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M28-21

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M28-21  [#permalink]

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New post 16 Sep 2014, 00:29
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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, 00:29
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2
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, 06:56
What about - 1/2^(1/3)? Would it be in the set?
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New post 17 Jan 2016, 04:12
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Re: M28-21  [#permalink]

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New post 28 Feb 2016, 19:40
1
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, 08:30
2
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, 20:07
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, 20:19
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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Re: M28-21  [#permalink]

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New post 12 Nov 2017, 22:08
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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Re: M28-21  [#permalink]

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New post 11 Mar 2018, 12:40
I think the wording of this makes it very unclear
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Re: M28-21  [#permalink]

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New post 28 Aug 2018, 10:13
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, 10:22
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Re: M28-21  [#permalink]

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New post 06 Nov 2018, 08:34
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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.


-1/8 exists, so does -x^2, so we have -1*(-1/8)^2 = -1 * 1/64 = -1/64
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Re: M28-21  [#permalink]

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New post 23 Dec 2018, 10:41
As per me, a set by definition is a collection of finite numbers. If we go by the explanations, the given set can go on to become a collection of infinite numbers. Please correct me if my understanding is flawed.
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New post 23 Dec 2018, 11:12
Aastha2807 wrote:
As per me, a set by definition is a collection of finite numbers. If we go by the explanations, the given set can go on to become a collection of infinite numbers. Please correct me if my understanding is flawed.


Yes, this particular set will be infinite. The question asks which of the following MUST be in the set and both \(-\frac{1}{64}\) and \(\frac{1}{64}\) will definitely be in the set, though \(\frac{1}{\sqrt[3]{2}}\) will not necessarily be there.
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Re: M28-21 &nbs [#permalink] 23 Dec 2018, 11:12
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