Last visit was: 31 Oct 2024, 00:24 It is currently 31 Oct 2024, 00:24

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.

Close

Request Expert Reply

Confirm Cancel
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3200 [10]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
Most Helpful Community Reply
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Own Kudos [?]: 3200 [3]
Given Kudos: 172
Location: India
WE:Education (Education)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 03 Aug 2021
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Intern
Intern
Joined: 14 Jun 2021
Posts: 5
Own Kudos [?]: 3 [1]
Given Kudos: 1
Send PM
The figure above shows 2 circles. The larger circle has center A [#permalink]
1
Great!

Originally posted by venkyappu19981999 on 04 Aug 2021, 11:49.
Last edited by venkyappu19981999 on 07 Aug 2021, 09:20, edited 1 time in total.
avatar
Intern
Intern
Joined: 26 Jul 2021
Posts: 5
Own Kudos [?]: 4 [1]
Given Kudos: 58
Send PM
Re: The figure above shows 2 circles. The larger circle has center A [#permalink]
1
KarunMendiratta wrote:
Explanation:

(Refer to the figure below)

Since, AM = ME = \(R\)
AME is an isosceles right triangle i.e. AE = \(\sqrt{2}R\)

AE = \(R + 2r\)
\(\sqrt{2}R = R + 2r\)
\((\sqrt{2}-1)R = 2r\)

Col. A: \(R\)
Col. B: \(2(2r)\)

Col. A: \(R\)
Col. B: \(2(\sqrt{2}-1)R\)

Dividing both sides by \(R\);

Col. A: \(1\)
Col. B: \(2(\sqrt{2}-1)\)

Col. A: \(1\)
Col. B: \(0.828\)

Hence, option A


How is AE = R + 2r? It should be AE = R + r + \sqrt{r} as 2r is the diameter of the smaller circle and it doesn't touch the vertex of the square.
Intern
Intern
Joined: 01 May 2021
Posts: 49
Own Kudos [?]: 37 [1]
Given Kudos: 2
Send PM
Re: The figure above shows 2 circles. The larger circle has center A [#permalink]
1
Simply, assign value to R.
Suppose R=8
Then R +2r is diagonal of square.
Diagonal value can be found by using Pythagoras theorem.
Then subtract R from diagobal value to get 2r and multiply that answer with 2 to get 4r.
Which shows Qty A is greater than Qty. B.

Posted from my mobile device
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4999
Own Kudos [?]: 73 [0]
Given Kudos: 0
Send PM
Re: The figure above shows 2 circles. The larger circle has center A [#permalink]
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
Re: The figure above shows 2 circles. The larger circle has center A [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
222 posts

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