MathRevolution 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 > 0If 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 < 0Since 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 < 0If
possibility #2 is true, then x is NEGATIVE and y is POSITIVE, which means
x²y = (NEGATIVE²)(POSITIVE) = POSITIVE. So,
x²y > 0Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Answer =
Cheers,
Brent
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