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# Is xyz > 0 ? (1) xy > 0 (2) yz > 0

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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
==>In the original condition, the answer is highly likely to be E with 3 variables (x,y,z). If you do 1)&2), you get x=y=z=1 yes, and x=y=z=-1, hence no, and not suffi. Thus, the answer is E.

- Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
St. 1: 2 cases---> x,y +ve or x,y -ve No info abt z--- not sufficient
St. 2: 2 Cases---> y,z +ve or y,z -ve No info abt x--- not sufficient
Combining-------> all 3 can be +ve all 3 can be -ve--- So both answers possible---> Not sufficient
Hence E
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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
We want to know whether the product xyz is positive or not.

Statement 1. xy > 0. Product of xy is positive.
Now if z is positive, product xyz will be positive, but if z is negative, product xyz will be negative. so Insufficient.

Statement 2. yz > 0. Product of yz is positive.
Now if x is positive, product xyz will be positive, but if x is negative, product xyz will be negative. so Insufficient.

Combining the two statements: xy > 0 and yz > 0. So x and y have same sign (both positive or both negative) and y,z also have same sign (both positive or both negative). This means all three x,y,z have same sign (all three positive or all three negative).

Now if all three are positive, product xyz will be positive.
But if all three are negative, product xyz will be negative.
Insufficient still.

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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
Top Contributor
MathRevolution wrote:
Is xyz > 0?

1) xy > 0
2) yz > 0

Target question: Is xyz > 0?

Jump straight to....

Statements 1 and 2 combined
There are several values of x and y that satisfy BOTH statements. Here are two:
Case a: x = 1, y = 1 and z = 1, in which case xyz = (1)(1)(1) = 1. So, xyz > 0
Case b: x = -1, y = -1 and z = -1, in which case xyz = (-1)(-1)(-1) = -1. So, xyz < 0
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Cheers,
Brent
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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
MathRevolution wrote:
Is xyz>0?
1) xy>0
2) yz>0

We need to determine whether xyz > 0.

Statement One Alone:

xy>0

We see that either:

x = negative and y = negative

OR

x = positive and y = positive

However, without any information regarding z, we cannot answer the question.

Statement Two Alone:

yz > 0

We see that either:

y = negative and z = negative

OR

y = positive and z = positive

However, without any information regarding x, we cannot answer the question.

Statements One and Two Together:

Using our statements together, we see:

When x = negative, y = negative, and z = negative, xyz is less than zero.

When x = positive, y = positive, and z = positive, xyz is greater than zero.

So, statements one and two together are not sufficient to answer the question.

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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
==> In the original condition, there are 3 variables (x,y,z) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if x=y=z=1, yes, but if x=y=x=-1, no, hence it is not sufficient.

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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
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Re: Is xyz > 0 ? (1) xy > 0 (2) yz > 0 [#permalink]
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