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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 01 Oct 2008, 05:50
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Aaron will jog from home at x miles per hour and then walk back at y miles an hour on the same route. 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]
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Re: PS: OG 11 - D24 [#permalink] New post 01 Oct 2008, 05:55
is answer is C :

average speed will be xy/(x+y) , now total time is t so simply multiple t with xy/x+y
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Re: PS: OG 11 - D24 [#permalink] New post 01 Oct 2008, 06:03
vr4indian wrote:
is answer is C :

average speed will be xy/(x+y) , now total time is t so simply multiple t with xy/x+y


How do you calculate average speed as xy/(x+y)??
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Re: PS: OG 11 - D24 [#permalink] New post 01 Oct 2008, 06:33
not sure but its goes like this:
d = distance

total time t = d/x + d/y = d (1/x +1/y) = d(x+y/xy)

so distance will be xy/x+y * t
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Re: PS: OG 11 - D24 [#permalink] New post 01 Oct 2008, 06:33
amitdgr wrote:
vr4indian wrote:
is answer is C :

average speed will be xy/(x+y) , now total time is t so simply multiple t with xy/x+y


How do you calculate average speed as xy/(x+y)??


Say S miles from home

S/x + S/y = t

=> S = xyt(x+y)

Answer:C
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Re: PS: OG 11 - D24 [#permalink] New post 01 Oct 2008, 06:42
i remember this question and then and now i have no idea what i am being asked to calculate

need a translator :)
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Re: PS: OG 11 - D24   [#permalink] 01 Oct 2008, 06:42
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Aaron will jog from home at x miles per hour and then walk

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