Bunuel wrote:
Is \(x^2y^3w^4z^1 < 0\) ?
(1) yz < 0
(2) z < 0
Hi,
there are two sets of answer-
1) A as the answer --- Not speaking about any of the variables being 0..
2) E as the answer--- Correctly taking 0 as an option for variables ..lets see the equation first and what is required..
\(x^2y^3w^4z^1 < 0\)
1) when all are non-zero integers..\(x^2w^4\) will be > 0, we have to check if y and z are of opposite sign..
2) if any of the integer is 0..if any integer is zero, entire equation will be 0
AND our answer to "Is \(x^2y^3w^4z^1 < 0\) ?" will be NO..
and suff..
lets see the statements-
(1) yz < 0this shows that y and z are of opposite sign, so ans is YES..
But say x is 0, then ans will be NO..
Insuff
HAD the q been \(x^2y^3w^4z^1 > 0\), this statement would have been suff as ans would be NO in both cases, but not NOW
(2) z < 0Clearly insuff
Combined,still insuff as the cases as shown in statement 1 still remain
E
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