Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

78% (02:52) correct
22% (02:38) wrong based on 100 sessions

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 home at X miles per hour and then walk back [#permalink]
10 Apr 2012, 03:21

3

This post received KUDOS

Expert's post

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 home at x miles per hour and then walk back [#permalink]
22 Apr 2012, 09:30

1

This post received KUDOS

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. _________________

Re: Aaron will jog home at x miles per hour and then walk back [#permalink]
22 Apr 2012, 20:20

1

This post received KUDOS

Expert's post

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}

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}.

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

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 home at x miles per hour and then walk back [#permalink]
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}

The Cambridge open day wasn’t quite what I was used to; no sample lectures, no hard and heavy approach; and it even started with a sandwich lunch. Overall...

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...