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aaron will jog from home at x miles per hour and then walk

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aaron will jog from home at x miles per hour and then walk [#permalink] New post 26 Aug 2007, 10:02
aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?

a. (xt)/y
b. (x+t)/(xy)
c. (xyt)/(x+y)
d. (x+y+t)/(xy)
e. ((y+t)/(x)) - (t/y)

help guys, i don't get the OA for this one.
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Re: distance problem... [#permalink] New post 26 Aug 2007, 10:21
chronolinkz wrote:
aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?

a. (xt)/y
b. (x+t)/(xy)
c. (xyt)/(x+y)
d. (x+y+t)/(xy)
e. ((y+t)/(x)) - (t/y)

help guys, i don't get the OA for this one.


I get C.
First find average mph:
Set d = distance one way.
Jog total hour = d/x
Walk total hour = d/y
Ave mph = Total Distance / Total Time = (2d) / ((d/x) + (d/y))
= 2xy / (x+y)

Therefore, in t time, Arron will cover: 2xyt / (x+y) miles
You want to find d in term of x, y, and t, so you have:
2xyt / (x+y) = 2d
d = xyt / (x+y)
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Re: distance problem... [#permalink] New post 26 Aug 2007, 10:24
chronolinkz wrote:
aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?

a. (xt)/y
b. (x+t)/(xy)
c. (xyt)/(x+y)
d. (x+y+t)/(xy)
e. ((y+t)/(x)) - (t/y)

help guys, i don't get the OA for this one.


d/x+d/y = t
dy/xy+dx/xy = t
d(x+y) = txy
d= txy/(x+y)

ans c
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 [#permalink] New post 26 Aug 2007, 10:33
r1 = x
r2 = y
total time = t = t1+t2
d = equal both directions

t1 = d/x
t2 = d/y

t = t1+t2 = d/x + d/y

combine so you can simplify and solve for d --> d(x+y)/xy = t

solve for d --> txy/(x+y) = d

Answer: C
  [#permalink] 26 Aug 2007, 10:33
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aaron will jog from home at x miles per hour and then walk

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