If m>0 and n>0, is m+x/n+x > m/n?

1) m< n

2) x > 0

What is the better approach ? plluging numbers or solving ?

I think the answer should be 'C'

There is no need to plug numbers . It is given m<n in A and both m & n >0 so now we need to know about X , it says X>0 .

IMO:C

No need to plug in numbers...solve it like this

m+x/n+x -m/n > 0

x*(n-m)/n*(n+x) > 0

for this to be positive, we need
either numerator and denominator both +ve or both -ve.

we know only one thing: n = +ve
we need sign of X, (N-M), and (N+X) ?

A) m< n ==> N- M == +ve
still we need two more sign.

B) X > 0 ==> X is +ve and N+X is +ve as N is +ve.
But again we dont know the sign of N-M.

combine these two, we get all we want. So answer is C.

Good approach, Sunny. Let me know if anything's unclear or if you have any other questions relating to this problem.

_________________
Ben Ku
Instructor
ManhattanGMAT

can u please tell me, when i do i decide to take the numbers to other side? like sunny did here ??

should always do it ??

When you have an algebraic equation or inequality, it's usually helpful to have everything on one side, because then you can compare it to zero.

Here are three examples:

If you have x^2 - 6 = x, placing everything on one side gives you x^2 - x - 6 = 0. We can factor it to get (x - 3)(x + 2) = 0. either x - 3 = 0 or x + 2 = 0, so x = -2 or 3.

If you have x^3 -6x = x^2, it's tempting to divide everything by x. However, you should never divide equations by a variable (unless you know it's not zero). The approach here is also bring everything to one side: x^3 - x^2 - 6x = 0, and simplify from there: x(x - 3)(x + 2) = 0, so x = -2, 0, or 3.

If you have x^2 - 6 > x, we can place everything on one side getting: x^2 - x - 6 > 0 or (x - 3)(x + 2) > 0. Here, we see that because (x - 3)(x + 2) is positive, then either (x-3) and (x + 2) are both positive or both negative. Only numbers less than -2 or greater than 3 will work.

Hope that helps.

_________________
Ben Ku
Instructor
ManhattanGMAT