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Hayden began walking from F to G, a distance of 40 miles, at the same
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Updated on: 28 Aug 2018, 05:21
Question Stats:
69% (01:50) correct 31% (01:59) wrong based on 375 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 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) \(\frac{40(xy)}{x+y}\) (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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Originally posted by enigma123 on 04 Mar 2012, 12:14.
Last edited by Bunuel on 28 Aug 2018, 05:21, 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
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04 Mar 2012, 15:05




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Re: Hayden began walking from F to G, a distance of 40 miles, at the same
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04 Mar 2012, 13:21
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
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04 Mar 2012, 15:07
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, at the same
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Re: Hayden began walking from F to G, a distance of 40 miles, at the same
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07 Sep 2013, 18:52
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
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07 Sep 2013, 20:54
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
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09 Sep 2013, 02:06
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, at the same
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22 Apr 2014, 06:10
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
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24 Apr 2014, 00:26
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, at the same
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24 Apr 2014, 00:27
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
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18 Nov 2016, 21:33
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, at the same
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23 Nov 2016, 05:30
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, at the same
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23 Nov 2016, 21:09
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, at the same
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06 Oct 2018, 09:23
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) \(\frac{40(xy)}{x+y}\) (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? 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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