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# M28-21

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Math Expert
Joined: 02 Sep 2009
Posts: 44627

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16 Sep 2014, 01:29
Expert's post
4
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Difficulty:

75% (hard)

Question Stats:

47% (01:34) correct 53% (01:29) wrong based on 38 sessions

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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 $$III$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44627

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16 Sep 2014, 01:29
Expert's post
1
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BOOKMARKED
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 $$\frac{1}{\sqrt[3]{2}}$$.

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Joined: 01 Sep 2015
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16 Jan 2016, 07:56
What about - 1/2^(1/3)? Would it be in the set?
Math Expert
Joined: 02 Sep 2009
Posts: 44627

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17 Jan 2016, 05:12
MaxOb12 wrote:
What about - 1/2^(1/3)? Would it be in the set?

No, we cannot say that -1/2^(1/3) must be in the set.
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Joined: 05 Aug 2015
Posts: 48

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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 $$\frac{1}{\sqrt[3]{2}}$$.

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

Math Expert
Joined: 02 Sep 2009
Posts: 44627

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29 Feb 2016, 09:30
1
KUDOS
Expert's post
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 $$\frac{1}{\sqrt[3]{2}}$$.

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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Joined: 26 Sep 2016
Posts: 8
Location: India
GMAT 1: 620 Q49 V26
GPA: 1.32
WE: Engineering (Energy and Utilities)

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 44627

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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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Joined: 18 Mar 2015
Posts: 103
Location: India
Schools: ISB '19
GMAT 1: 600 Q47 V26
GPA: 3.59

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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
Manager
Joined: 26 Feb 2018
Posts: 67
Location: United Arab Emirates
GMAT 1: 710 Q47 V41
GMAT 2: 770 Q49 V47

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11 Mar 2018, 13:40
I think the wording of this makes it very unclear
Re: M28-21   [#permalink] 11 Mar 2018, 13:40
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# M28-21

Moderators: chetan2u, Bunuel

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