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If, (x^2 + 8)yz < 0, wz> 0 and xyz<0 then which of

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If, (x^2 + 8)yz < 0, wz> 0 and xyz<0 then which of

by GMATinsight » Wed Oct 14, 2015 8:55 am
If, (x^2 + 8)yz < 0, wz> 0 and xyz<0 then which of the following must be true?

(I) x < 0
(II) wy < 0
(III) yz < 0

(A)II only
(B)III only
(C)I and III only
(D)II and III only
(E)I, II and III all

Answer: Option D
Last edited by GMATinsight on Thu Oct 15, 2015 2:11 am, edited 1 time in total.
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by GMATinsight » Wed Oct 14, 2015 8:56 am
GMATinsight wrote:If, (x^2 + 8)yz < 0, wz> 0 and xyz<0 then which of the following must be true?

(I) x < 0
(II) wy < 0
(III) yz < 0

(A)II only
(B)III only
(C)I and III only
(D)II and III only
(E)I, II and III all

Answer: Option D
(x^2 + 8)yz < 0
(x^2 + 8) must be positive (greater than or equal to 8 as x^2 can't be Negative)
i.e. yz < 0
i.e. y and z have opposite signs (One of y and z is Negative and other is positive)

wz> 0
i.e. w and z have Same signs (Both Positive or Both Negative)

xyz < 0
Since yz is Negative so x must be Positive

also If w and z are Negative then y is Positive or Vice Versa

(I) x < 0 is Definitely Incorrect
(II) wy < 0 is CORRECT as w and y have opposite signs
(III) yz < 0 is CORRECT

Answer: option D
Last edited by GMATinsight on Thu Oct 15, 2015 2:12 am, edited 1 time in total.
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by Matt@VeritasPrep » Thu Oct 15, 2015 12:13 am
Strictly speaking, (III) isn't quite correct. yz < 0 is a consequence of (x² + 8)*yz < 0 ... it has nothing to do with the sign of w.
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