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# For a certain set of numbers, if x is in the set, then both

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Joined: 26 Apr 2013
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For a certain set of numbers, if x is in the set, then both  [#permalink]

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Updated on: 20 Oct 2013, 22:38
3
16
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Difficulty:

85% (hard)

Question Stats:

45% (01:52) correct 55% (01:58) wrong based on 301 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 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
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.

Originally posted by tk1tez7777 on 20 Oct 2013, 16:54.
Last edited by Bunuel on 20 Oct 2013, 22:38, edited 1 time in total.
Edited the question.
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Re: For a certain set of numbers, if x is in the set, then both  [#permalink]

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20 Oct 2013, 22:24
tk1tez7777 wrote:
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√3

I only
II only
III only
I and II only
I, II and II

OA is D.
Source: Gmatclub test

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.

Maybe the data in the question is wrong... 1/2 can be either -x^3 or x in the set. Filling up the other values, you do not get any of the other answers. If it were 1/4 you would perhaps get a -1/64

If the third option were -1/cb.root(3) , it would also be correct.

Either that or I am understanding the question wrong
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For a certain set of numbers, if x is in the set, then both  [#permalink]

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20 Oct 2013, 22:41
2
tk1tez7777 wrote:
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
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).

Similar questions to practice:
a-set-of-numbers-has-the-property-that-for-any-number-t-in-t-98829.html
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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Re: For a certain set of numbers, if x is in the set, then both  [#permalink]

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20 Oct 2013, 22:45
Bunuel wrote:
tk1tez7777 wrote:
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
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).

Similar questions to practice:
a-set-of-numbers-has-the-property-that-for-any-number-t-in-t-98829.html
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.

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  [#permalink]

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25 Jun 2019, 16:40
Bunuel wrote:
tk1tez7777 wrote:
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
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).

Similar questions to practice:
http://gmatclub.com/forum/a-set-of-numb ... 98829.html
http://gmatclub.com/forum/k-is-a-set-of ... 03005.html
http://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
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Posts: 58404
Re: For a certain set of numbers, if x is in the set, then both  [#permalink]

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04 Jul 2019, 01:17
aliakberza wrote:
Bunuel wrote:
tk1tez7777 wrote:
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
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).

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

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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Re: For a certain set of numbers, if x is in the set, then both   [#permalink] 04 Jul 2019, 01:17
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