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Expert replies
by fangtray » Mon Apr 02, 2012 5:54 am
Aaron will jog from home at x miles per hr and then walk back home by the same route at y miles per hr. How many miles from home can Aaron jog so that he spends a total of t hrs jogging and walking?

a. (xt)/y
b. (x+t)/xy
c. (xyt)/x+y
d. (x+y+t)/(xy)
e. [(x+t)/x] - (t/y)

I plugged in values for x and y and t, and then tried to match the values from the answer choices. I got the right answer, but that method took more than 2 minutes, and for these types of questions, it is difficult at first to see what numbers to plug in for X,Y, and T so that they are "smart" numbers. I was wondering if there was a simpler algebraic method, or some way for us to know what numbers to pick so that the problems could be done faster.
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by GMATGuruNY » Mon Apr 02, 2012 6:06 am
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by Anurag@Gurome » Mon Apr 02, 2012 7:16 pm
fangtray wrote:Aaron will jog from home at x miles per hr and then walk back home by the same route at y miles per hr. How many miles from home can Aaron jog so that he spends a total of t hrs jogging and walking?

a. (xt)/y
b. (x+t)/xy
c. (xyt)/x+y
d. (x+y+t)/(xy)
e. [(x+t)/x] - (t/y)

I plugged in values for x and y and t, and then tried to match the values from the answer choices. I got the right answer, but that method took more than 2 minutes, and for these types of questions, it is difficult at first to see what numbers to plug in for X,Y, and T so that they are "smart" numbers. I was wondering if there was a simpler algebraic method, or some way for us to know what numbers to pick so that the problems could be done faster.
Using algebraic method:

Let total distance covered by Aaron = D
Time taken to walk away from home = t1 = D/x hours
Time taken to go home = t2 = D/y hours
Total time = t = D/x + D/y
t = D(1/x + 1/y)
t = D(x + y)/xy
D = txy/(x + y)

The correct answer is C.
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