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

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New post 04 Mar 2019, 00:58
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

79% (01:13) correct 21% (01:12) wrong based on 67 sessions

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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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New post 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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New post 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.

B is the correct answer.
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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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New post 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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New post 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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Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

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New post 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.

Image
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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New post 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.

Answer: B
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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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