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26 May 2017, 04:41
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If \(wx > 0, xy > 0\), and \(yz < 0\), is \(w^{3}z^{2} < 0\)?
(1) \(z > 0\)
(2) \(x < 0\)
(1) \(z > 0\)
(2) \(x < 0\)
26 May 2017, 04:41
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Bookmarks
Official Solution:
You can take the 1st step of the variable approach and modify the original condition and the question, it becomes If \((wx)(xy)(yz) < 0\), or \(wx^{2}y^{2}z < 0\), and since even exponents are ignored for inequalities, only the numbers that have the odd number exponents are left. Thus, you get only \(wz < 0\). Also, the question is \(w^{3}z^{2} < 0\)?, and if you ignore the even exponents here, it becomes the same as is \(w < 0\)?. If you look at it as If \(wz < 0\), it becomes \(w < 0\)? \(\rightarrow\) is \(z > 0\)?
In the case of con 1), \(z > 0\) hence yes, it is sufficient. 2) If it is also \(x < 0\), from the original condition, \(wx > 0\), \(w < 0\) becomes yes, and so it is sufficient. Therefore, the answer is D.
Answer: D
You can take the 1st step of the variable approach and modify the original condition and the question, it becomes If \((wx)(xy)(yz) < 0\), or \(wx^{2}y^{2}z < 0\), and since even exponents are ignored for inequalities, only the numbers that have the odd number exponents are left. Thus, you get only \(wz < 0\). Also, the question is \(w^{3}z^{2} < 0\)?, and if you ignore the even exponents here, it becomes the same as is \(w < 0\)?. If you look at it as If \(wz < 0\), it becomes \(w < 0\)? \(\rightarrow\) is \(z > 0\)?
In the case of con 1), \(z > 0\) hence yes, it is sufficient. 2) If it is also \(x < 0\), from the original condition, \(wx > 0\), \(w < 0\) becomes yes, and so it is sufficient. Therefore, the answer is D.
Answer: D
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