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# If z > 0 is z^(1/3) > z^(1/2) ?

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Intern
Joined: 16 May 2013
Posts: 20
If z > 0 is z^(1/3) > z^(1/2) ?  [#permalink]

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Updated on: 16 Aug 2014, 02:10
2
00:00

Difficulty:

25% (medium)

Question Stats:

79% (01:47) correct 21% (01:39) wrong based on 92 sessions

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If z > 0 is z^(1/3) > z^(1/2) ?

(1) z^(1/2) > z
(2) z^(1/3) > z^(2/3)

Originally posted by pran21 on 15 Aug 2014, 23:29.
Last edited by Bunuel on 16 Aug 2014, 02:10, edited 1 time in total.
Edited the question
Manager
Joined: 21 Sep 2012
Posts: 213
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
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Re: If z > 0 is z^(1/3) > z^(1/2) ?  [#permalink]

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16 Aug 2014, 02:20
1
if z>0 is z^(1/3) > z^(1/2) ?
since z>0, raise both the sides to the power of 6
z^(1/3*6)>z^(1/2*3)
z^2>z^3
it is possible only when z<1 (given that z>0)
so we need to find whether z<1

statement 1:

z^(1/2) > z
it is possible only when x<1
sufficient

statement 2:

z^(1/3) > z^(2/3)
given that x>0, it is possible only when x<1
sufficient

Ans - D
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Joined: 02 Sep 2009
Posts: 54371
Re: If z > 0 is z^(1/3) > z^(1/2) ?  [#permalink]

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16 Aug 2014, 02:26
If z > 0 is z^(1/3) > z^(1/2) ?

For $$\sqrt[3]{z}$$ to be greater than $$\sqrt{z}$$, z must be between 0 and 1. If z is more than or equal to 1, then the cube root from z will obviously be less than or equal to the square root of z. For example:

$$(\sqrt[3]{\frac{1}{64}}=\frac{1}{4})>(\sqrt{\frac{1}{64}}=\frac{1}{8})$$.

$$(\sqrt[3]{64}=4)<(\sqrt{64}=8)$$.

(1) z^(1/2) > z --> $$\sqrt{z}>z$$ --> 0 < z < 1. Sufficient.

(2) z^(1/3) > z^(2/3) --> $$\sqrt[3]{z}> (\sqrt[3]{z})^2$$ --> since z > 0, then $$0 < \sqrt[3]{z} < 1$$ --> 0 < z < 1. Sufficient.

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Re: If z > 0 is z^(1/3) > z^(1/2) ?  [#permalink]

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25 Jul 2017, 03:40
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Re: If z > 0 is z^(1/3) > z^(1/2) ?   [#permalink] 25 Jul 2017, 03:40
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