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Hayden began walking from F to G, a distance of 40 miles

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

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04 Mar 2012, 13:14
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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 Haydens walking speed was x miles per hour and Avas 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) $$40x-\frac{y}{x+y}$$
(C) $$\frac{(x-y)}{x+y}$$
(D) $$\frac{40y}{x+y}$$
(E) $$\frac{40x}{x+y}$$

[Reveal] Spoiler:
Ok. This is how I am solving this.

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?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Sep 2013, 05:06, edited 1 time in total.
Edited the question.

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Re: Hayden & Eva [#permalink]

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04 Mar 2012, 14:21
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.
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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04 Mar 2012, 16:05
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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 Haydens walking speed was x miles per hour and Avas 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.

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

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04 Mar 2012, 16:07
Thanks Bunuel. But why we are only calculating it for Hayden? That's where I am getting confused buddy.
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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04 Mar 2012, 16:14
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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.
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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07 Sep 2013, 19:52
This is a confusing problem - especially if someone assumes x = y, then choice D is also valid.

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

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07 Sep 2013, 21:54
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)
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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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 Haydens walking speed was x miles per hour and Avas 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

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

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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

Hope i make it clear
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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24 Apr 2014, 01:26
Refer diagram below;

We require to find d (in terms of x & y) (how many miles away from F were they)

Setting up the equation

$$\frac{d}{x} = \frac{40-d}{y}$$

(x+y)d = 40x

$$d = \frac{40x}{x+y}$$

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

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24 Apr 2014, 01:27
minkathebest wrote:
This is a confusing problem - especially if someone assumes x = y, then choice D is also valid.

Please refer diagram above. There should be no confusion after
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]

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

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09 Nov 2016, 09:30
Hello from the GMAT Club BumpBot!

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

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18 Nov 2016, 22:33
F------------------------M-------------------------------G
Hayden-----> <------Ava

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.

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

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23 Nov 2016, 06:30
Can we get this answer by plugging in some values

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

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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.

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

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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.

Thanks

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Re: Hayden began walking from F to G, a distance of 40 miles   [#permalink] 23 Nov 2016, 22:44
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