x<0, Is x < -3?
st(1) implies |x| > |3|. Since x<0 we have Sufficiency here and x < -3 (the mode of x is greater than 3 but has negative signs, i.e. located to the left from -3.
st(2) implies x < -cubic.root(9) Not Sufficient, as we don't know exactly the value of x

a

kishokbabu wrote:If x is negative, is x < -3 ?

(1) x^2 > 9
(2) x^3 < -9

explanation reqd for this qn

IMO A

(1) x^2 > 9 : From this we know X > 3 or X < -3. But question stem says x is negative. So X > 3 is ruled out and we are left with only X < -3. SUFFICIENT.

(2) X^3 < -9 : INSUFFICIENT. We know (-2)^3 = -8 and (-3)^3 = -27. So for some value of X between -2 and -3 we would have X^3 < -9. And for all values of X < -3, X^3 is < -9. For example take -2.9, it is slightly greater than -3. And (-2.9)^3 is definitely less than -9, since it is closer to -27. So we can find values of X < -3 and X > -3 that satisfy (2). Hence Insufficient.

Hope that helps.

Yes according to me the answer is A...
kishokbabu, wat is the OA?

Statement 1: x^2 > 9

x > 3 or x < -3

But the question says that x is negative and so x should be less than -3.


Statement 2: x^3 < -9

x < cube root of -3

but, cube root of -3 is not less than -3 but it is a bit more than that..


Hence the answer should be A...