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Hayden began walking from F to G, a distance of 40 miles, at
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27 Aug 2018, 09:01
27 Aug 2018, 09:01
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Question Stats:
69% (01:58) correct
30% (02:13) wrong based on 39 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) \(\frac{40(x-y)}{x+y}\)
(B) \(40x-\frac{y}{x+y}\)
(C) \(\frac{(x-y)}{x+y}\)
(D) \(\frac{40y}{x+y}\)
(E) \(\frac{40x}{x+y}\)
(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) \(\frac{40x}{x+y}\)
Re: Hayden began walking from F to G, a distance of 40 miles, at
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12 Dec 2018, 14:34
12 Dec 2018, 14:34
1
As usual, when two objects are approaching eachother, we can add the velocities and set the distance traversed equal to the sum of the velocities times the time that they take.
Then the 40 = (x+y)t. If we solve for time, we get t = 40/(x+y).
But now, measuring from Hayda to get the number of miles away from F they are, we use the distance equation again (distance = velocity * time): distance = x*t = x * 40/(x+y).
Thus, the distance from F when they meet is 40x/(x+y).
Then the 40 = (x+y)t. If we solve for time, we get t = 40/(x+y).
But now, measuring from Hayda to get the number of miles away from F they are, we use the distance equation again (distance = velocity * time): distance = x*t = x * 40/(x+y).
Thus, the distance from F when they meet is 40x/(x+y).
Re: Hayden began walking from F to G, a distance of 40 miles, at
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16 Jun 2021, 03:20
16 Jun 2021, 03:20
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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