VJesus12 wrote:Is xy< 0 ? $$\left(1\right)\ xy^2 < y^2$$ $$\left(2\right)\ y^2 < y$$ The OA is the option E.
Could someone give me some help here? I think that the first statement is sufficient. <i class="em em-confused"></i>
We are given that xy < 0.
If xy < 0, one of x and y must be postive and the other must be negative.
Let's see each statenment one by one.
(1) xy^2 < y^2
Since y^2 is a positive quantity, we can cancel it.
=> x < 1
=> x can be positive (for example, x = 1/2), 0, or negative (for example, x = -1). We are sure about the nature of x as well as y. Insufficient.
(2) y^2 < y
Since y^2 is a positive quantity and y is greater than y^2, y must be a positive quantity. However, we do not know about x. Insufficient.
(1) and (2) together
Even after combining the statements, we cannot be sure wether xy < 0 as x can be positive or negative.
The correct answer:
E
Hope this helps!
-Jay
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