mariah wrote:Is x<1?
1). X^(-1/3)<1
2). X^(-2)<1
Analyze the question stem: "is x<1". The possible answers here are "yes, x is smaller than 1" or "NO, x is not smaller than 1". For DS yes/no questions, the best initial approach is usually to set out and try to "break" the question - show that the statement(s) allow both a yes and a no answer, and thus are insufficient.
Stat. (1) x^-1/3 is 1 / x^1/3, or "1 over cube root of x". If one over cube root of x is smaller than 1, then there are one of two scenarios, exemplified by x=+/-8:
1) Either cube root of x is greater than 1: if x=8, then cube root of 8 = 2, so that 1/2 is indeed smaller than 1. In this case, the answer to the question stem is no, as 8 is greater than 1.
OR
2) cube root of x is negative, in which case for any x 1/cube root x is negative, and thus by definition smaller than 1. Thus, x could be -8, so that 1/cube root(-8) = 1/-2 < 1. But in this case the answer to the question stem is yes, since x=-8 is smaller than 1.
Thus, stat. (1) is insufficient.
Stat. (2) The same initial process with stat. (1) - deal with the negative exponent first: x^-2 = .
x^2 is always positive (or at least non negative), so here we can multiply the inequality by x^2 on both sides:
1/x^2 < 1
1 < x^2
x can be both greater than 1, or smaller than -1 (exampele, x can still be 8 or -8, and both will satisfy 1<x^2.
Since both statements allow x to be either 8 or -8, for which we have both a yes and a no answer to the question "is x<1", then even combined is still insufficient, and the answer is E.
