Fatehdeep Singh wrote:x > y^2 ?
I) x>y+5
II)x^2 - y^2 =0
OA after some discussions?
Solution:
Consider first I alone.
Let x = 5 and y = -1.
Here (x = 5) > (y+5 = -1+5 = 4) and (x = 5) > (y^2 = 1).
Next let x = -1/3 and y = -6.
Now, -1/3 = x > -6+5 = -1.
But x = -1/3 < y^2 = 36.
So, nothing definite can be said from (1) alone.
Next consider II) alone.
Let x = -1 and y = 1.
So, x = -1 < y^2 = 1.
Next, let x = 1/2 and y = 1/2.
So, x = ½ > y^2 = ¼.
Again, nothing definite can be said from (2) alone.
Lastly, combine both statements and check.
If x > y+5, we get that x > y.
Or x is not equal to y.
Now II means that (x+y)(x-y) = 0.
This means that x = -y.
Now, x > y+5 means that -y > y+5 or y < -5/2.
Now x > y^2 means that -y > y^2 or y^2+y < 0 or -1<y<0.
This is not possible because y <-5/2.
So, the answer to the main question is "no".
Or, both statements together are sufficient to answer the question.
The correct answer is (C).