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
Brent Hanneson - Creator of GMATPrepNow.com
