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Nivranshu
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Asked: Is \(x^2yz\) negative?

x^2yz is negative when yz is negative since x^2 is non-negative.

(1) x < 0
NOT SUFFICIENT

(2) y < 0
If z>0; x^2yz is 0 or negative.
But if z<0; x^2yz is 0 or positive.
NOT SUFFICIENT

(1) + (2)
If z>0; x^2yz is 0 or negative.
But if z<0; x^2yz is 0 or positive.
NOT SUFFICIENT

IMO E
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Nivranshu
Is \(x^2yz\) negative?

(1) x < 0
(2) y < 0
This is a case in which a little visualizing ahead of time can ensure that you arrive at an accurate answer efficiently. If any of the variables equal 0, then it is game over: the product will be 0. Apart from that scenario, we know the following:

1) \(x^2\) will be positive (so, provided x ≠ 0, we only need to look at y and z);

2) The combinations of negatives and positives are as follows, but they are each dependent on knowing the values of y and z, relative to 0:

yzProduct
+++
+--
-+-
--+

Statement (1) tells us about x. Since it is not 0, we know squaring it will yield a positive value. But we know nothing of y or z. Not sufficient.

Statement (2) tells us about y: it is negative. But we know nothing of z, and we can see from our table above that the product could go either way. Not sufficient.

Putting the two statements together still leaves us wondering about z. Hence, the answer must be (E).

Happy studies.

- Andrew
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Nivranshu
Is \(x^2yz\) negative?

(1) x < 0
if y>0 , z>0 then the equation is positive
y<0, z>0 , the equation is negative
Clearly insufficient

(2) y < 0
for z>0 the eqn is negative
for z<0 it's positive
Clearly insufficient

When 1 and 2 is combined we get
We still can get 2 possibilities Clearly insufficient

Therefore IMO E
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#1
x<0
Sign of x would have no effect on x^2yz being + or -ve as either of y or z has to be -ve
Insufficient
#2
Y<0
Value of z if <0 will make target + and if z>0 then -ve
Insufficient
From. 1&2
We cannot determine conclusively as z sign is not known
Insufficient
Option E

Nivranshu
Is \(x^2yz\) negative?

(1) x < 0
(2) y < 0

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