Bunuel wrote:

If which x < y < 0, of the following inequalities must be true?

A. y + 1 < x

B. y - 1 < x

C. xy^2 < x

D. xy < y^2

E. xy < x^2

We can see that both x and y are negative, and y is greater than x. So A is not true. B is not true, either, since we don’t know how much y is greater than x. If we divide both sides by x in the inequality in C (and switch the inequality sign since x is negative), we have:

y^2 > 1

However, since we don’t know the value of y, we can’t determine whether its square is in fact greater than 1. So C might not be true. If we divide both sides by y in the inequality in D (and switch the inequality sign since y is negative), we have:

x > y

This is not true since we know y > x. So the correct answer must be E. However, let’s verify it by dividing both sides of the inequality by x:

y > x

This is true since we know y is indeed greater than x.

Answer: E

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Scott Woodbury-Stewart

Founder and CEO

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