GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 19:23

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58402
If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

04 Mar 2019, 00:58
00:00

Difficulty:

25% (medium)

Question Stats:

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

### HideShow timer Statistics

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

_________________
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 5014
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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
VP
Joined: 31 Oct 2013
Posts: 1464
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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.

Director
Joined: 06 Jan 2015
Posts: 689
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution
Intern
Joined: 19 Mar 2018
Posts: 4
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58402
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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 $$x=\sqrt{9}=3$$.
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8086
Location: United States (CA)
Re: If w = 1/16^(1/2), x = 1/1000^(1/3) and y = (1/4)^(-2) then  [#permalink]

### Show Tags

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

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