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For a certain set of numbers, if x is in the set, then both -x^2 Updated on: 13 Sep 2025, 01:19
Originally posted by Bunuel on 20 Oct 2013, 16:54.
Last edited by Bunuel on 13 Sep 2025, 01:19, edited 2 times in total.
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46% (01:53) correct 54% (02:00) wrong based on 1034 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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For a certain set of numbers, if x is in the set, then both -x^2 Updated on: 13 Sep 2025, 01:20
Originally posted by Bunuel on 20 Oct 2013, 22:41.
Last edited by Bunuel on 13 Sep 2025, 01:20, edited 1 time in total.
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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 \(\frac{1}{2}\) is in the set, the following must also be in the set:

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

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

Since \(-\frac{1}{4}\) is in the set, the following must also be in the set:

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

Since \(-\frac{1}{8}\) is in the set, the following must also be in the set:

\(-x^2 =-\frac{1}{64}\)

The only number we cannot obtain is \(\frac{1}{\sqrt[3]{2}}\).

Answer: D

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https://gmatclub.com/forum/a-set-of-num ... 98829.html
https://gmatclub.com/forum/k-is-a-set-o ... 03005.html
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For a certain set of numbers, if x is in the set, then both -x^2 20 Oct 2013, 22:45
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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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct . But how can -1/64 be in the set.

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 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

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 = -1/4;
-x^3 = -1/8.

Since -1/4 is in the set, the so must be:
-x^3 = 1/64;

Since -1/8 is in the set, the so must be:
-x^2 = -1/64.

The only number we cannot get is 1/2^(1/3).

Answer: D.

Similar questions to practice:
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k-is-a-set-of-integers-such-that-if-the-integer-r-is-in-k-103005.html
for-a-certain-set-of-numbers-if-x-is-in-the-set-then-x-136580.html

P.S. Please hide the OA under the spoiler.
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Ah Thanks ! I stopped with the first iteration! Should have checked for -1/4 and -1/8 !
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For a certain set of numbers, if x is in the set, then both -x^2 25 Jun 2019, 16:40
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Bunuel
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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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct . But how can -1/64 be in the set.

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 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

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, then so must be:
-x^2 = -1/4;
-x^3 = -1/8.

Since -1/4 is in the set, then so must be:
-x^3 = 1/64;

Since -1/8 is in the set, then so must be:
-x^2 = -1/64.

The only number we cannot get is 1/2^(1/3).

Answer: D.

Similar questions to practice:
https://gmatclub.com/forum/a-set-of-numb ... 98829.html
https://gmatclub.com/forum/k-is-a-set-of ... 03005.html
https://gmatclub.com/forum/for-a-certain ... 36580.html

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Hi Bunuel

Could you please confirm whether \((\frac{-1}{2})^\frac{1}{3}\) would be in the set? Given that \(\frac{1}{2}=-x^3\) ==>> \(x=-(\frac{1}{2})^\frac{1}{3}\). Thanks
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For a certain set of numbers, if x is in the set, then both -x^2 04 Jul 2019, 01:17
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Bunuel
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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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct . But how can -1/64 be in the set.

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 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

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, then so must be:
-x^2 = -1/4;
-x^3 = -1/8.

Since -1/4 is in the set, then so must be:
-x^3 = 1/64;

Since -1/8 is in the set, then so must be:
-x^2 = -1/64.

The only number we cannot get is 1/2^(1/3).

Answer: D.

Similar questions to practice:
https://gmatclub.com/forum/a-set-of-numb ... 98829.html
https://gmatclub.com/forum/k-is-a-set-of ... 03005.html
https://gmatclub.com/forum/for-a-certain ... 36580.html

P.S. Please hide the OA under the spoiler.

Hi Bunuel

Could you please confirm whether \((\frac{-1}{2})^\frac{1}{3}\) would be in the set? Given that \(\frac{1}{2}=-x^3\) ==>> \(x=-(\frac{1}{2})^\frac{1}{3}\). Thanks
Show more

No, \(-(\frac{1}{2})^\frac{1}{3}\) is not necessarily in the set. We are not told that if x is in the set then \(-\sqrt[3]{x}\) is in the set.
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For a certain set of numbers, if x is in the set, then both -x^2 17 Jun 2020, 08:14
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Bunuel

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 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Could you please confirm whether \((\frac{-1}{2})^\frac{1}{3}\) would be in the set? Given that \(\frac{1}{2}=-x^3\) ==>> \(x=-(\frac{1}{2})^\frac{1}{3}\). Thanks

No, \(-(\frac{1}{2})^\frac{1}{3}\) is not necessarily in the set. We are not told that if x is in the set then \(-\sqrt[3]{x}\) is in the set.
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VeritasKarishma

I have a doubt in the highlighted part. we are told that if x is there then \(-x^3\) will also be there. By this if 1/2 is taken as -x^3, then we'll get \((\frac{-1}{2})^\frac{1}{3}\) as x

Can you pls clarify?
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For a certain set of numbers, if x is in the set, then both -x^2 18 Jun 2020, 22:39
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GDT
Bunuel

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 1/2 is in the set , which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III

Could you please confirm whether \((\frac{-1}{2})^\frac{1}{3}\) would be in the set? Given that \(\frac{1}{2}=-x^3\) ==>> \(x=-(\frac{1}{2})^\frac{1}{3}\). Thanks

No, \(-(\frac{1}{2})^\frac{1}{3}\) is not necessarily in the set. We are not told that if x is in the set then \(-\sqrt[3]{x}\) is in the set.

VeritasKarishma

I have a doubt in the highlighted part. we are told that if x is there then \(-x^3\) will also be there. By this if 1/2 is taken as -x^3, then we'll get \((\frac{-1}{2})^\frac{1}{3}\) as x

Can you pls clarify?
Show more

Are you given that if -x^3 is there in the set, then x is also there? No. This means if I know that -8 is in the set, then I cannot say whether 2 is in the set or not. But I can say that -64 and 512 are in the set. We can only go forward with what we know exists.
Let's say I start building the set and I put 1/2 in it. Then I must put -1/4 and -1/8 in it too. Then I must put -1/16, 1/64, - 1/64 and 1/512 in it too and so on...
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For a certain set of numbers, if x is in the set, then both -x^2 11 Jul 2020, 17:10
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tk1tez7777
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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct. But how can -1/64 be in the set.
Show more

If x is 1/2, then (-x)^2 will be (-1/2)*(-1/2)= 1/4
and (-x)^3 will be (-1/2)*(-1/2)*(-1/2) = -1/8

isn't it?

And A set could contain only [x, (-x)^2, (-x)^3], as the question didn't mention any further where it goes, how can I end up with 1/64 or -1/64?

Could you please Bunuel ScottTargetTestPrep VeritasKarishma, clear my thought.
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For a certain set of numbers, if x is in the set, then both -x^2 15 Jul 2020, 00:30
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kamrulhasan26043
tk1tez7777
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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct. But how can -1/64 be in the set.

If x is 1/2, then (-x)^2 will be (-1/2)*(-1/2)= 1/4
and (-x)^3 will be (-1/2)*(-1/2)*(-1/2) = -1/8

isn't it?

And A set could contain only [x, (-x)^2, (-x)^3], as the question didn't mention any further where it goes, how can I end up with 1/64 or -1/64?

Could you please Bunuel ScottTargetTestPrep VeritasKarishma, clear my thought.
Show more


The way the set is defined, it HAS to be a set of infinite elements.
"if x is in the set, then both -x^2 and -x^3 are also in the set."

If 1/4 is in the set, it is a MUST that -1/16 and -1/64 will be in the set. Going further, since -1/16 is in the set, ...

By its very definition, the set will continue to be formed.
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For a certain set of numbers, if x is in the set, then both -x^2 10 Sep 2020, 07:18
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kamrulhasan26043
tk1tez7777
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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III


Source: Gmatclub test
Show Spoiler
Explanation given is this that if -1/8 is in the set then 1/64 will also be in the set. Which is correct. But how can -1/64 be in the set.

If x is 1/2, then (-x)^2 will be (-1/2)*(-1/2)= 1/4
and (-x)^3 will be (-1/2)*(-1/2)*(-1/2) = -1/8

isn't it?

And A set could contain only [x, (-x)^2, (-x)^3], as the question didn't mention any further where it goes, how can I end up with 1/64 or -1/64?

Could you please Bunuel ScottTargetTestPrep VeritasKarishma, clear my thought.
Show more


Solution:

Notice that the exact statement in the question is “if x is in the set, then -x^2 and -x^3 are also in the set”. This is different from saying that “the set consists of x, -x^2 and -x^3 where x = 1/2”. In the latter case, the set does indeed consist of only 1/2, -(1/2)^2 = -1/4 and -(1/2)^3 = -1/8. However, the former sentence tells us that for any element of the set, the negative square and the negative cube of that element are also in the set. Taking x to be 1/2, we can conclude that -1/4 is in the set since -1/4 = -(1/2)^2 and -1/8 is in the set since -1/8 = -(1/2)^3. Now, we can let x be -1/4 or -1/8 and apply the same sentence. Letting x = -1/4, we find that -(-1/4)^3 = 1/64 is in the set. Letting x = -1/8, we find that -(-1/8)^2 = -1/64 is in the set. We can actually let x be 1/64 or -1/64 and continue the above process to find even more elements which are necessarily in the set.
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For a certain set of numbers, if x is in the set, then both -x^2 22 Oct 2020, 05:42
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If x is in the set, then -x^2 and -x^3 are in the set.

If 1/2 is in the set, then -1/4 and -1/8 are in the set.

If we know -1/4 is in the set, then (1/4)^3 = 1/64 is in the set.

If we know -1/8 is in the set, then -(-1/8)^2 = - 1/64 is in the set.

We don't know for sure if III. 1/2^(1/3) is in the set.

Answer is D.
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For a certain set of numbers, if x is in the set, then both -x^2 20 Dec 2020, 06:52
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tk1tez7777
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 1/2 is in the set, which of the following must also be in the set ?

I. -1/64
II. 1/64
III. 1/2^(1/3)

A. I only,
B. II only,
C. III only,
D. I and II only
E. I, II and III
Show more

\(\frac{1}{2}\) is in the set, so \((-\frac{1}{2})^2\)=\(\frac{1}{4}\) is in the set, Then \((-\frac{1}{4})^3=-\frac{1}{64}\) is in the set.

\(\frac{1}{2}\) is in the set, (-\(\frac{1}{2}\))^3 is in the set=\(-\frac{1}{8}\) is in the set, Then (-\(\frac{1}{8})^2=\frac{1}{64}\)

\(I\) and \(II\) both are answers

The answer is \(D.\)
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For a certain set of numbers, if x is in the set, then both -x^2 29 Nov 2024, 03:53
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