vinni.k wrote:If x and y are both negative and xy < y^2, which of the following must be true?

a) x < y < x^2 < y^2
b) x < y < y^2 < x^2
c) y < x < x^2 < y^2
d) x^2 < y^2 < y < x
e) y^2 < x^2 < y < x

OA is C

Since y<0, dividing each side by y requires that we FLIP the inequality:
xy < y²
xy/y > y²/y
x > y.
Eliminate A and B, which indicate x<y.

The remaining answer choices describe the relationship between x² and y².
Since x<0 and y<0, squaring each side of x>y implies multiplying each side by a negative value.
Thus, when we square each side, we must once again FLIP the inequality:
x > y
x² < y².
Eliminate E, which indicates that x²>y².

Since x<0, x²>0, implying that x<x².
Eliminate D, which indicates that x>x².

The correct answer is C.

Wow, two amazing answers by two wonderful people. Appreciate you replies.

Thanks & Regards
Vinni

vinni.k wrote:If x and y are both negative and xy < y^2, which of the following must be true?

a) x < y < x^2 < y^2
b) x < y < y^2 < x^2
c) y < x < x^2 < y^2
d) x^2 < y^2 < y < x
e) y^2 < x^2 < y < x

We see that in all the answer choices we have to compare the following four quantities: x, y, x^2 and y^2. We are given that x and y are both negative and xy < y^2.

We see that both xy and y^2 are positive since the product of two negative quantities is positive and the square of a nonzero quantity is always positive. However, we divide both sides of the inequality by y (a negative quantity), we have:

x > y

Since x^2 and y^2 are both positive, we see that y is the smallest of the four quantities. The only answer choice that has y as the smallest quantity is choice C. Thus, it's the correct answer.

(Note: We don't have to analyze, in this case, which is the larger quantity between x^2 and y^2 since y has to be the smallest quantity. However, if we have to, it is always true that if y < x < 0, then y^2 > x^2 > 0. For example, -3 < -2, but (-3)^2 > (-2)^2 since 9 > 4.)

Answer: C