the same amount of time to cover their respective distances [FM miles by Hayden and GM miles by Ava]. Therefore, the ratio of their speeds is equal to that of the distances they cover until they meet. T/4, FM:GM=x:y. Since [FM+GM]=40 miles, FM=40x/[x+y]. Ans:E.

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

23 Nov 2016, 22:09

ashishahujasham Yes, we can. Let's assume x=20 mph and y=60 mph. Then, FM:GM=20:60=1:3. That is, FM (distance covered by Hayden) is 1/4th the total distance (40 miles). T/4, FM=10 miles. Now, lets plug in the assumed values of 'x' &'y' in FM=40x/(x+y): 40x20/(20+60)=800/80=10 miles. I am not sure from your question if that is what you needed explained. If so, glad to have been able to help.

Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

Show Tags

23 Nov 2016, 22:44

effatara wrote:

ashishahujasham Yes, we can. Let's assume x=20 mph and y=60 mph. Then, FM:GM=20:60=1:3. That is, FM (distance covered by Hayden) is 1/4th the total distance (40 miles). T/4, FM=10 miles. Now, lets plug in the assumed values of 'x' &'y' in FM=40x/(x+y): 40x20/(20+60)=800/80=10 miles. I am not sure from your question if that is what you needed explained. If so, glad to have been able to help.