Official Guide Explanation:
Problem Solving #117




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Solution and Metadata

Question: 117
Page: 76
Difficulty: 5 (Moderate)
Category 1: Algebra > Inequalities > other
Category 2: Arithmetic > Real Numbers >

Explanation: If xy > 0, either both are positive or both are negative. If yz < 0, then one of the variables is positive and one is negative. Thus, we have two possibilities:

(i) x is positive, y is positive, and z is negative

(ii) x is negative, y is negative, and z is positive

Consider each choice:

(A)    With (i), xyz is negative. With (ii), it's positive.

(B)    With (i), xyz2 is positive. With (ii), it's also positive.

(C)    With (i), xy2z is negative. With (ii), it's also negative.

(D)    With (i), xy2z2 is positive. With (ii), it's negative.

(E)    With (i), x2y2z2 is positive. With (ii), it's also positive.

Choice (C) is correct. Note that you don't have to go through all of that work: If you find a scenario in which one of the choices is positive, you know it isn't ALWAYS negative, and you can move on to the next choice.

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