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

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29 Oct 2005, 16:08

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A

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C

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Difficulty:

15% (low)

Question Stats:

78% (02:52) correct
22% (03:19) wrong based on 339 sessions

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Aaron will jog from 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

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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29 Oct 2005, 16:33

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

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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29 Oct 2005, 16:38

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

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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10 Apr 2012, 04:21

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Expert's post

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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.

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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22 Apr 2012, 10:30

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Expert's post

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. _________________

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 \(\frac{xyt}{x+y}\).

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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26 Jan 2015, 08:24

Hay i always hear that there are small tricks gmac puts in to help figure out problems..

the trick i used for this problem: In all the other answer miles/hour is being added to hours. since they dont have the same unit you cant add them together, only multiply therefore all the answers with (x + t) or ( y + t) can not be correct and there is only one that is in the correct format

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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02 Feb 2015, 23:21

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aikido_fudoshin wrote:

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

Re: Aaron will jog from home at x miles per hour and then walk [#permalink]

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20 Apr 2015, 08:30

Hey Bunuel,

I have a query.. i solved this question by checking units.. Please tell is this the right way.. Through this was i solved it in 7-8 seconds..else question might take more time.. plz tell if my process is right or wrong?

Aaron will jog from 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

As i saw only in C we have (speed^2 )* time / speed = distance..

No other option comply the rule of unit..

gmatclubot

Re: Aaron will jog from home at x miles per hour and then walk
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20 Apr 2015, 08:30

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