Max@Math Revolution wrote:If xyz ≠0, is x³y�z� > 0?
1) xz > 0
2) xyz > 0
Given: xyz ≠0
Target question: Is x³y�z� > 0?
This is a great candidate for
rephrasing the target question.
Since we know that x² must be POSITIVE, we can safely take the inequality
x³y�z� > 0 and divide both sides by x² to get:
xy�z� > 0
Similarly, since y� is POSITIVE, we can safely divide both sides by y� to get:
xz� > 0
Finally, since z� is POSITIVE, we can safely divide both sides by z� to get:
xz > 0
REPHRASED target question: Is xz > 0?
Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
Statement 1: xz > 0
PERFECT!
The answer to the REPHRASED target question is
YES, xz IS greater than 0
Since we can answer the
REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: xyz > 0
There are several values of x, y and z that satisfy statement 2. Here are two:
Case a: x = 1, y = 1 and z = 1. In this case, xz = (1)(1) = 1. So, the answer to the REPHRASED target question is
YES, xz IS greater than 0
Case b: x = 1, y = -1 and z = -1. In this case, xz = (1)(-1) = -1. So, the answer to the REPHRASED target question is
NO, xz is NOT greater than 0
Since we cannot answer the
REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
