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Last Problem in OG12 Diagnostic

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by okigbo » Fri Sep 04, 2009 11:00 pm
24. 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)


OA: [spoiler]C[/spoiler]
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Re: Last Problem in OG12 Diagnostic

by maihuna » Sat Sep 05, 2009 8:00 am
okigbo wrote:24. 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 he takes p hours going and t-p hours coming

px = (t-p)*y
=>px+py = ty so p = ty/x+y

so px = txy/x+y
Charged up again to beat the beast :)
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Re: Last Problem in OG12 Diagnostic

by doclkk » Sat Sep 05, 2009 8:42 am
okigbo wrote:24. 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)


OA: C
Pick numbers

X = 4
Y = 2
T = 1 hour running, 2 hours walking = 3.

How many hours can Aaron spend running. Well, if he spends one hour jogging and 2 hours walking then he will just jog a total of 4 miles.

So find 4.

A. 12/ 2 = 6 - NO
B. 7 / 6 = NO
C. 24/ 6 = YES
D = 9 / 6 = NO
E = 5 / something that isn't 1.25

C Wins.
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by stormier » Thu Dec 23, 2010 9:59 am
OK, perhaps the easiest way to answer is to look at the units of the options.

Question is - how many miles, so the answer must have units of distance.

(a) has unit of time
(b) is adding distance and time in the numerator, which cannot be true
(c) has unit of distance - (speed*speed*time/speed = speed*time = distance)
(d) same explanation as b
(e) two quantities have different units, on top of the presence of errors of choice b
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by fskilnik@GMATH » Thu Dec 23, 2010 11:22 am
Hi there!

Let z be the number we are looking for: # of miles jogging or walking ("... same route").

From Units Control, we know that miles over mph gives hour, therefore the "hour-unit equation " is:

z/x + z/y = t , therefore z(1/x + 1/y)=t hence z = (txy)/(x+y) and we are done.

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by gmat7202011 » Thu Dec 23, 2010 11:43 am
The average speed for Aaron's entire trip is 2xy/(x+y).

The total distance traveled d, at an average speed is 2xy/(x+y) * t

d = 2xyt/(x+y).

The question asks what is the distance Aaron has to travel jogging, one way from his home, which is basically d/2

Hence d/2 = xyt/(x+y)

Since i have considered the average speed, it does not matter jogging or walking.

Hope this helps.
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by rene26 » Fri Dec 02, 2011 6:37 am
I really like how you all have tackled this problem in a different way... it really highlights the agility of mathematical minds!!

I wanted to ask you, Fabio, how you went from z(1/x +1/y) = t to z = (txy)/(x+y). I got stuck trying to carry that algebra out. Thank-you!
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by GMATGuruNY » Fri Dec 02, 2011 6:52 am
okigbo wrote:24. 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)


OA: C
Let the distance = 10 miles.
Let x = 5 miles per hour and y = 2 miles per hour.
Then t = total travel time = 10/5 + 10/2 = 7.
Since the question asks for the distance, our target is 10.
Now we plug x=5, y=2 and t=7 into the answers to see which yields our target of 10.

Only answer choice C works:
(xyt)/(x+y) = (5*2*7)/(5+2) = 10.

The correct answer is C.
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by fskilnik@GMATH » Fri Dec 02, 2011 1:55 pm
rene26 wrote:I really like how you all have tackled this problem in a different way... it really highlights the agility of mathematical minds!!

I wanted to ask you, Fabio, how you went from z(1/x +1/y) = t to z = (txy)/(x+y). I got stuck trying to carry that algebra out. Thank-you!
Hi, rene26, thanks for your interest in my solution.

Please note that:

(1) if (say) A = 1/x + 1/y = (x+y)/xy , then

(2) 1/A = (1/x + 1/y)^(-1) = xy/(x+y) ,

(3) therefore z = t/A = t*(1/A) = (txy)/(x+y) and we are done.

Regards,
Fábio.
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by AsadAbu » Tue Apr 16, 2019 2:54 am
okigbo wrote:24. 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)
OA: C
Here you go for this solution.

https://gmatclub.com/forum/aaron-will-j ... l#p2260928
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by Brent@GMATPrepNow » Tue Apr 16, 2019 5:27 am
okigbo wrote:24. 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)
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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by Scott@TargetTestPrep » Wed Apr 17, 2019 5:43 pm
okigbo wrote:24. 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)


OA: C
We can let d = the distance traveled on walking or 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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