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M28-21

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M28-21  [#permalink]

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New post 16 Sep 2014, 01:29
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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 \(III\)

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Re M28-21  [#permalink]

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New post 16 Sep 2014, 01:29
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
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Re: M28-21  [#permalink]

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New post 16 Jan 2016, 07:56
What about - 1/2^(1/3)? Would it be in the set?
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New post 17 Jan 2016, 05:12
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Re: M28-21  [#permalink]

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New post 28 Feb 2016, 20:40
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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Re: M28-21  [#permalink]

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New post 29 Feb 2016, 09:30
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happyface101 wrote:
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!


The rule says that if a number is in the set, then minus that numbered squared is also in the set and minus that number cubed is also in the set.

So, if 1/2 is in the set, then so must be:

\(-x^2 = -\frac{1}{4}\);

\(-x^3 = -\frac{1}{8}\).

By the same logic, if \(-\frac{1}{4}\) is in the set, then so must be:

\(-x^3 =\frac{1}{64}\)

By the same logic, if \(-\frac{1}{8}\) is in the set, then so must be:

\(-x^2 =-\frac{1}{64}\)
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Re M28-21  [#permalink]

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New post 21 Sep 2017, 21:07
I think this is a high-quality question and I don't agree with the explanation. Since -1/8 exits, 1/64 exist but not -1/64.
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New post 21 Sep 2017, 21:19
Pm091989 wrote:
I think this is a high-quality question and I don't agree with the explanation. Since -1/8 exits, 1/64 exist but not -1/64.


It would be interesting to know your reasoning. Also, reading the solution carefully and going through the entire thread might help.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M28-21  [#permalink]

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New post 12 Nov 2017, 23:08
i selected E for root 2 value must be around 0.0161 so it is in set where largest value is 1/2
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New post 11 Mar 2018, 13:40
I think the wording of this makes it very unclear
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New post 28 Aug 2018, 11:13
Bunuel Why cant we go backwards.. like say if 1/2 exists in the set then (1/\sqrt{2}) and 1/[cube_root]2[/cube_root] also would be existing for 1/2 to exist in the set?
Pls explain
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New post 28 Aug 2018, 11:22
Re: M28-21 &nbs [#permalink] 28 Aug 2018, 11:22
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