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integer!

by Ozlemg » Mon Jul 04, 2011 1:08 pm
If xyz < 0 and yz > 0, which of the following must be positive?
(A) xy
(B) xz
(C) (x^2)yz
(D) x(y^2)z
(E) xy(z^2)
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by MBA.Aspirant » Mon Jul 04, 2011 2:35 pm
Ozlemg wrote:If xyz < 0 and yz > 0, which of the following must be positive?
(A) xy
(B) xz
(C) (x^2)yz
(D) x(y^2)z
(E) xy(z^2)
xyz < 0 and yz > 0, means either x is -ve or y & z are both -ve

the only scenario when the product must be +ve, is (x^2)yz

if x is -ve, squaring will make it +ve, and yz is already +ve
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by edvhou812 » Mon Jul 04, 2011 3:29 pm
xyz < 0

yz > 0

I picked x to be positive and both y and z to be negative, which makes the above statements true.

After going through the options, and seeing how each works with the rules of positive and negative integers, (x^2)yz is the only one that works because x^2 remains a positive, and yz becomes negative*negative=positive, so (x^2)yz becomes a positive*positive, which is always positive.

Answer: C
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