Bunuel wrote:
If x, y, and z are positive integers, is y > x?
(1) y^2 = xz
(2) z - x > 0
Kudos for a corrector solution.
Is y>x?
Statement 1, y^2=xz ---> if x=y=z =1 ,t hen "no" but if y=3, x=1, z=3, then "yes". This statement is not sufficient.
Statement 2, z>x, not sufficient.
Combining, y^2=xz and z>x. The only way this is possible is by y>x
(y=z=3, x=1)C is the correct answer.
The highlighted part seems some mistake. @y=z, y and z will also be equal to x which is not possible because it's given that z>x
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