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

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VP
Joined: 30 Jun 2008
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Aaron will jog from home at x miles per hour and then walk [#permalink]

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01 Oct 2008, 06:50
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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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Manager
Joined: 22 Sep 2008
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Re: PS: OG 11 - D24 [#permalink]

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01 Oct 2008, 06:55
is answer is C :

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

Kudos [?]: 48 [0], given: 0

VP
Joined: 30 Jun 2008
Posts: 1034

Kudos [?]: 705 [0], given: 1

Re: PS: OG 11 - D24 [#permalink]

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01 Oct 2008, 07:03
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)??
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Kudos [?]: 705 [0], given: 1

Manager
Joined: 22 Sep 2008
Posts: 121

Kudos [?]: 48 [0], given: 0

Re: PS: OG 11 - D24 [#permalink]

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01 Oct 2008, 07:33
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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Manager
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Re: PS: OG 11 - D24 [#permalink]

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01 Oct 2008, 07:33
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)

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Senior Manager
Joined: 04 Aug 2008
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Re: PS: OG 11 - D24 [#permalink]

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01 Oct 2008, 07:42
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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Kudos [?]: 37 [0], given: 1

Re: PS: OG 11 - D24   [#permalink] 01 Oct 2008, 07:42
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# Aaron will jog from home at x miles per hour and then walk

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