beat_gmat_09 wrote:This if from OG
If x,y, and z are positive numbers, is x > y > z ?

1) xz > yz
2) yx > yz

x,y and z are positive integers(numbers) and we have to determine if x > y and y > z
From 1) : as z is positive the inequality can be divided by z on both sides, thus x > y
Not sufficient.
From 2) : again as y is positive x > z
Not sufficient.
1&2 both :
one way to combine - x > y and x > z, we are still not able to find whether y > z. Not sufficient. OA for this is E and OE is similar to above.
My doubt is regarding statements 1&2 combined.
I think we can also divide the inequalities here.
So (1)/(2) = xz/yx > yz/yz => z / y > 1 => as z and y are integers z > y
This is sufficient to determine that x,z > y, at least we know z > y and y is not > z thus it is sufficient to answer the question.
Combining inequalities is even done in OG explanation. Is the division method wrong to follow?
Please share your thoughts.
Thanks.

goyalsau wrote:Great work guys,

Beat_gmat_09, Has shown a very good point. Anshu Has given a very good explanation.

I just wan't to ask one thing.?

If x > z & p > q

then we can add the inequality and subtract , Irrespective of there signs.

We can do x + p > z + q & x - p > z - q or p - x > q - z

we can not divide but Can we do the multiplication of as well.

anshumishra wrote:
You can't do that, here is why :

6 > 5
7 > 5

Can I do 6/7 > 5/5
or 6/7 > 1
No, Right ? Because in an inequality we don't know by how much LHS can be greater/smaller than RHS.

OR
7/6 > 5/5
7/6 > 1 (Here we are just lucky, that the inequality still holds).