Last visit was: 16 Nov 2024, 05:42 It is currently 16 Nov 2024, 05:42
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

x > 0

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 29962
Own Kudos [?]: 36246 [0]
Given Kudos: 25911
Send PM
x > 0 [#permalink] New post   17 Dec 2017, 12:52
Expert Reply
00:00

Question Stats:

66% (00:55) correct 33% (01:06) wrong based on 74 sessions

Hide Show timer Statistics

\(x > 0\)

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.
ShowHide Answer
Official Answer
A
avatar
Popo
Manager
Manager
Joined: 25 Nov 2017
Posts: 51
Own Kudos [?]: 67 [0]
Given Kudos: 0
Send PM
Re: x > 0 [#permalink] New post   Updated on: 18 Dec 2017, 18:41
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

Originally posted by Popo on 18 Dec 2017, 01:59.
Last edited by Popo on 18 Dec 2017, 18:41, edited 1 time in total.
User avatar
D
Carcass
Verbal Expert
Joined: 18 Apr 2015
Posts: 29962
Own Kudos [?]: 36246 [1]
Given Kudos: 25911
Send PM
Re: x > 0 [#permalink] New post   18 Dec 2017, 11:48
1
Expert Reply
The answer is A
avatar
Popo
Manager
Manager
Joined: 25 Nov 2017
Posts: 51
Own Kudos [?]: 67 [0]
Given Kudos: 0
Send PM
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?
User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4813
Own Kudos [?]: 11166 [0]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: x > 0 [#permalink] New post   19 Dec 2017, 13:35
1
Expert Reply
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.
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2273 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: x > 0 [#permalink] New post   19 Dec 2017, 21:30
1
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
avatar
Emike56
Intern
Intern
Joined: 08 Apr 2018
Posts: 44
Own Kudos [?]: 20 [0]
Given Kudos: 0
Send PM
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
User avatar
sandy
Retired Moderator
Joined: 07 Jun 2014
Posts: 4813
Own Kudos [?]: 11166 [0]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Re: x > 0 [#permalink] New post   05 Jun 2018, 12:51
1
Expert Reply
@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.
avatar
AE
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 133 [0]
Given Kudos: 0
Send PM
Re: x > 0 [#permalink] New post   19 Jan 2019, 14:30
Ans A
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 470 [0]
Given Kudos: 0
Send PM
Re: x > 0 [#permalink] New post   19 Jan 2019, 18:53
Expert Reply
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
User avatar
bumpbot
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 5014
Own Kudos [?]: 74 [0]
Given Kudos: 0
Send PM
Re: x > 0 [#permalink] New post   30 Aug 2021, 03:06
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
gmatclubot
Re: x > 0 [#permalink]
30 Aug 2021, 03:06
Moderators:
adewale223
GRE Instructor
78 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1111 posts
VibhuAnurag
GRE Instructor
234 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact