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Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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Bunuel wrote: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectAaron 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)/xt/y Diagnostic Test Question: 24 Page: 23 Difficulty: 650 GMAT Club is introducing a new project: The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide for GMAT® Review, 13th Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! The correct answer should have units of length (distance). Also, we cannot add/subtract velocities and time, it has no meaning. Therefore, we can immediately eliminate answers B, D and E. The expression in A has units of time, so it cannot be the correct answer. Only C is left, and it is the correct answer. It also has the correct units of length. Formally, if we denote by D the distance we are looking for, we can write the equation: D/x + D/y = t, from which, solving for D we get D =xyt/(x+y). (time spent jogging + time spent walking = total time) Answer: C Note: The answer should not depend on whether Aaron first jogs and then walks, or the other way around. This means that the expression for the distance should be symmetrical in x and y (meaning, it doesn't change if we switch between x and y). The only expressions on the list of answers which fulfill this condition are C and D. But D has velocities added to time, therefore it cannot be correct. This is just another way to easily pinpoint the correct answer.
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Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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SOLUTIONAaron 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)/xt/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 plugin 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 from home at x miles per hour and then walk back home [#permalink]
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20 Jul 2012, 06:57
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Bunuel wrote: SOLUTION
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)/xt/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 plugin 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. This is a question for which an algebraic approach would be much faster than number picking. When working with numbers, one easily forgets about units and meanings, and for example, we have no problem in adding speeds to time. Such operations have no meaning in the real, physical world. Once you see x + t or y + t, don't even bother to check that answer. So, don't be afraid of letters and a little algebra. And first of all, think of the meaning!!!
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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[quote="Bunuel"] The Official Guide for GMAT® Review, 13th Edition  Quantitative Questions ProjectAaron 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)/xt/y A note here: When one travels at speed x for time t and at a speed y for time t (same amount of time), the average speed for the journey is (x+y)/2 i.e. arithmetic mean of x and y. This is because average speed = Total Distance/Total time = (xt + yt)/2t = (x + y)/2 When one travels at speed x for distance d and at a speed y for distance d (same distance), the average speed for the journey is 2xy/(x+y) This is so because average speed = Total Distance/Total time = 2d/(d/x + d/y) = 2xy/(x + y) You don't really need to learn these formulas but knowing that there are these special cases will increase your speed. Once you use them a couple of times, you will automatically remember these. In this question, if you know that average speed will be 2xy/(x+y), you multiply it by t to get total distance traveled which is 2xyt/(x+y). So distance one way must be xyt/(x+y)
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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21 Sep 2013, 06:10
I got confused, I thought this question is saying t hrs for jogging and t hours for walking are separate. Got answer something like. D=2dx/(x+y) Where D is miles after t hours he started walking. d one side distance. sufficient to guess look alike option, but I was damm wrong in understanding this question ..... silly++ mistake.
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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21 Sep 2013, 07:32
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?
In other words, what is the combined time of him jogging and walking (t1+t2)
x=d/t 5/10/2
y=d/t 2=10/5
total time = 7
5*2*7/(7) 70/7 = 10
d=10
ANSWER: (C) xyt/(x+y)



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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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17 Dec 2013, 09:20
Bunuel wrote: SOLUTION
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)/xt/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 plugin 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. Hi, I do not understand how you did this step: \(d(\frac{1}{x}+\frac{1}{y})=t\) > \(d=\frac{xyt}{x+y}\) How do you go from the leftward equation to the right? Can you explain in more detail? Thanks in advance.



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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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18 Dec 2013, 00:54
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aeglorre wrote: Bunuel wrote: SOLUTION
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)/xt/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 plugin 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. Hi, I do not understand how you did this step: \(d(\frac{1}{x}+\frac{1}{y})=t\) > \(d=\frac{xyt}{x+y}\) How do you go from the leftward equation to the right? Can you explain in more detail? Thanks in advance. \(d(\frac{1}{x}+\frac{1}{y})=t\); \(d(\frac{y+x}{xy})=t\); \(d=\frac{xyt}{x+y}\). Hope it's clear.
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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24 Dec 2013, 04:37
Alternate method
Aaron will jog x miles in 1 hour and will return by walk x miles in x/y hour. So, Aaron will take (1 + x/y)hr to jog and walk x miles Or, Aaron will take 1 hr to jog and walk 1/(1 + x/y) miles Or, Aaron will take t hr to jog and walk t*(xy/(x + y)).
Therefore, answer is (C).



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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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If you go by the units approacj in a problem like this..It takes hardly 10 seconds....You spend a min just understanding wat the question is So option C = (m/sec* m/sec*sec)/m/sec= m U cannot add distance & time or speed & time..3m+4s= This approach is also easily applicable in area & volume cases to eliminate options.. Hope this helps
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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02 Jul 2015, 04:36
Alternate Method
Let us consider
x = 4
y = 5
The question states that the distance is same
therefore xt = yt
if we say 4 m/hr * 5 hr = 20 m
then 5 m/hr * 4 hr = 20 m
total time is 9 hr
on substituting x = 4 , y = 5 and t = 9 in the options we can get
C) 4 * 5 * 9 / 9 = 20 miles



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Bunuel wrote: 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)/xt/y We can let d = the distance traveled when walking and jogging. Thus, the jogging time is d/x and the walking time is d/y. Since the total time is t: d/x + d/y = t Multiply the entire equation by xy: dy + dx = txy d(y + x) = txy d = txy/(y + x) Answer: C
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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Always Average is equal to total distance by total time. In this case (D/x+D/Y) = t Hence D = xyt/(x+y) Hence answer C
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Re: Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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28 Nov 2017, 10:54
JusTLucK04 wrote: If you go by the units approacj in a problem like this..It takes hardly 10 seconds....You spend a min just understanding wat the question is : So option C = (m/sec* m/sec*sec)/m/sec= m U cannot add distance & time or speed & time..3m+4s= :wtf: This approach is also easily applicable in area & volume cases to eliminate options.. Hope this helps Hi, I like this approach but I am not sure how you get distance, So, (m/h*m/h*h)/(m/h+m/h)= speed*speed*time/speed+speed= distance/speed=time?Could someone explain it please.



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Aaron will jog from home at x miles per hour and then walk back home [#permalink]
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\(rate * time = distance\)
\(x * \frac{a}{x} = a\)
\(y * \frac{a}{y} = a\)
\(\frac{a}{x}+\frac{a}{y}=t\)
\(\frac{1}{x} + \frac{1}{y} =\frac{t}{a}\)
\(\frac{x+y}{xy}=\frac{t}{a}\)
\(a=\frac{txy}{x+y}\)




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