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PS - VIC

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Senior Manager
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New post 04 Dec 2008, 18:38
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A
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D
E

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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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New post 05 Dec 2008, 00:17
I pick A using the plug and chuck method
X=6 mph Y=2mph if Aaron ran for 1 hour, he would spend 3 hours walking back. So t=4 hr

A - 12
B doesn't make sense dimensionally (adding time with velocity)
C - 6
D doesn't make sense dimensionally (adding time with velocity)
E doesn't make sense dimensionally (adding time with velocity)

I'd love to see the derivation for the formula tho.

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New post 05 Dec 2008, 04:06
I really cant see a good way to solve this one. Vksunder, Can you pls post the QA annd the explanation ?

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New post 05 Dec 2008, 05:51
The OA - C. The question is from OG 11 under the diagnostic questions.

I still havent figured out a way to solve this by plugging in numbers. Can someone pl[ease help. Thanks!

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Re: PS - VIC [#permalink]

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New post 05 Dec 2008, 06:01
Hi,

Since the distance is same, let's say S, then the equation that will give us the total time is

S/y + S/x=t
You can figure out what to do next

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New post 05 Dec 2008, 12:04
\(\frac{Sx}{xy}\) + \(\frac{Sy}{xy}\) = t
\({S(x+y)}/{xy}\) = t
S=\({xyt}/{x+y}\)

what got me at first was that the dimensions didn't work out.

x & y has mi/hr t has hr
so xyt would end u with mi
\({xyt}/{x+y}\) would have units of hr , but we are measuring distance, but somehow this works. I don't know why.

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Re: PS - VIC [#permalink]

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New post 05 Dec 2008, 12:34
GMAT TIGER wrote:
vksunder wrote:
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


c. (xyt)/(x+y)


I just plugged in but there should be a better way to solve this:

suppose x = 10 and y = 5.
total distance = 100

* time to complete the distance from home to jog = 50/10 = 5hours
* time to complete the distance from jog to home = 50/5 = 10 hours
so t = 15 hours.

lets plu-in (I am doing only in C):

C: (xyt)/(x+y) = (10x5x15)/(10+5) = 50

so this is the distance from home to jog.
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New post 05 Dec 2008, 21:42
NO need to plugin the numbers for this,

Average Speed S = (D + D)/T ;

we already know, average speed = 2xy/(x+y) if distance of 2 variable speeds(x,y) are same.
this i can derive it for you, if you really need so

so 2D/t = 2xy/(x+y) ; so D = xyt/(x+y)

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Re: PS - VIC   [#permalink] 05 Dec 2008, 21:42
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