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# M28-21

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Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [0], given: 12794

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16 Sep 2014, 00:29
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

43% (01:39) correct 57% (01:25) wrong based on 35 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

_________________

Kudos [?]: 139549 [0], given: 12794

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [0], given: 12794

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16 Sep 2014, 00:29
Expert's post
1
This post was
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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Kudos [?]: 139549 [0], given: 12794

Intern
Joined: 01 Sep 2015
Posts: 2

Kudos [?]: 2 [0], given: 3

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16 Jan 2016, 06:56
What about - 1/2^(1/3)? Would it be in the set?

Kudos [?]: 2 [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [0], given: 12794

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17 Jan 2016, 04:12
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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Kudos [?]: 139549 [0], given: 12794

Manager
Joined: 05 Aug 2015
Posts: 55

Kudos [?]: 42 [0], given: 36

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28 Feb 2016, 19:40
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!
_________________

Working towards 25 Kudos for the Gmatclub Exams - help meee I'm poooor

Kudos [?]: 42 [0], given: 36

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [1], given: 12794

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29 Feb 2016, 08:30
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:

$$-x^2 =-\frac{1}{64}$$
_________________

Kudos [?]: 139549 [1], given: 12794

Intern
Joined: 26 Sep 2016
Posts: 6

Kudos [?]: 5 [0], given: 17

Location: India
GPA: 1.32
WE: Engineering (Energy and Utilities)

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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.

Kudos [?]: 5 [0], given: 17

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [0], given: 12794

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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.
_________________

Kudos [?]: 139549 [0], given: 12794

Manager
Joined: 18 Mar 2015
Posts: 103

Kudos [?]: 4 [0], given: 113

Location: India
Schools: ISB '19
GMAT 1: 600 Q47 V26
GPA: 3.59

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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

Kudos [?]: 4 [0], given: 113

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

Moderators: chetan2u, Bunuel

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