If m, k, x, and y are positive numbers, is mx + ky > kx + my ?

Question asks: Is mx + ky > kx + my
or Is mx - kx > my - ky
or Is (m-k)x > (m-k)y

(1) If m > k, then m-k is positive, but we dont know that which is greater out of x and y. So we cant say whether the product of (m-k)x is greater or the product of (m-k)y is greater. So this statement is not sufficient.

(2) x > y. Now if m > k, then m-k is positive and the product of same positive number with x will yield a higher number than the product of same positive number with y: and in this case the answer to the question would be YES. But if m < k, then m-k is negative and the answer to the question would be NO. So this statement is not sufficient.

Combining the two statements, m > k so m-k is positive and x > y so the product of (m-k)x is greater than the product of (m-k)y. Answer to the question is YES. sufficient.

Hence C answer

Simplify the Q:
mx + ky > kx + my ?
m (x-y) > K(x-y)
(x-y)(m-k)>0?

1) x - y > 0. Don't know anything about (m - k). Not sufficient
2) m - k > 0. Don't know anything about (x - y). Not sufficient

1+2) Sufficient

ANSWER: C

First thing I see is that the difference between the sides of the inequality is the position of m,k
This makes me think about MIN-MAX. Also, we don't know the signs of the variables and inequalities are often about what is + or -

1) m > k
This means that: BIG*x + small*y > small*x + BIG*y
We don't have any info about k,m so insufficient

2) x > y
This means that m*BIG + k*small > k*BIG + m*small
Again, no info about the values of x,y so insufficient

3) If we combine:
BIG*BIG + small*small > small*BIG + BIG*small
We can see that the right hand side is bigger, so both together are sufficient, C

Testing values:
x>y
3>2
m>k
3>2

3*3+2*2 > 2*3+2*3
13>12

x>y
-1>-2
m>k
3>2

-1*3 + -2*2 > -2*3 + -1*2
-7>-8

x>y
-1>-2
m>k
-2>-3

-1*-2 + -2*-3 > -2*-2 + -1*-3
8>7

\(MX+KY>KX+MY \implies M(X-Y)>K(X-Y)\)

The second statement tells us that \((X-Y) > 0\) therefore we can divide by \(X-Y\).

The question then becomes \(M>K?\)

This is exactly the information we get form statement 1. Therefore, C is the answer
_________________

I tried simplifying the question to: mx-my > kx - ky and then factored out from that: m[x-y] > k[x-y] ---> is m>k?

How come that reasoning is wrong?

Because you cannot just divide by \((x-y)\) unless you know that \(x-y \neq 0\) or that \(x \neq y\)
_________________

Hi Im sorry i don't understand the logic behind why we need to know that x-y = 0 or not. Could you pls elaborate?

Because if it were 0, then we could not say anything about m or k as both sides are = 0 if x=y or x-y=0 (hence left side = right side) and m>k if not x=y or x-y=0

Posted from my mobile device
_________________

Bunuel Positive numbers can we consider only positive integers or we can consider fractions also ? please help in this

Positive numbers consists of positive integers, positive fractions, and positive irrational numbers.
_________________

If m, k, x, and y are positive numbers, is \(mx + ky > kx + my\) ?
Details:
mx + ky > kx + my?
=> (m-k)x-(m-k)y>0?
=> (m-k)(x-y)>0?
+ + ?
or - - ?
=> m > k & x> y ?
or m<k & x< y ?

(1) \(m > k\)--> insuff: we don't know whether x>y or x< y?
(2) \(x > y\)--> insuff: we don't know whether m>k or m<k?
combining (1) & (2) we can say that > k & x> y i.e. mx + ky > kx + my --> suff
Answer: C

Posted from my mobile device

Asked: If m, k, x, and y are positive numbers, is \(mx + ky > kx + my\) ?

Let m, k, x, y be 2,1,4,3

(1) \(m > k\)
Since x & y are unknown
NOT SUFFICIENT

(2) \(x > y\)
Since m & k are unknown
NOT SUFFICIENT

(1) + (2)
(1) \(m > k\)
(2) \(x > y\)
mx + ky > my + kx
m(x-y) > k(x-y)
Since x-y> 0
m > k which is given
mx + ky > kx + my
For example: -
If (m,k,x,y) = (2,1,4,3) : mx + ky = 2*4 + 1*3 = 11; kx + my = 1*4 + 2*3 = 10; mx + ky > kx + my
If (m,k,x,y) = (4,1,3,2) : mx + ky = 4*3 + 1*2 = 13; kx + my = 1*3 + 4*2 = 11; mx + ky > kx + my
SUFFICIENT

IMO C

given
\(mx + ky > kx + my\)
which can be written as
m(x-y)>k(x-y)
#1
\(m > k\)
no info about x & y as y>x or x>y ; insufficient
#2
\(x > y\)
no info about relation of m & k insufficient
from 1 &2
we can say that m(x-y)>k(x-y) ; true
sufficient
OPTION C

before i looked at (1) & (2), i did this: mx + ky > kx + my -> mx-my>kx-ky-> m*(x-y)>k*(x-y) -> m>k

why is that not correct? Because x-y could be 0 and I may never divide by 0?