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# If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then

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Joined: 02 Sep 2009
Posts: 53658
If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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04 Mar 2019, 00:58
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15% (low)

Question Stats:

91% (01:10) correct 9% (01:58) wrong based on 33 sessions

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If $$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$ then

A. w < x < y
B. x < w < y
C. y < x < w
D. y < w < x
E. x < y < w

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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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04 Mar 2019, 01:01
1
Bunuel wrote:
If $$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$ then

A. w < x < y
B. x < w < y
C. y < x < w
D. y < w < x
E. x < y < w

solving expression we get
$$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$
w= 1/4 = .25
x=1/10= 0.1
y= 16
so y>w>x
IMO B
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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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04 Mar 2019, 01:08
Bunuel wrote:
If $$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$ then

A. w < x < y
B. x < w < y
C. y < x < w
D. y < w < x
E. x < y < w

w = 1/4

x = 1/10

y = 16..

y>w>x.

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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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04 Mar 2019, 02:49
Bunuel wrote:
If $$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$ then

A. w < x < y
B. x < w < y
C. y < x < w
D. y < w < x
E. x < y < w

$$w = \sqrt{\frac{1}{16}}$$= $$\frac{1}{4}$$

$$x = \sqrt[3]{\frac{1}{1000}}$$ = $$\frac{1}{10}$$

$$y = (\frac{1}{4})^{(-2)}$$ = 16

Hence B
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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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05 Mar 2019, 00:20
Hi Bunuel,

Why didn't we have the negative root of w?
In that case w will be +1/4 and -1/4.
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Posts: 53658
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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05 Mar 2019, 00:26
basabidas wrote:
Hi Bunuel,

Why didn't we have the negative root of w?
In that case w will be +1/4 and -1/4.

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;
Similarly $$\sqrt{\frac{1}{16}} = \frac{1}{4}$$, NOT +1/4 or -1/4.

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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06 Mar 2019, 19:55
Bunuel wrote:
If $$w = \sqrt{\frac{1}{16}}$$, $$x = \sqrt[3]{\frac{1}{1000}}$$ and $$y = (\frac{1}{4})^{(-2)}$$ then

A. w < x < y
B. x < w < y
C. y < x < w
D. y < w < x
E. x < y < w

Simplifying, we have:

w = 1/4

x = 1/10

y = 4^2 = 16

Thus, x < w < y.

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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then   [#permalink] 06 Mar 2019, 19:55
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