Brent@GMATPrepNow wrote:If y < x < z and xyz < 0, is xy > 0?

1) yz > 0
2) xz > 0

Target question: Is xy > 0?

Given: y < x < z and xyz < 0
If xyz < 0, then there are only TWO POSSIBLE CASES:
case a: all 3 numbers are negative (x, y and z are all NEGATIVE)
case b: 1 number is negative and the other 2 numbers are positive. Since y < x < z, then it must be the case that y is NEGATIVE, and x and z are POSITIVE

Statement 1: yz > 0
Let's compare this information with our given information.
The statement 1 information satisfies the conditions in case a. Reason: In case a, y and z are both negative, so yz > 0. So, case a IS possible
Conversely, this same information does not satisfy the conditions in case b. Reason: In case b, y is negative and z is positive, so yz < 0. So, case b is NOT possible
Since only case a is possible, we can be certain that x, y and z are all NEGATIVE, which means xy > 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: xz > 0
We'll compare this information with our given information.
The statement 2 information satisfies the conditions in case a. Reason: In case a, y and z are both negative, so xz > 0. So, case a IS possible
This information ALSO satisfies the conditions in case b. Reason: In case b, x and z are both positive, so xz > 0. So, case b IS possible
So, BOTH cases are possible.
In case a, x, y and z are all NEGATIVE, which means xy > 0
In case b, y is negative and x and z are positive, which means xy < 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT


Answer: A

Cheers,
Brent