rainbownlife wrote:is x -ve?
1) x^3(1-x^2)<0
2) x^2 - 1 <0

(1) x^3(1-x^2)<0

Since the product is negative, either:

x^3 < 0 or (1 - x^2) < 0

For the first expression to be negative, x would be negative.

For the second expression to be negative, x could be positive or negative (since x^2 will always be positive or 0).

Therefore, x could be positive or negative: insufficient.

(2) x^2 - 1 <0

x^2 < 1

Since x^2 is always positive (or 0), x could be a negative fraction, positive fraction or 0: insufficient.

Combined:

Let's use our statement (2) limitations (since they're better defined) and see how they interact with statement (1).

If x is a negative fraction, x^3 would be negative. So, x could still be a negative fraction.

If x is 0, neither x^3 nor (1 - x^2) would be negative, so x can no longer be 0.

If x is a positive fraction, neither x^3 nor (1 - x^2) would be negative, so x can no longer be a positive fraction.

Therefore, x must be a negative fraction - a definite "YES" answer: choose (C).

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There really isn't a much quicker way to attack this question. You just need to be really comfortable with number properties to get through tough DS yes/no questions in a reasonable amount of time.

Of course, you could also have done it with picking numbers, but there's no guarantee that that method will be any quicker - it's more hit and miss for timing.

thanks Stuart. Great explanation!