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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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E

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45% (medium)

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64% (02:44) correct
36% (01:57) wrong based on 222 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?

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?

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

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, 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

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

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

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14 Jun 2015, 09:25

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