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

[A4Fever]

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

[RR]

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

[Guest]

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.

[RR]

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

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

[A4Fever]

Thanks RR - appreciate the response and quick replies.

Good luck!

[dom]

so whats the OA?

[RonPurewal]

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.

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.

[supratim7]

going further..

Is mx + ky > kx + my?
(1) m > k
(2) x > y

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

this can hold in two situations
(A) (k - m) > 0 AND (y - x) > 0
(B) (k - m) < 0 AND (y - x) < 0

Only when we combine Stmt (1) & (2) it confirms case (B)
So, ANS: C

Modifying above question..

Is mx + ky = kx + my?
(1) m = k
(2) x = y

mx + ky = kx + my ??
(similar manipulation yields)
(k - m)(y - x) = 0 ??

this can hold in three situations
(A) (k - m) = 0
(B) (y - x) = 0
(C) (k - m) = 0 AND (y - x) = 0

Stmt (1) confirms case (A)
Stmt (2) confirms case (B)
So, ANS: D

So the takeaway is logic/behavior/soln changes quiet a bit between ">/≥/</≤" and "=".
Am I right??

Many thanks | Supratim

[RonPurewal]

Modifying above question..

Is mx + ky = kx + my?
(1) m = k
(2) x = y

your analysis of this new item is, of course, correct. but, if you were to see this thing on the exam, hopefully you wouldn't do that much work.
since these are now EQUATIONS, you can just do SUBSTITUTION (which is generally not an available option with inequalities).
viz.:

statement (1)
just plug in m for all the k's.
--> Is mx + my = mx + my?
clearly "yes". so, sufficient.

statement (2)
just plug in x for all the y's.
--> Is mx + kx = kx + mx?
clearly "yes". so, sufficient.

that's all the work you'd need to do for that one; hopefully you'd do that.
(in this situation, it's understandable that you used the more roundabout method, because you were working off a problem that couldn't be solved in this simpler way.)

So the takeaway is logic/behavior/soln changes quiet a bit between ">/≥/</≤" and "=".
Am I right??

you are absolutely right"”and, in fact, this is the only reason why inequalities even appear on the gmat exam in the first place! if you ever see an inequality problem, ALWAYS be aware of how the inequalities work DIFFERENTLY from equations.

the gmat exam almost never has problems with inequalities that work exactly as equations would"”since, in that case, they would just give you equations!

[supratim7]

Thank you for the detailed explanation/illustration Ron. Appreciate it :)

[jnelson0612]

:-)

[supratim7]

going further..

Is mx + ky > kx + my?
(1) m > k
(2) x > y

Process#1
mx + ky > kx + my ??
ky - my > kx - mx ??
y(k - m) > x(k - m) ??
y(k - m) - x(k - m) > 0 ??
(k - m)(y - x) > 0 ??

Process#2
mx + ky > kx + my ??
mx - kx + ky - my > 0 ??
x(m - k) + y(k -m) > 0 ??
Couldn't factor further

Why am I getting stuck in Process#2? After all it's just another (valid) way of manipulating inequalities...

[tim]

Yes this is another valid way of manipulating inequalities. Just remember that not every valid operation you can perform on a math question is helpful. This is one such case. :)

[supratim7]

Yes this is another valid way of manipulating inequalities.

Thought as much

Just remember that not every valid operation you can perform on a math question is helpful.

This too is evident. I was expecting, however, some explanation/elucidation for this (apparent) mathematical enigma. Thanks anyways.

[tim]

Unfortunately, in most cases where an unhelpful calculation has been performed I don't think there's really any explanation that can be given; it's just the wrong calculation.

However, on further reflection, I notice that your calculation actually wasn't as unhelpful as we both thought. Observe:

Process#2
mx + ky > kx + my ??
mx - kx + ky - my > 0 ??
x(m - k) + y(k -m) > 0 ??
x(m - k) - y(m - k) > 0 ??
(x - y)(m - k) > 0 ??

Notice how you had an (m-k) factor as well as a (k-m) factor. When you see this, you should factor out a -1 from one of these terms (it doesn't matter which one) so that you have two like terms and can factor further.