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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] New post 04 Mar 2012, 12:14
00:00
A
B
C
D
E

Difficulty:

  25% (low)

Question Stats:

67% (02:27) correct 32% (01:41) wrong based on 53 sessions
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, 04:06, edited 1 time in total.
Edited the question.
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Re: Hayden & Eva [#permalink] New post 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 [#permalink] New post 04 Mar 2012, 15:05
Expert's post
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 [#permalink] New post 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 [#permalink] New post 04 Mar 2012, 15:14
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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] New post 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 [#permalink] New post 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 [#permalink] New post 07 Sep 2013, 23:26
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) \frac{40(x-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?


I think there is a typo, A and B are the same.
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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink] New post 08 Sep 2013, 04:08
Expert's post
Skag55 wrote:
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) \frac{40(x-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?


I think there is a typo, A and B are the same.


Edited option B. Thank you.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Hayden began walking from F to G, a distance of 40 miles [#permalink] New post 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 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
Re: Hayden began walking from F to G, a distance of 40 miles   [#permalink] 09 Sep 2013, 02:06
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Hayden began walking from F to G, a distance of 40 miles

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