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

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

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New post Updated on: 24 Oct 2014, 02:40
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

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

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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.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 16 Oct 2010, 16:03
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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

Answer B
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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

Answer B




Hi Pls explain how come r2,23,r4 values are 2,3,4 respectively.I can understand that r1=1 though.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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

Answer B




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.
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 19 Oct 2010, 05:42
thanks for your explanation .....
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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...
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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

1.5
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 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!!!!!
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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New post 24 Oct 2014, 02:41
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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 \(\frac{30\pi}{20\pi} = 1\frac{1}{2}\).


Answer: B
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Re: Four concentric circles share the same center. The smallest circle has  [#permalink]

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Re: Four concentric circles share the same center. The smallest circle has   [#permalink] 29 Nov 2018, 00:25
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