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# Aaron will jog from home at X miles per hour and then walk

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Intern
Joined: 29 Mar 2008
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Aaron will jog from home at X miles per hour and then walk [#permalink]  20 Apr 2008, 16:22
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 joggin and walking?

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

Are there alternative ways of solving this question?? I understood the answer, but i just didnt like the way they did it?

thx
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Joined: 10 Jun 2007
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Re: OG 11 Diagnostic #24 [#permalink]  20 Apr 2008, 17:45
Victor81 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 joggin and walking?

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

Are there alternative ways of solving this question?? I understood the answer, but i just didnt like the way they did it?

thx

Which way is the OE?

Set d = total distance, and this is what we are looking for.
time going there + time going tack = t
d/x + d/y = t
d = txy / (x+y)

Ans = C
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Re: OG 11 Diagnostic #24 [#permalink]  22 Apr 2008, 07:42
Thanks man, yours is a much better way of solving. I had the right idea, just couldnt fully put it together. The OG answer is as follows:

Let J be the number of hours Aaron spends jogging
Let T -J bet the total number of hours he spends walking
So Aaron jogs XJ Miles and walks y(t-J) miles
Because Aaron travels the same route, XJ = y(t-J)

J = yt/x+Y
the number of miles he can jog is xj, so substitute for J

xJ = Yt/x+Y
x(yt/x+Y) = xyt/x+y
Re: OG 11 Diagnostic #24   [#permalink] 22 Apr 2008, 07:42
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