GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 05 Aug 2020, 12:17

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.

Close

Request Expert Reply

Confirm Cancel

The figure above shows 2 circles. The larger circle has center A, radi

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Senior Manager
Senior Manager
avatar
P
Joined: 04 Sep 2017
Posts: 318
The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 21 Sep 2019, 05:57
1
1
31
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

48% (02:57) correct 52% (03:17) wrong based on 297 sessions

HideShow timer Statistics

Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)


PS75571.01


Attachment:
2019-09-21_1753.png
2019-09-21_1753.png [ 27.97 KiB | Viewed 5414 times ]
Most Helpful Community Reply
Senior Manager
Senior Manager
User avatar
P
Joined: 31 May 2018
Posts: 432
Location: United States
Concentration: Finance, Marketing
CAT Tests
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 28 Sep 2019, 04:14
13
7
side of the square = 2R
diagonal of the square = 2√2R
AE = half of the diagonal of the square (A, B, C, and E are collinear and lies on the diagonal of the square)
AE = √2R
AE = AB+BC+CE
√2R = R+r+√2r (BC = r--radius of small circle CE = diagonal of small square)

R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{√2+1}{√2-1}\)

D is the answer
Attachments

ps4.png
ps4.png [ 13.75 KiB | Viewed 4636 times ]

General Discussion
SVP
SVP
avatar
V
Joined: 20 Jul 2017
Posts: 1504
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 21 Sep 2019, 06:24
1
1
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

PS75571.01


Attachment:
2019-09-21_1753.png


Definitely NOT a 2 min Question!

From the figure, AC = R + r
Draw a horizontal line from A and vertical line from C. Let both the lines meet at point H

Note that, triangle ACH is an isosceles right angled triangle, with AH = CH = R - r
--> \(AC^2 = AH^2 + CH^2\)
--> \((R + r)^2 = (R - r)^2 + (R - r)^2\)
--> \(R^2 + 2Rr + r^2 = 2R^2 - 4Rr + 2r^2\)
--> \(R^2 - 6Rr + r^2 = 0\)
--> \((R/r)^2 - 6R/r + 1 = 0\)
Let R/r = x
--> \(x^2 - 6x + 1 = 0\)
--> \(x = (6 ± \sqrt{36 - 4*1*1})/(2*1)\)
--> \(x = (6 ± \sqrt{32})/2\)
--> \(x = (6 ± 4\sqrt{2})/2\)
--> \(x = (3 ± 2\sqrt{2}\)
--> \(x = 3 + 2\sqrt{2}\) [Since R>r --> x > 1]

\(x = (3 + 2\sqrt{2})\) is equivalent to Option D
--> \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\) (Multiply numerator & Denominator by \(\sqrt{2} + 1\) )
--> \((\sqrt{2} + 1)^2 / (\sqrt{2}^2 - 1^2)\)
--> \((2 + 1 + 2\sqrt{2})/1\)
--> \(3 + 2\sqrt{2}\)

IMO Option D

Pls Hit kudos if you like the solution
Manager
Manager
User avatar
P
Joined: 14 Apr 2017
Posts: 79
Location: Hungary
GMAT 1: 760 Q50 V42
WE: Education (Education)
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 21 Sep 2019, 10:09
7
1
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)


PS75571.01


The big and the small squares are similar to each other, so we can set up an equation between the ratio of their side lengths and the ratio of their diagonals.

\(\frac{2R}{r}=\frac{2(R+r+r\sqrt{2})}{r\sqrt{2}}\)

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

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

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

Answer: D
_________________
My book with my solutions to all 230 PS questions in OG2018:
Zoltan's solutions to OG2018 PS
Intern
Intern
avatar
B
Joined: 24 Oct 2016
Posts: 9
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 24 Sep 2019, 23:27
logically we can tell that
AE = AB+ BC + CE
AE = root(2) R
AB =R
Bc =r
Ce= root(2)r
equat the following values

Posted from my mobile device
Manager
Manager
avatar
S
Joined: 18 Apr 2019
Posts: 78
Location: India
GMAT 1: 720 Q48 V40
GPA: 4
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 28 Sep 2019, 01:16
1
Let's see..

As soon as i read "If points A, B, C, and E are collinear" i had a feeling the question will probably revolve around the length of AE. Further on seeing the fig it's quite obvious we can form a right triangle. S, with these things in mind, let's proceed.

What's AE ?
AE = R + r + rRoot2 (i.e the diagonal of CFDE)

Sorry for not providing any visual representation but just try to visualise with me.

Now, if we take the right triangle with AE as the hypotenuse.
R^2 + R^2 = (R+r+rRoot2)^2
2R^2 = (R+r+rRoot2)^2
Taking root (since dis can't be -ve)
(Root2)R = R+r+rRoot2
Solve this and you'll get D

If you still have problem solving it. Let me know, i'll post the detailed soln
Manager
Manager
User avatar
B
Joined: 11 Apr 2020
Posts: 77
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 08 Jul 2020, 02:49
[quote="gmatt1476"]Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)
_________________
Veritas Free Mock 1 : 660 (Q49/V31)
Veritas Free Mock 2 : 680 (Q50/V33)
E-GMAT SigmaX Free Mock : 610 (Q47/V26)
GMAT Official Practice 1 : 670 (Q50/V31)
Kaplan Free Test : 650 (Q48/V31)
Intern
Intern
avatar
B
Joined: 21 Nov 2018
Posts: 22
Location: Nigeria
Concentration: Finance, Entrepreneurship
GPA: 4
WE: Analyst (Investment Banking)
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 04:20
MikeScarnScott wrote:
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)



Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks
Manager
Manager
User avatar
B
Joined: 11 Apr 2020
Posts: 77
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 05:48
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)



Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks



The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.
_________________
Veritas Free Mock 1 : 660 (Q49/V31)
Veritas Free Mock 2 : 680 (Q50/V33)
E-GMAT SigmaX Free Mock : 610 (Q47/V26)
GMAT Official Practice 1 : 670 (Q50/V31)
Kaplan Free Test : 650 (Q48/V31)
Intern
Intern
avatar
B
Joined: 21 Nov 2018
Posts: 22
Location: Nigeria
Concentration: Finance, Entrepreneurship
GPA: 4
WE: Analyst (Investment Banking)
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 06:07
MikeScarnScott wrote:
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)



Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks



The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.



Thanks a lot for responding.

I am still unclear about this ' Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

When i calculate the square root of \(\sqrt{8R^2}\) i get 2R\(\sqrt{2}\)

Do you mind clarifying please?

Also is there a reason why the 45-45-90 triangle rule doesn't work here?
Manager
Manager
User avatar
B
Joined: 11 Apr 2020
Posts: 77
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 06:17
1
OmotayoH wrote:
MikeScarnScott wrote:
OmotayoH wrote:
MikeScarnScott wrote:
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)

Let me offer my explanation, it took me less than a minute to solve this one

Very important to note that the half of the diagonal of the bigger sqaure i.e AE which is equal to \(\sqrt{2}\)R is made up of 3 elements R+2r+ a small element x (small line joining the end of small circle to E)

Now to find x:
We know that the small square inside the smaller circle has a side of length 'r'
the diagonal of this square would be \(\sqrt{2}\)r

x= \(\sqrt{2}\)r - r

Thus,
\(\sqrt{2}\)R= R+2r+\(\sqrt{2}\)r-r

Thus,
R(√2-1) = r(√2+1)

\(\frac{R}{r}\) = \(\frac{(√2+1)}{(√2-1)}\)



Thanks for this.

Could you please explain how you got this - ' the diagonal of the bigger sqaure i.e AE which is equal to 2√2R'

My understanding is that half a square will form an isosceles triange with angles 45-45-90 and sides x-x-x\sqrt{2} (xroot2). Thus i am at a loss to how you got 2√2R

Tanks



The side of the bigger square would be 2R (diameter of bigger cirlce is 2R which will be equal to side of big square)

(2R)^2 + (2R)^2 = (diagonal of big sqaure)^2
4R^2 + 4R^2 = (diagonal of big sqaure)^2
8R^2 = (diagonal of big sqaure)^2
Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

Hope this helps.



Thanks a lot for responding.

I am still unclear about this ' Therefore, (diagonal of big sqaure) = \(\sqrt{8R^2}\) = 2\(\sqrt{2}\)R

When i calculate the square root of \(\sqrt{8R^2}\) i get 2R\(\sqrt{2}\)

Do you mind clarifying please?

Also is there a reason why the 45-45-90 triangle rule doesn't work here?


R is outside the square-root and 2R\(\sqrt{2}\) is same as 2\(\sqrt{2}\)R
45-45-90 will also work
1:1:\(\sqrt{2}\) = 2R:2R:2R\(\sqrt{2}\)
_________________
Veritas Free Mock 1 : 660 (Q49/V31)
Veritas Free Mock 2 : 680 (Q50/V33)
E-GMAT SigmaX Free Mock : 610 (Q47/V26)
GMAT Official Practice 1 : 670 (Q50/V31)
Kaplan Free Test : 650 (Q48/V31)
PS Forum Moderator
User avatar
P
Joined: 18 Jan 2020
Posts: 1533
Location: India
GPA: 4
Premium Member CAT Tests
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 06:21
1
OmotayoH
Your 90-45-45 right angle Isosceles triangle is nothing but two halves of square. We have considered side of bigger square as 2R
That's why the diagonal is 2√2R
You can also write it like 2R, 2R and 2√2R (45°, 45°, 90° triangle)
No such difference. Both are two sides of same coin.

Posted from my mobile device
Manager
Manager
User avatar
B
Joined: 24 May 2018
Posts: 103
Location: India
Concentration: General Management, Human Resources
WE: Engineering (Energy and Utilities)
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 07:32
Please refer to the pic uploaded,

This is how I have done.

Posted from my mobile device
Attachments

DSC_1368~2.JPG
DSC_1368~2.JPG [ 1021.01 KiB | Viewed 915 times ]

CEO
CEO
User avatar
V
Joined: 03 Jun 2019
Posts: 3362
Location: India
GMAT 1: 690 Q50 V34
WE: Engineering (Transportation)
Reviews Badge CAT Tests
Re: The figure above shows 2 circles. The larger circle has center A, radi  [#permalink]

Show Tags

New post 10 Jul 2020, 10:10
gmatt1476 wrote:
Image
The figure above shows 2 circles. The larger circle has center A, radius R cm, and is inscribed in a square. The smaller circle has center C, radius r cm, and is tangent to the larger circle at point B and to the square at points D and F. If points A, B, C, and E are collinear, which of the following is equal to R/r ?

A. \(\frac{2}{\sqrt{2} + 1}{\)

B. \(\frac{2}{\sqrt{2} - 1}{\)

C. \(\frac{2}{2\sqrt{2} + 1}{\)

D. \(\frac{\sqrt{2} + 1}{\sqrt{2} - 1}{\)

E. \(\frac{2\sqrt{2} + 1}{2\sqrt{2} - 1}{\)



PS75571.01


Attachment:
2019-09-21_1753.png


Along a diagonal;
\(2(R + r + r\sqrt{2}) = 2R\sqrt{2}\)
\(R (\sqrt{2} - 1) = r(\sqrt{2}+1)\)
\(\frac{R}{r} = \frac{\sqrt{2}+1}{\sqrt{2} - 1}\)

IMO D
_________________
Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
GMAT Club Bot
Re: The figure above shows 2 circles. The larger circle has center A, radi   [#permalink] 10 Jul 2020, 10:10

The figure above shows 2 circles. The larger circle has center A, radi

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





cron

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