Bunuel wrote:

If x and y are non-negative, is (x + y) greater than xy?

(1) x = y

(2) x + y is greater than x^2 + y^2

As we're not asked about general logical properties (such as positive/negative), well go for an equation-driven approach.

This is a Precise methodology.

We'll write an 'inequality' with a question mark:

\(x + y (?) xy\)

(1) Substituting into our question stem gives \(2x (?) x^2\) --> \(0 (?) x^2 - 2x\) --> \(0 (?) x(x - 2).\)

Since we don't know if \(x(x - 2)\) is positive or not, this is insufficient.

(2) gives \(x + y > x^2 + y^2\). Looking at this expression we should be reminded of the core identity: \((x + y)^2 = x^2 + 2xy + y^2\)

Using this, we can write:

\(x + y - 2xy > x^2 + y^2 - 2xy\) which gives \(x + y - 2xy > (x - y)^2\) and then \(x + y > (x - y)^2 + 2xy\)

This essentially says that (x + y) is larger than a positive number plus 2xy.

Then (x + y) is larger than 2xy meaning it is larger than xy.

Sufficient!

(B) is our answer.

_________________

David

Senior tutor at examPAL

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