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Aron

Expert replies
by j_shreyans » Sat Apr 04, 2015 9:51 pm
Aron 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 Aron 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

OAC
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Source: — Problem Solving |

by GMATGuruNY » Sun Apr 05, 2015 2:50 am
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)/x] - (t/y)
Let the distance = 10 miles.
Let x = 5 miles per hour and y = 2 miles per hour.
Time for Aaron to jog 10 miles at a rate of 5 miles per hour = d/r = 10/5 = 2 hours.
Time for Aaron to walk 10 miles at a rate of 2 miles per hour = d/r = 10/2 = 5 hours.
t = total time spent jogging and walking = 2+5 = 7 hours.

The question stem ask for the distance that Aaron jogs (10 miles).
This is our target.
Now plug x=5, y=2 and t=7 into the answer choices to see which yields our target of 7.
Only C works:
(xyt)/(x+y) = (5*2*7)/(5+2) = 10.

The correct answer is C.
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by Brent@GMATPrepNow » Sun Apr 05, 2015 7:17 am
j_shreyans wrote:Aron 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 Aron 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

OAC
As with all VIACs (Variables In the Answer Choices questions), we can solve this via the INPUT-OUTPUT approach (as Mitch has done) or via an ALGEBRAIC approach.
Typically, when the answer choices look complex (as they do here), I find the INPUT-OUTPUT approach easier.
However, the algebraic approach isn't too bad.

Let's let d = the number of miles (distance) that Aaron jogs.
This means that d also = the distance that Aaron walks.

Let's start with a WORD EQUATION:
total time = (time spent jogging) + (time spent walking)
In other words: t = (time spent jogging) + (time spent walking)
Since time = distance/speed, we can write: t = d/x + d/y [our goal is to solve this equation for d]
The least common multiple of x and y is xy, so we can eliminate the fractions by multiplying both sides by xy. When we do so, we get...
txy = dy + dx
Factor right side to get: txy = d(x + y)
Divide both sides by (x+y) to get: txy/(x+y) = d
So, the correct answer is C

Cheers,
Brent
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