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If m, k, x and y are positive numbers...

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A4Fever
 Post subject: If m, k, x and y are positive numbers...
 Post Posted: Wed Nov 12, 2008 4:00 pm 
is mx + ky > kx + my?

(1) m > k

(2) x > y

Hi all - this is from the GMAC practice test. Can anyone help with regards to the best way to approach this number? Picking numbers seems to be a dawnting task for this particular problem.

Thanks
A4Fever
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RR
 Post subject:
 Post Posted: Fri Nov 14, 2008 9:16 am 
I am not sure, but will try

mx + ky > kx + my
mx - kx > my - ky
(m-k)x > (m - k)y
(m-k)x - (m-k)y > 0
(m-k)(x-y) > 0
ie m > k and x > y

Answer C
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Guest
 Post subject:
 Post Posted: Fri Nov 14, 2008 1:21 pm 
you got it right - answer is C.

I also rephrased like you did however I factored it out differently. The only place where you lost me is in the following manipulations:

(m-k)x - (m-k)y > 0
(m-k)(x-y) > 0

How did you get to the last step? Pretty simple arithmetic but I just don't see it.

Here's how I broke it down:

Question: mx+ky > kx+my

Breakdown:
mx-my > kx - ky
m(x-y) > k(x-y)

Then divided both sides by (x-y) and chose A.

Thanks for your help.
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RR
 Post subject:
 Post Posted: Sat Nov 15, 2008 3:48 am 
Quote:
I also rephrased like you did however I factored it out differently. The only place where you lost me is in the following manipulations:

(m-k)x - (m-k)y > 0
(m-k)(x-y) > 0


I am basically taking (m - k) outside as common
(m-k)x - (m-k)y > 0
To simplify, assume m - k = a
Therefore,
ax - ay > 0
a(x - y) > 0
Replace a with (m-k)
(m - k)(x - y) > 0

Quote:
Here's how I broke it down:

Question: mx+ky > kx+my

Breakdown:
mx-my > kx - ky
m(x-y) > k(x-y)

Then divided both sides by (x-y) and chose A.


You cannot divide by (x - y) because you do not know the sign of (x - y). If (x - y) is negative, the inequality might change. For eg : If you have the inequality
2x > 3x
If you divide by x on both sides, you will get
2 > 3
which doesn't make sense. It should be solved as
2x - 3x > 0
-x > 0
x < 0
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A4Fever
 Post subject:
 Post Posted: Sat Nov 15, 2008 11:58 am 
Thanks RR - appreciate the response and quick replies.

Good luck!
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dom
 Post subject:
 Post Posted: Sun Nov 16, 2008 4:35 am 
so whats the OA?
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RonPurewal
 Post subject:
 Post Posted: Thu Nov 20, 2008 8:09 am 
Offline
ManhattanGMAT Staff


Posts: 7146
rr, EXCELLENT workup.

there's only one thing that you didn't point out (which, unfortunately, may leave a few readers hanging):
REPHRASE:
if you know that x > y, then you know that x - y is positive, and vice versa.
if you know that x < y, then you know that x - y is negative, and vice versa.

it's not hard to manipulate to get these statements; for instance, merely subtracting y from both sides of x > y will give x - y > 0.
but that's not the point; the point is to recognize, INSTANTLY, that knowing the status of the inequality involving x and y (i.e., whether x > y or x < y) is equivalent to knowing the sign of x - y.

dom wrote:
so whats the OA?


well, the question prompt is:
is (m - k)(x - y) > 0?

based on the considerations above, statement #1 gives us the sign of the expression (m - k), and statement #2 gives us the sign of the expression (x - y).
if we have both of these signs, then we can figure out the sign of their product, so both statements together are sufficient.
(note that we don't even have to figure out the actual signs; it's good enough to realize that we can find them)

so, should be (c), unless, of course, the oa is wrong.
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