Data Sufficiency

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.

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.

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).

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.

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.

Lets take an example :

5 > -5
-1 > -5

Multiply the inequalities :
-5 > 25 (Not right)...
So you can't always multiply. Check the sign if you do.

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).

Picking numbers was my last resort, actually i tended avoid it to save time for this question, but its better to go with numbers and check. One should be careful when multiplying/dividing inequalities.

@goyalsau: anshumishra is right on multiplying. Its safe to add/subtract inequalities, but not safe to divide/multiply.
In this question all the three variables were +ve and division was no harm, still the combination didn't work.