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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8150
GMAT 1: 760 Q51 V42 GPA: 3.82
Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 58% (01:42) correct 42% (01:47) wrong based on 57 sessions

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[GMAT math practice question]

Let $$x=(\frac{1}{3})^{\frac{-1}{2}}$$, $$y=(\frac{1}{2})^{-\frac{1}{3}}$$, and $$z=(\frac{1}{4})^{\frac{-1}{4}}$$. Which of the following is true?

$$A. x<y<z$$
$$B. z<y<x$$
$$C. x<z<y$$
$$D. y<z<x$$
$$E. y<x<z$$

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8150
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.  [#permalink]

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2
=>
We raise each of $$x, y$$, and $$z$$ to the exponent $$12$$ as this will yield integers that are easily compared.
$$x^{12} = ((\frac{1}{3})^{\frac{-1}{2}})^{12} = ((3)^{\frac{1}{2}})^{12} = 3^6 = 729$$
$$y^{12} = ((\frac{1}{2})^{\frac{-1}{3}})^{12} = ((2)^{\frac{1}{3}})^{12} = 2^4 = 16$$
$$z^{12} = ((\frac{1}{4})^{\frac{-1}{4}})^{12} = ((4)^{\frac{1}{4}})^{12} = 4^3 = 64$$

We have $$y^{12} < z^{12} < z^{12}$$. Since $$x, y$$, and $$z$$ are all positive, this tells use that $$y < z < x.$$

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RC Moderator V
Joined: 24 Aug 2016
Posts: 782
GMAT 1: 540 Q49 V16 GMAT 2: 680 Q49 V33 Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.  [#permalink]

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1
MathRevolution wrote:
[GMAT math practice question]

Let $$x=(\frac{1}{3})^\frac{-1}{2}$$, $$y=(\frac{1}{2})^-\frac{1}{3}$$, and $$z=(\frac{1}{4})^\frac{-1}{4}$$. Which of the following is true?

$$A. x<y<z$$
$$B. z<y<x$$
$$C. x<z<y$$
$$D. y<z<x$$
$$E. y<x<z$$

$$x=(\frac{1}{3})^\frac{-1}{2}$$
= $$1/\frac{1}{3}^\frac{1}{2}$$
= $$1/\frac{1}{\sqrt{3}}$$
= $$\sqrt{3}$$ = 1.7 approx

Similarly , $$y=(\frac{1}{2})^-\frac{1}{3}$$ = $$\sqrt{2}$$ = 1.2 approx

And, and $$z=(\frac{1}{4})^\frac{-1}{4}$$ = $$\sqrt{4}$$ = $$\sqrt{2}$$ =1.4 approx

Hence , y<z<x ans D
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Thanks in appreciation.
Intern  B
Joined: 13 May 2017
Posts: 2
Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.  [#permalink]

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As nobody could really calculate the squares you just have to simplify them:

x = (1/3)^(-1/2) = 3^(1/2)
y = (1/2)^(-1/3) = 2^(1/3)
z = (1/4)^(-1/4) = 4^(1/4) = 2^(1/2)

I think now it's easy to sort them by size. Re: Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.   [#permalink] 28 Mar 2018, 12:00
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# Let x=(1/3)^-1/2, y=(1/2)^-1/3, and z=(1/4)^-1/4.  