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

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Aaron will jog home at x miles per hour and then walk back [#permalink] New post 29 Oct 2005, 15:08
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Aaron will jog home at x miles per hour and then walk back home on 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Apr 2012, 02:59, edited 1 time in total.
Edited the question and added the OA
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Re: Aaron will jog home at X miles per hour and then walk back [#permalink] New post 10 Apr 2012, 03:21
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mymbadreamz wrote:
I'm lost with this question too. Can someone please explain? thanks!


Aaron will jog home at x miles per hour and then walk back home on 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

Algebraic approach:

Say the distance Aaron jogs is d miles, notice that the distance Aaron walks back will also be d miles (since he walks back home on the same route).

Next, total time t would be equal to the time he spends on jogging plus the time he spends on walking: \frac{d}{x}+\frac{d}{y}=t --> d(\frac{1}{x}+\frac{1}{y})=t --> d=\frac{xyt}{x+y}.

Answer: C.

Number picking approach:

Say the distance in 10 miles, x=10 mile/hour and y=5 mile/hour (pick x and y so that they will be factors of 10).

So, Aaron spends on jogging 10/10=1 hour and on walking 10/5=2 hours, so total time t=1+2=3 hours.

Now, we have that x=10, y=5 and t=3. Plug these values into the answer choices to see which gives 10 miles. Only answer choice C fits: \frac{xyt}{x+y}=\frac{10*5*3}{10+5}=10.

Answer: C.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Hope it helps.

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Re: Aaron will jog home at x miles per hour and then walk back [#permalink] New post 22 Apr 2012, 09:30
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ENAFEX wrote:
I got the Distance as \frac{2XYt}{X+Y}

The logic I used was to find the average rates and then multiple it by time,t.

Average rates = \frac{2ab}{a+b}

where, a and b are the two given rates X and Y.

So, D= \frac{2XY}{X+Y}*t

What is the mistake here?


Not sure what are you doing in you solution. Why do you have 2 there? How did you get that the average rate is 2xy(x+y)?

Here is proper algebraic approach once more:

Say the distance Aaron jogs is d miles, notice that the distance Aaron walks back will also be d miles (since he walks back home on the same route).

Next, total time t would be equal to the time he spends on jogging plus the time he spends on walking: \frac{d}{x}+\frac{d}{y}=t --> d(\frac{1}{x}+\frac{1}{y})=t --> d=\frac{xyt}{x+y}.

Please study it and tell me if you see a problem there.

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Re: Aaron will jog home at x miles per hour and then walk back [#permalink] New post 22 Apr 2012, 20:20
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ENAFEX wrote:
Average Rate=\frac{total distance}{total time}
Total Time= \frac{d}{a}+\frac{d}{b}
Average Rate= \frac{d+d}{\frac{d}{a}+\frac{d}{b}}
= \frac{2d}{\frac{d(a+b)}{ab}}
=\frac{2ab}{a+b}


something similar to the problem below.

http://gmatclub.com/forum/average-rate-on-round-trip-85218.html


I see. The point is that \frac{2xyt}{x+y} is the total distance, meaning that it's the distance for the round trip and we are asked to find only one way distance (how many miles from home can Aaron jog), which would be half of this value so \frac{xyt}{x+y}.

Hope it's clear.

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 [#permalink] New post 29 Oct 2005, 15:33
i dont know is this is a right approach...but try by plugging numbers
Let the total distance be : 100 km
Hypothetical
Speed Time (hrs)
Jogs : 50 kmph (X) 2
Walks : 20 kmph (Y) 5
------
7 (T)

try plugging in numbers...the ans should be 100
only XYT/(X + Y) = 100

hey! am just too absurd in my reasoning !!!!
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 [#permalink] New post 29 Oct 2005, 15:38
JUST BEING A BIT MORE CLEAR HERE...
i dont know is this is a right approach...but try by plugging numbers
Let the total distance be : 100 km
Hypothetical
Speed
Jogs : 50 kmph (X)
Walks : 20 kmph (Y)

Time
Jogs : 2 hrs
Walks : 5 hrs
------
Total : 7 hrs (T)

try plugging in numbers...the ans should be 100
only XYT/(X + Y) = 100

hey! am just too absurd in my reasoning !!!!
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 [#permalink] New post 29 Oct 2005, 17:48
Is there another way to do this?? How does OG solve it ?

Thanks
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 [#permalink] New post 30 Oct 2005, 05:50
OA is C.

Here's how the OG solves it:

Add in the extra variable J to represent the # of hours Aaron spends jogging. Therefore,

XJ=distance jogged
Y(T-J)=distance walked

XJ=Y(T-J)
XJ=YT-JY
XJ + JY= YT
J(X+Y)=YT
J=YT/(X+Y)

# of miles he can jog is XJ therefore sub in YT/(X+Y) for J

=XYT/(X+Y)

I like the picking #'s approach better. Thanks!
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Re: Aaron will jog home at X miles per hour and then walk back [#permalink] New post 09 Apr 2012, 20:01
I'm lost with this question too. Can someone please explain? thanks!

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Re: Aaron will jog home at x miles per hour and then walk back [#permalink] New post 22 Apr 2012, 07:23
I got the Distance as \frac{2XYt}{X+Y}

The logic I used was to find the average rates and then multiple it by time,t.

Average rates = \frac{2ab}{a+b}

where, a and b are the two given rates X and Y.

So, D= \frac{2XY}{X+Y}*t

What is the mistake here?

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Re: Aaron will jog home at x miles per hour and then walk back [#permalink] New post 22 Apr 2012, 16:35
Average Rate=\frac{total distance}{total time}
Total Time= \frac{d}{a}+\frac{d}{b}
Average Rate= \frac{d+d}{\frac{d}{a}+\frac{d}{b}}
= \frac{2d}{\frac{d(a+b)}{ab}}
=\frac{2ab}{a+b}


something similar to the problem below.

http://gmatclub.com/forum/average-rate-on-round-trip-85218.html

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Re: Aaron will jog home at x miles per hour and then walk back   [#permalink] 22 Apr 2012, 16:35
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