Max@Math Revolution wrote:If xy < 0, is x²y < 0?

1) x > 0
2) x³y < 0

Target question: Is x²y < 0 ?

Given: xy < 0
If the product xy is NEGATIVE, then there are only 2 possibilities
Possibility #1: x is positive and y is negative
Possibility #2: x is negative and y is positive

Statement 1: x > 0
If x is positive, then we can eliminate Possibility #2, leaving only Possibility #1
This means that x is POSITIVE and y is NEGATIVE
So, x²y = (POSITIVE²)(NEGATIVE) = NEGATIVE
In other words, we can conclude that x²y < 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x³y < 0
What can we conclude from this?
Well, we can conclude that EITHER possibility #1 is true, OR EITHER possibility #2 is true

If possibility #1 is true, then x is POSITIVE and y is NEGATIVE, which means x²y = (POSITIVE²)(NEGATIVE) = NEGATIVE. So, x²y < 0
If possibility #2 is true, then x is NEGATIVE and y is POSITIVE, which means x²y = (NEGATIVE²)(POSITIVE) = POSITIVE. So, x²y > 0

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent