It is currently 19 Sep 2020, 14:42
My Tests

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

x > 0

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Founder
Founder
User avatar
Joined: 18 Apr 2015
Posts: 13306
Followers: 286

Kudos [?]: 3369 [0], given: 12171

CAT Tests
x > 0 [#permalink] New post 17 Dec 2017, 12:52
Expert's post
00:00

Question Stats:

66% (01:00) correct 33% (01:02) wrong based on 59 sessions
\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Manager
Manager
Joined: 25 Nov 2017
Posts: 51
Followers: 0

Kudos [?]: 48 [0], given: 5

Re: x > 0 [#permalink] New post 18 Dec 2017, 01:59
The diagonal of a square can be found by the following formula : Squroot 2 * side
Since we already know the length = Squroot 2 * side = Squroo 2 * x
We can divide the square root and we have side = x. So the area is x^2

For Quantity B we know that the diameter is x, so the radii must be x/2 and for the area it is multiplied by pi.

Lets plug some values.
For x = 2.
QA = Area is equal to 4
QB= Area is 3,14
QA is greater than B.

Lets try x =36
QA = 36
QB = 9,42

But what is bizarre it that when I plug 1 B is greater

Last edited by Popo on 18 Dec 2017, 18:41, edited 1 time in total.
Founder
Founder
User avatar
Joined: 18 Apr 2015
Posts: 13306
Followers: 286

Kudos [?]: 3369 [0], given: 12171

CAT Tests
Re: x > 0 [#permalink] New post 18 Dec 2017, 11:48
Expert's post
The answer is A
_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Manager
Manager
Joined: 25 Nov 2017
Posts: 51
Followers: 0

Kudos [?]: 48 [0], given: 5

Re: x > 0 [#permalink] New post 18 Dec 2017, 18:41
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?
1 KUDOS received
Retired Moderator
User avatar
Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 171

Kudos [?]: 2914 [1] , given: 394

Re: x > 0 [#permalink] New post 19 Dec 2017, 13:35
1
This post received
KUDOS
Expert's post
Popo wrote:
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?


When you plug 1. The Quanity A is \(x^2=1\)
and Quantiy B is \(\frac{3.14}{4}x^2 = 0.785\).

Hence quanity A is always greater.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

1 KUDOS received
VP
VP
Joined: 20 Apr 2016
Posts: 1302
WE: Engineering (Energy and Utilities)
Followers: 22

Kudos [?]: 1308 [1] , given: 251

Re: x > 0 [#permalink] New post 19 Dec 2017, 21:30
1
This post received
KUDOS
Carcass wrote:
\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.



Here let us put statement 1

Area of the square = \(\frac{{diagonal^2}}{2}\)

or \((\sqrt{2}x)^2\) * \(\frac{1}{2}\)
or \(\frac{2x^2}{2}\)

From statement 2

Area of circle = \(\pi * r^2\)

or \(\frac{22}{7}\) * \((\frac{Diameter}{2})^2\)

or \(\frac{22x}{28}\)


From the two results statement 1 > statement 2 for any positive value of x
_________________

If you found this post useful, please let me know by pressing the Kudos Button


Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Intern
Intern
Joined: 08 Apr 2018
Posts: 44
Followers: 0

Kudos [?]: 16 [0], given: 15

Re: x > 0 [#permalink] New post 04 Jun 2018, 19:58
sandy wrote:
Popo wrote:
Carcass wrote:
The answer is A


Can I ask you what is the right reasoning?


When you plug 1. The Quanity A is \(x^2=1\)
and Quantiy B is \(\frac{3.14}{4}x^2 = 0.785\).

Hence quanity A is always greater.

For quantity B, why are we dividing by 4? When a diameter is given, we divide by 2 to get the radius, we should divide x by 2 to get the radius, square it and then multiply by the value of pi
1 KUDOS received
Retired Moderator
User avatar
Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 171

Kudos [?]: 2914 [1] , given: 394

Re: x > 0 [#permalink] New post 05 Jun 2018, 12:51
1
This post received
KUDOS
Expert's post
@Emike56

You are right if \(x\) is the diameter then radius of circle is \(\frac{x}{2}\).

Area of the circle is \(\pi \times radius^2= \pi \times (\frac{x}{2})^2=\pi \times \frac{x^2}{2^2}= \pi \times \frac{x^2}{4}\).

Hope this explains the 4.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Director
Director
Joined: 09 Nov 2018
Posts: 505
Followers: 0

Kudos [?]: 55 [0], given: 1

Re: x > 0 [#permalink] New post 19 Jan 2019, 14:30
Ans A
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Followers: 10

Kudos [?]: 176 [0], given: 4

Re: x > 0 [#permalink] New post 19 Jan 2019, 18:53
Expert's post
Carcass wrote:
\(x > 0\)

Quantity A
Quantity B
The area of a square region with diagonal of length \(\sqrt{2} x\)
The area of a circular region with diameter of length x


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.


The area of a square region with diagonal of length \(\sqrt{2} x\).
If the diagonal is \(x\sqrt{2}\), the sides are x as \(side^2+side^2=(x\sqrt{2})^2=2x^2=2side^2....side=x\)
so area = \(side^2=x^2=A\)

The area of a circular region with diameter of length x
so radius = x/2.
area = B = \(\pi*(\frac{x}{2})^2=\pi*\frac{x^2}{4}=\frac{22}{28}*x^2=\frac{11}{14}*A\).

Thus A is greater
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: x > 0   [#permalink] 19 Jan 2019, 18:53
Display posts from previous: Sort by

x > 0

  Question banks Downloads My Bookmarks Reviews Important topics  


GRE Prep Club Forum Home| About| Terms and Conditions and Privacy Policy| GRE Prep Club Rules| Contact

Powered by phpBB © phpBB Group

Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.