Bunuel wrote:
Official Solution:
For a certain set of numbers, if \(x\) is in the set, then both \(-x^2\) and \(-x^3\) are also in the set. If the number \(\frac{1}{2}\) is in the set , which of the following must also be in the set?
I. \(-\frac{1}{64}\)
II. \(\frac{1}{64}\)
III. \(\frac{1}{\sqrt[3]{2}}\)
A. \(I\) only
B. \(II\) only
C. \(III\) only
D. \(I\) and \(II\) only
E. \(I\), \(II\) and \(III\)
Since 1/2 is in the set, the so must be:
\(-x^2 = -\frac{1}{4}\);
\(-x^3 = -\frac{1}{8}\).
Since \(-\frac{1}{4}\) is in the set, the so must be:
\(-x^3 =\frac{1}{64}\)
Since \(-\frac{1}{8}\) is in the set, the so must be:
\(-x^2 =-\frac{1}{64}\)
The only number we cannot get is \(\frac{1}{\sqrt[3]{2}}\).
Answer: D
I don't quite understand... the problem says if X is in the set, then -X^2 and -X^3 are in the set, but it doesn't say that -X^4 are also in the set ... which is what we are assuming to get from -1/4 to -1/64. Please help - thank you!
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Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor