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

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

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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) \(\frac{40x}{x+y}\)


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?

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Originally posted by enigma123 on 04 Mar 2012, 13:14.
Last edited by Bunuel on 28 Aug 2018, 06:21, edited 2 times in total.
Edited the question.
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same  [#permalink]

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New post 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.

Answer: E.
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 04 Mar 2012, 16:07
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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, at the same  [#permalink]

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

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New post 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, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 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

ANSWER: E
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 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

\(d = \frac{40x}{x+y}\)

Answer = E
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 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, at the same  [#permalink]

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New post 23 Nov 2016, 22:09
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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, at the same  [#permalink]

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New post 06 Oct 2018, 10:23
enigma123 wrote:
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) \(\frac{40x}{x+y}\)



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?




Since H and E are walking in opposite direction, relative speed will be summation i.e. X+Y
Time required by both of them to meet= 40/Relative speed= 40/(X+Y)
Now from F what distance they will be= Speed*time=X*40/(X+Y)
Ans: E
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same   [#permalink] 06 Oct 2018, 10:23
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