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If xyz < 0 and yz > 0, which of the following must be positi

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If xyz < 0 and yz > 0, which of the following must be positi [#permalink] New post 20 Feb 2013, 15:41
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C
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E

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94% (01:45) correct 6% (00:20) wrong based on 77 sessions
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Re: If xyz < 0 and yz > 0, which of the following must be positi [#permalink] New post 20 Feb 2013, 16:17
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If yz>0 and xyz<0, lets get first yz>0:

a) or y is + and z is +

If this is true, we get xyz<0 and we know: x has to be -

b) or y is - and z is -

If this is true, we get xyz<0 and we know: x has to be -

Therefore, we have OR option A: y+ z+ x- ; OR option B: y- z- x- . Which is positive?

(A) xy: is positive only in option B
(B) xz: is positive only in option B
(C) (x^2)yz: is positive for both option A and B
(D) x(y^2)z: is positive only in option B
(E) xy(z^2): is positive only in option B

Solution MUST be C
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Re: If xyz < 0 and yz > 0, which of the following must be positi [#permalink] New post 20 Feb 2013, 20:06
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carcass 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)



Given that yz is positive and xyz is negative, you can say that x MUST be negative. As for y and z either they are both positive or both negative.

Option (C) has x^2 which is positive and yz which is positive. So (C) must be positive.
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Re: If xyz < 0 and yz > 0, which of the following must be positi [#permalink] New post 21 Feb 2013, 02:01
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Re: If xyz < 0 and yz > 0, which of the following must be positi [#permalink] New post 21 Feb 2013, 06:28
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Re: If xyz < 0 and yz > 0, which of the following must be positi   [#permalink] 21 Feb 2013, 06:28
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