a good algebra question

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a good algebra question

by sana.noor » Mon May 13, 2013 3:45 am
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 HaydenÂ’s walking speed was x miles per hour and AvaÂ’s was y miles per hour, how many miles away from F were they, in terms of x and y, when they met?
(A) 40(x-y)/x+y
(B) 40x-y/x+y
(C) x-y/x+y
(D) 40y/x+y
(E) 40x/x+y

OA is E
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by mkdureja » Mon May 13, 2013 4:58 am
* As Hayden and Ava are approaching each other, distance between them is getting reduced by the rate of their total speed, i.e. 'x+y' miles/hr in this case.
* They will meet when their distance becomes zero. It will take Distance/Speed = 40/(x+y) hrs.
* How far away they are from F = distance traveled by Hayden =
Hayden's speed * time = x * 40/(x+y) miles.

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by GMATGuruNY » Mon May 13, 2013 5:11 am
sana.noor 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 HaydenÂ’s walking speed was x miles per hour and AvaÂ’s was y miles per hour, how many miles away from F were they, in terms of x and y, when they met?
(A) 40(x-y)/x+y
(B) 40x-y/x+y
(C) x-y/x+y
(D) 40y/x+y
(E) 40x/x+y

OA is E
Let x = 2 miles per hour (Hayden's rate) and y = 3 miles per hour (Ava's rate).

When people travel TOWARD each other, they WORK TOGETHER to cover the distance between them.
Thus, we ADD their rates.
Combined rate for Hayden and Ava walking toward each other = 2+3 = 5 miles per hour.
Time for Hayden and Ava to travel the 40 miles between them = d/r = 40/5 = 8 hours.
Distance from F = distance traveled by Hayden in 8 hours = r*t = 2*8 = 16 miles. This is our target.

Now we plug x=2 and y=3 into the answers to see which yields our target of 16.
Since x-y = 2-3 = -1, any answer choice that includes x-y can be eliminated.
Eliminate A, B and C.

Answer choice D: 40y/(x+y) = (40*3)/(2+3) = 24.
Doesn't work.
Eliminate D.

The correct answer is E.
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