Official Guide Explanation:
Problem Solving #117

Background

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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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