Inequalities - Wierd doubt

[sam2304]

If x, y and z are positive integers, is x > y > z ?

1. xz > yz => x > y - no info about z INSUFF.
2. yx > yz => x > z - no info about y INSUFF.

So ans choice is C or E.
I went this way using both.
Divide 1 by 2. we get z/y > 1 => z > y since all are positive. Already we have from 1 and 2 that x > y and x > z so combining the three i get x > z > y which is a NO (is x > y > z ?) using C. I don't know if i am being stupid but everything seems right. What's wrong in the above approach. Am i missing something ? Enlighten me please

OA is E.

[sl750]

I think you made a mistake when you combined both statements

In statement 1, we know x>z. In statement 2, we know x>y. However, we do not whether y>z or z>y. Therefore we cannot conclude sufficiently whether x>y>z or not

[sam2304]

By combining i have got z > y and hence x > z > y.

[Ian Stewart]

I went this way using both.
Divide 1 by 2. we get z/y > 1 => z > y since all are positive.

I've highlighted your mistake. You absolutely can NOT divide one inequality by another. You can see why this is not correct just using numerical examples. The following inequalities are both true:

4 > 2
3 > 1

but if we divide the first inequality by the second, the result is not true.

You certainly can divide one *equation* by another (though it isn't often useful to do that), at least as long as you're sure you're not dividing by zero. In fact, with equations we can combine them (add, subtract, multiply, divide, etc) in almost any way we like. That is certainly *not* the case for inequalities. With two inequalities, you can always safely add them together if they face the same way, though you'd only ever consider doing that in a question with at least 2 unknowns. You cannot safely subtract or divide (these are just wrong) inequalities, or even multiply them if negative numbers might be involved.

[varungoel]

does anyone have an under 2 minute solution to this problem?