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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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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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New post 04 Mar 2019, 01:01
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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 \(x=\sqrt{9}=3\).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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