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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