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Given that x ≠ 0, is x^(1/3) > x^(1/5)?

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Given that x ≠ 0, is x^(1/3) > x^(1/5)?  [#permalink]

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New post 13 Aug 2018, 03:45
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A
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C
D
E

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  65% (hard)

Question Stats:

53% (01:48) correct 47% (01:39) wrong based on 130 sessions

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Given that x ≠ 0, is x^(1/3) > x^(1/5)?  [#permalink]

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New post 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)?  [#permalink]

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New post 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 < 1

x may be 0.5 or -5 hence

NOT SUFFICIENT

Statement 2: x > -1

x may be -0.5 or 5 hence

NOT SUFFICIENT

Combining the two statements

-1 < 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)?  [#permalink]

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New post 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, \(\frac{x^2-1}{x^5}>0\)
Or, \(\frac{\left(x+1\right)\left(x-1\right)}{x^5}>0\)
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)?  [#permalink]

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New post 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, \(\frac{x^2-1}{x^5}>0\)
Or, \(\frac{\left(x+1\right)\left(x-1\right)}{x^5}>0\)
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)?  [#permalink]

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New post 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.


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Given that x ≠ 0, is x^(1/3) > x^(1/5)?  [#permalink]

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New post 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 -1

Statement 1: x < 1

x may be 0.5 or -5 hence

NOT SUFFICIENT

Statement 2: x > -1

x may be -0.5 or 5 hence

NOT SUFFICIENT

Combining the two statements

-1 < x < 1

Hence answer to the question is always NO hence

SUFFICIENT

Answer: Option C


Dear GMATinsight

The 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)?  [#permalink]

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New post 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 -1

Statement 1: x < 1

x may be 0.5 or -5 hence

NOT SUFFICIENT

Statement 2: x > -1

x may be -0.5 or 5 hence

NOT SUFFICIENT

Combining the two statements

-1 < x < 1

Hence answer to the question is always NO hence

SUFFICIENT

Answer: Option C


Dear GMATinsight

The 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)?   [#permalink] 03 Dec 2018, 00:15
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