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Rahul@gurome wrote:(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
: well in the question it gives you X<YRahul@gurome wrote:(1) x^2>y^2. Let us take examples for this.
If x = -4 and y = 3, it implies x < y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
If x = 4 and y = 3, it implies x > y but x^2 = 16 and y^2 = 9, so x^2 > y^2 in this case.
So, x may or may not be less than 0.
Hence, (1) is NOT SUFFICIENT.
(2) y > 1 implies x < 1 as x < y, but still we are not sure whether x < 0 as can take any other value between 0 and 1.
Hence, (2) is NOT SUFFICIENT.
Combining (1) and (2), we get that x < 0 always as x^2 > y^2.
The correct answer is (C).
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