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For a certain set of numbers, if x is in the set, then both
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Updated on: 20 Oct 2013, 21:38
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44% (01:49) correct 56% (01:56) wrong based on 328 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.
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Originally posted by tk1tez7777 on 20 Oct 2013, 15:54.
Last edited by Bunuel on 20 Oct 2013, 21:38, 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
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20 Oct 2013, 21:24
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
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20 Oct 2013, 21:41
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). Answer: D. Similar questions to practice: asetofnumbershasthepropertythatforanynumbertint98829.htmlkisasetofintegerssuchthatiftheintegerrisink103005.htmlforacertainsetofnumbersifxisinthesetthenx136580.htmlP.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
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20 Oct 2013, 21:45
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). Answer: D. Similar questions to practice: asetofnumbershasthepropertythatforanynumbertint98829.htmlkisasetofintegerssuchthatiftheintegerrisink103005.htmlforacertainsetofnumbersifxisinthesetthenx136580.htmlP.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
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25 Jun 2019, 15:40
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). Answer: D. Similar questions to practice: http://gmatclub.com/forum/asetofnumb ... 98829.htmlhttp://gmatclub.com/forum/kisasetof ... 03005.htmlhttp://gmatclub.com/forum/foracertain ... 36580.htmlP.S. Please hide the OA under the spoiler. Hi BunuelCould 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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Re: For a certain set of numbers, if x is in the set, then both
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04 Jul 2019, 00:17
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). Answer: D. Similar questions to practice: http://gmatclub.com/forum/asetofnumb ... 98829.htmlhttp://gmatclub.com/forum/kisasetof ... 03005.htmlhttp://gmatclub.com/forum/foracertain ... 36580.htmlP.S. Please hide the OA under the spoiler. Hi BunuelCould 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
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