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

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Director
Joined: 11 Jun 2007
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Aaron will jog from home at x miles per hour and then walk [#permalink]

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16 Sep 2007, 19:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

24. 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)

this is the way I did it but not sure if my logic is correct:

d = rt
r = 1/x + 1/y = xy/ x+y

d = (xy/ x+y) t
d = xyt / x+y (answer C)

Director
Joined: 17 Sep 2005
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Re: OG Diagnostic word problem [#permalink]

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16 Sep 2007, 19:32
beckee529 wrote:
24. 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)

this is the way I did it but not sure if my logic is correct:

d = rt
r = 1/x + 1/y = xy/ x+y

d = (xy/ x+y) t
d = xyt / x+y (answer C)

Let the one way distance is D miles.

Speed = Distance / Times

S1 (To Jog) = D/T1 where T1 = Time to Jog
=> T1 = D/x where S1 = x ------------------- (1)

S1 (To Walk) = D/T2 where T2 = Time to walk
=> T2 = D/y where S2 = y --------------------(2)

Total Time Taken = T1 + T2 = t (as per question)

Substitute the values of T1 and T2 from (1) and (2),
We get,

D/x + D/y = t
=> D = txy/(x+y)

Hence C

- Brajesh
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Joined: 07 Jul 2004
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Location: Singapore
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16 Sep 2007, 20:36
Let the distance be d

Then time spent jogging = d/x hours, time spent walking = d/y hours

t = d/x + d/y
t = dx + dy/xy
txy/(x+y) = d
16 Sep 2007, 20:36
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