VJesus12 wrote:If a, x, y, and z are integers greater than zero, are both x and y less than z?

(1) ax + xz + ay + yz < z^2 + az
(2) x < z

The OA is the option A.

I know the second statement is not sufficient. But, how can I show that the first one is sufficient? Thanks for all your help.

Hello Vjesus12.

I will show you why is sufficient the first statement.

(1) ax + xz + ay + yz < z^2 + az

We have $$ax+xz+ay+yz < z^2+az$$ $$x\left(a+z\right)+y\left(a+z\right) < z\left(z+a\right)$$ $$\left(x+y\right)\left(a+z\right)-z\left(z+a\right) < 0$$ $$\left(z+a\right)\left(\left(x+y\right)-z\right) < 0$$ Now, since "a" and "z" are positive then a+z is positive. Therefore, from the last equation we have that $$x+y-z < 0$$ $$\Rightarrow\ \ \ z+y < z$$ Since "x" and "y" are positive, then we can conlcude that both, x and y, are less than z.

Hence, the first statement is SUFFICIENT.

(2) x < z

This statement doesn't tell us anything about "y". NOT SUFFICIENT.

This is why the correct answer is the option A.