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

 It is currently 24 May 2019, 16:07

### 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

# Four concentric circles share the same center. The smallest circle has

Author Message
TAGS:

### Hide Tags

Manager
Joined: 11 Jul 2010
Posts: 182
Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

Updated on: 24 Oct 2014, 02:40
6
00:00

Difficulty:

55% (hard)

Question Stats:

70% (02:56) correct 30% (02:58) wrong based on 112 sessions

### HideShow timer Statistics

Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For $$n$$ greater than 1, the area of the $$n$$th smallest circle in square inches, $$A_n$$, is given by $$A_n = A_{n-1} + (2n - 1)\pi$$.

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

A. $$1$$
B. $$1\frac{1}{2}$$
C. $$2$$
D. $$2\frac{1}{2}$$
E. $$3$$

S98-11

Originally posted by gmat1011 on 16 Oct 2010, 07:13.
Last edited by Bunuel on 24 Oct 2014, 02:40, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Retired Moderator
Joined: 02 Sep 2010
Posts: 759
Location: London
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

16 Oct 2010, 16:03
2
gmat1011 wrote:
Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For n greater than 1, the area of the n th smallest circle in square inches, A_n , is given by A_n = A_n-1 + (2n - 1)

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

(C) 2008 GMAT Club / MGMAT challenge

* 1
* 1 1/2
* 2
* 2 1/2
* 3

I think this should say $$A_n=A_{n-1} + (2n-1)\pi$$
$$A_1 = \pi, r_1=1$$
$$A_2=A_1+3\pi=4\pi, r_2=2$$
$$A_3=A_2+5\pi=9\pi, r_3=3$$
$$A_4=A_3+7\pi=16\pi, r_4=4$$

The sum of circumferences is $$S=\pi(2+4+6+8)=20\pi$$
The sum of areas is $$A=\pi(1+4+9+16)=30\pi$$

Hence ratio is 1.5

_________________
Manager
Joined: 29 Sep 2008
Posts: 90
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

16 Oct 2010, 21:59
shrouded1 pointed out correctly otherwise it would be difficult to calculate the radius.it would require a lot more calculation
Manager
Joined: 11 Jul 2010
Posts: 182
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

16 Oct 2010, 22:20
Thanks - yeah you guys are right... but thats exactly how the question was worded in gmat club tests

there was no pie... it only makes sense if there is a pie and reading the explanation again the answer seems to take that into account. The question needs re-wording... thanks.
Manager
Joined: 08 Sep 2010
Posts: 163
Location: India
WE 1: 6 Year, Telecom(GSM)
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

17 Oct 2010, 08:29
mrinal2100 wrote:
shrouded1 pointed out correctly otherwise it would be difficult to calculate the radius.it would require a lot more calculation

With this wording it was almost impossible ...yesterday i tried 1 hour for this question and at last i had to leave it out of frustation.I thought i was missing something somewhere in this question.
Manager
Joined: 11 Jul 2010
Posts: 182
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

17 Oct 2010, 10:02
sorry there - i feel your pain!

if it makes any difference I felt the same way. thats how the question was worded in the first place.
Manager
Joined: 27 Mar 2010
Posts: 88
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

18 Oct 2010, 20:56
shrouded1 wrote:
gmat1011 wrote:
Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For n greater than 1, the area of the n th smallest circle in square inches, A_n , is given by A_n = A_n-1 + (2n - 1)

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

(C) 2008 GMAT Club / MGMAT challenge

* 1
* 1 1/2
* 2
* 2 1/2
* 3

I think this should say $$A_n=A_{n-1} + (2n-1)\pi$$
$$A_1 = \pi, r_1=1$$
$$A_2=A_1+3\pi=4\pi, r_2=2$$
$$A_3=A_2+5\pi=9\pi, r_3=3$$
$$A_4=A_3+7\pi=16\pi, r_4=4$$

The sum of circumferences is $$S=\pi(2+4+6+8)=20\pi$$
The sum of areas is $$A=\pi(1+4+9+16)=30\pi$$

Hence ratio is 1.5

Hi Pls explain how come r2,23,r4 values are 2,3,4 respectively.I can understand that r1=1 though.
Manager
Joined: 08 Sep 2010
Posts: 163
Location: India
WE 1: 6 Year, Telecom(GSM)
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

18 Oct 2010, 21:55
utin wrote:
shrouded1 wrote:
gmat1011 wrote:
Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For n greater than 1, the area of the n th smallest circle in square inches, A_n , is given by A_n = A_n-1 + (2n - 1)

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

(C) 2008 GMAT Club / MGMAT challenge

* 1
* 1 1/2
* 2
* 2 1/2
* 3

I think this should say $$A_n=A_{n-1} + (2n-1)\pi$$
$$A_1 = \pi, r_1=1$$
$$A_2=A_1+3\pi=4\pi, r_2=2$$
$$A_3=A_2+5\pi=9\pi, r_3=3$$
$$A_4=A_3+7\pi=16\pi, r_4=4$$

The sum of circumferences is $$S=\pi(2+4+6+8)=20\pi$$
The sum of areas is $$A=\pi(1+4+9+16)=30\pi$$

Hence ratio is 1.5

Hi Pls explain how come r2,23,r4 values are 2,3,4 respectively.I can understand that r1=1 though.

From given equation we are getting A2= 4 Pie
A2 is the area of 2nd circle,From formula for the area of a circle,
Pie .(r2)^2 = 4 Pie
hence r2=2
Same way for r3, and r4 .

If it helped u in some way consider giving KUDOS.
Manager
Joined: 22 Sep 2010
Posts: 76
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

19 Oct 2010, 05:42
Senior Manager
Joined: 18 Aug 2009
Posts: 329
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

12 May 2011, 04:08
Share your frustration guys, nowhere in the question stem the pie is mentioned. So, i guess, whoever wrote this question forgot to put it there. It needs to have pie for the problem to be solved...
_________________
Never give up,,,
Director
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 966
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

12 May 2011, 04:24
ratio = 30pi/20pi

1.5
Intern
Joined: 06 Dec 2010
Posts: 16
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

29 Dec 2011, 11:10
I hate this question, while doing the MGMAT challenge, I wasted 4-5 minutes on it unknowningly, threw away my whole game!!!!!
Math Expert
Joined: 02 Sep 2009
Posts: 55271
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

24 Oct 2014, 02:41
1
2
Official Solution:

Four concentric circles share the same center. The smallest circle has a radius of 1 inch. For $$n$$ greater than 1, the area of the $$n$$th smallest circle in square inches, $$A_n$$, is given by $$A_n = A_{n-1} + (2n - 1)\pi$$.

What is the sum of the areas of the four circles, divided by the sum of their circumferences, in inches?

A. $$1$$
B. $$1\frac{1}{2}$$
C. $$2$$
D. $$2\frac{1}{2}$$
E. $$3$$

First, we figure out the area of the smallest circle. $$A_1 = \pi r^2 = \pi 1^2 = \pi$$.

Now, we find the area of the second smallest circle $$(n = 2)$$. $$A_2 = A_1 + (2(2) - 1) \pi = \pi + 3 \pi = 4 \pi$$. This means that the radius of the second smallest circle is 2 (since the area is $$\pi r^2$$).

The third smallest circle has area $$A_3 = A_2 + (2(3) - 1) \pi = 4 \pi + 5 \pi = 9 \pi$$. This means that the radius of this circle is 3.

Finally, the fourth smallest circle (that is, the largest circle) has area $$A_4 = A_3 + (2(4) - 1) \pi = 9 \pi + 7 \pi = 16 \pi$$. This means that the radius of this circle is 4.

The sum of all the areas is $$\pi + 4\pi + 9\pi + 16\pi = 30\pi$$.

The sum of all the circumferences is $$2\pi$$ times the sum of all the radii. The sum of all the radii is $$1 + 2 + 3 + 4 = 10$$, so the circumferences sum up to $$20\pi$$.

Thus, the sum of all the areas, divided by the sum of all the circumferences, is $$\frac{30\pi}{20\pi} = 1\frac{1}{2}$$.

_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 11012
Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

### Show Tags

29 Nov 2018, 00:25
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________
Re: Four concentric circles share the same center. The smallest circle has   [#permalink] 29 Nov 2018, 00:25
Display posts from previous: Sort by