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 possibleConversely, 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 possibleSince only case a is possible, we can be certain that

x, y and z are all NEGATIVE, which means

xy > 0Since we can answer the

target question with certainty, statement 1 is SUFFICIENT

Statement 2: xz > 0We'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 possibleThis 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 possibleSo, BOTH cases are possible.

In case a,

x, y and z are all NEGATIVE, which means

xy > 0In case b,

y is negative and x and z are positive, which means

xy < 0Since we cannot answer the

target question with certainty, statement 2 is NOT SUFFICIENT

Answer:

Cheers,

Brent

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Brent Hanneson – GMATPrepNow.com

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