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Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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13 Aug 2018, 03:45
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53% (01:48) correct 47% (01:39) wrong based on 130 sessions
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Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)? (1) x < 1 (2) x > –1
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Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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Updated on: 13 Aug 2018, 08:28
Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question stem: Is \(\sqrt[3]{x} > \sqrt[5]{x}\)? St1: x < 1 a) when x=1/2, answer to question stem is No.b) When, x=1/2, answer to question stem is Yes.c) When, x=1, answer to question stem is No. d) For any value of x less than 1, answer to question stem is No. Question stem is inconsistent. St2: x > –1 ]a) When, x=1/2, answer to question stem is Yes. b) when x=1/2, answer to question stem is No.c) When, x=1, answer to question stem is No. d) For any value of x greater than 1, answer to question stem is yes. Question stem is inconsistent. Insufficient. Combined, we have 1<x<1. Refer the highlighted points of x, Question stem is still inconsistent. Insufficient. Ans. (E)
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Originally posted by PKN on 13 Aug 2018, 06:24.
Last edited by PKN on 13 Aug 2018, 08:28, edited 1 time in total.



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Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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Updated on: 03 Dec 2018, 00:14
Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question : is \(\sqrt[3]{x} > \sqrt[5]{x}\)?\(\sqrt[3]{x} > \sqrt[5]{x}\)? only if x is either greater than 1 or less than 1 Statement 1: x < 1x may be 0.5 or 5 hence NOT SUFFICIENT Statement 2: x > 1x may be 0.5 or 5 hence NOT SUFFICIENT Combining the two statements1 < x < 1 Hence answer to the question is NO for 0.5 and yes for 0.5 hence NOT SUFFICIENT Answer: Option E
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Originally posted by GMATinsight on 13 Aug 2018, 06:57.
Last edited by GMATinsight on 03 Dec 2018, 00:14, edited 1 time in total.



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Re: Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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13 Aug 2018, 07:10
PKN wrote: Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question stem: Is \(\sqrt[3]{x} > \sqrt[5]{x}\)? Let \(\sqrt[3]{x} > \sqrt[5]{x}\) Or, \(\frac{1}{x^3}\frac{1}{x^5}>0\) Or, \(\frac{x^21}{x^5}>0\) Or, \(\frac{\left(x+1\right)\left(x1\right)}{x^5}>0\) So, 1<x<0 or, x>1(1) St1: x < 1 Question stem is inconsistent. (consistent in the range 1<x<0 and inconsistent in the range (0,1)) Insufficient. St2: x > –1 Question stem is inconsistent. (consistent in the range 1<x<0 or, x>1 and inconsistent in the range (0,1)) Insufficient. Combined, we have 1<x<1.(2) Question stem is inconsistent. (consistent in the range 1<x<0 or, x>1 and inconsistent in the range (0,1)) Insufficient. Ans. (E) \(\sqrt[3]{x}\) does NOT mean 1/x^3 instead that means x^(1/3)
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Re: Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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13 Aug 2018, 07:21
GMATinsight wrote: PKN wrote: Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question stem: Is \(\sqrt[3]{x} > \sqrt[5]{x}\)? Let \(\sqrt[3]{x} > \sqrt[5]{x}\) Or, \(\frac{1}{x^3}\frac{1}{x^5}>0\) Or, \(\frac{x^21}{x^5}>0\) Or, \(\frac{\left(x+1\right)\left(x1\right)}{x^5}>0\) So, 1<x<0 or, x>1(1) St1: x < 1 Question stem is inconsistent. (consistent in the range 1<x<0 and inconsistent in the range (0,1)) Insufficient. St2: x > –1 Question stem is inconsistent. (consistent in the range 1<x<0 or, x>1 and inconsistent in the range (0,1)) Insufficient. Combined, we have 1<x<1.(2) Question stem is inconsistent. (consistent in the range 1<x<0 or, x>1 and inconsistent in the range (0,1)) Insufficient. Ans. (E) \(\sqrt[3]{x}\) does NOT mean 1/x^3 instead that means x^(1/3)I completely misread the question. Thanks for noticing GMATinsight.
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Re: Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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13 Aug 2018, 13:00
Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Given: \(\sqrt[3]{x} > \sqrt[5]{x}\) We can raise the inequality to an odd power, hence we have \(x^{5/3} > x\) Statement 1: \(x < 1\) For x = 1/8, we get \(x^{5/3} = (1/8)^{5/3}\) = \((1/2)^{5}\) \(< 1/8\)......Hence NO For x = 1/8, we get \(x^{5/3} = (1/8)^{5/3}\) = \((1/2)^{5}\) \(> 1/8\)......Hence YES Statement 1 is Not Sufficient. Statement 2: \(x > –1\) We can use same examples as above & can show Statement 2 is Not Sufficient. Combining both statements we get \(1 < x < 1\) We can use same same examples as above & can show that combining is also Not Sufficient. Answer E. Thanks, GyM
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Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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02 Dec 2018, 11:11
GMATinsight wrote: Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question : is \(\sqrt[3]{x} > \sqrt[5]{x}\)?\(\sqrt[3]{x} > \sqrt[5]{x}\)? only if x is either greater than 1 or less than 1Statement 1: x < 1x may be 0.5 or 5 hence NOT SUFFICIENT Statement 2: x > 1x may be 0.5 or 5 hence NOT SUFFICIENT Combining the two statements1 < x < 1Hence answer to the question is always NO hence SUFFICIENT Answer: Option C Dear GMATinsightThe highlighted is not the same. Also, 0.5 & 0.5 satisfies both cases, giving different answer to the question. 0.5 gives no \(\sqrt[3]{0.5}\)=0.79 & \(\sqrt[5]{0.5}\)=0.83 0.5 gives yes \(\sqrt[3]{0.5}\)=0.79 & \(\sqrt[5]{0.5}\)=0.83 Answer: E



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Re: Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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03 Dec 2018, 00:15
Mo2men wrote: GMATinsight wrote: Bunuel wrote: Given that x ≠ 0, is \(\sqrt[3]{x} > \sqrt[5]{x}\)?
(1) x < 1
(2) x > –1 Question : is \(\sqrt[3]{x} > \sqrt[5]{x}\)?\(\sqrt[3]{x} > \sqrt[5]{x}\)? only if x is either greater than 1 or less than 1Statement 1: x < 1x may be 0.5 or 5 hence NOT SUFFICIENT Statement 2: x > 1x may be 0.5 or 5 hence NOT SUFFICIENT Combining the two statements1 < x < 1Hence answer to the question is always NO hence SUFFICIENT Answer: Option C Dear GMATinsightThe highlighted is not the same. Also, 0.5 & 0.5 satisfies both cases, giving different answer to the question. 0.5 gives no \(\sqrt[3]{0.5}\)=0.79 & \(\sqrt[5]{0.5}\)=0.83 0.5 gives yes \(\sqrt[3]{0.5}\)=0.79 & \(\sqrt[5]{0.5}\)=0.83 Answer: E Thank you... made correction.. I don't know how I made this mistake while same values are taken in statement 2...
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Re: Given that x ≠ 0, is x^(1/3) > x^(1/5)?
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03 Dec 2018, 00:15






