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# M28-21

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Math Expert
Joined: 02 Sep 2009
Posts: 52392

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16 Sep 2014, 00:29
12
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Difficulty:

95% (hard)

Question Stats:

43% (01:13) correct 57% (01:30) wrong based on 214 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 $$III$$

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16 Sep 2014, 00:29
3
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 $$\frac{1}{\sqrt[3]{2}}$$.

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Joined: 01 Sep 2015
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16 Jan 2016, 06:56
What about - 1/2^(1/3)? Would it be in the set?
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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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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 $$\frac{1}{\sqrt[3]{2}}$$.

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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Math Expert
Joined: 02 Sep 2009
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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 $$\frac{1}{\sqrt[3]{2}}$$.

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
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GMAT 1: 620 Q49 V26
GPA: 1.32
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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.
Math Expert
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Posts: 52392

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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.
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Schools: ISB '19
GMAT 1: 600 Q47 V26
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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
Manager
Joined: 26 Feb 2018
Posts: 79
Location: United Arab Emirates
GMAT 1: 710 Q47 V41
GMAT 2: 770 Q49 V47

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11 Mar 2018, 12:40
I think the wording of this makes it very unclear
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Joined: 16 Jun 2018
Posts: 10

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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
Math Expert
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Posts: 52392

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28 Aug 2018, 10:22
psych77 wrote:
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

If x then y does not mean if y then x.
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Joined: 05 Oct 2018
Posts: 33
Location: United States
Schools: Wharton Exec '21
GMAT 1: 770 Q49 V47
GPA: 3.95
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06 Nov 2018, 08:34
1
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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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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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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# M28-21

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