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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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65% (02:47) correct
35% (02:01) wrong based on 310 sessions
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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 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) \(\frac{40(xy)}{x+y}\) (B) \(40x\frac{y}{x+y}\) (C) \(\frac{(xy)}{x+y}\) (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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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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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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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 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
ANSWER: E



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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 \(\frac{d}{x} = \frac{40d}{y}\) (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 [#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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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink]
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18 Nov 2016, 22:33
FMG 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 Sent from my A1601 using GMAT Club Forum mobile app



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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 Sent from my A1601 using GMAT Club Forum mobile app




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