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Four concentric circles share the same center. The smallest circle has
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Updated on: 24 Oct 2014, 01:40
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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 \(A_n = A_{n1} + (2n  1)\pi\). 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. \(3\) S9811
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Originally posted by gmat1011 on 16 Oct 2010, 06:13.
Last edited by Bunuel on 24 Oct 2014, 01:40, 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
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16 Oct 2010, 15:03
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_n1 + (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 \(A_n=A_{n1} + (2n1)\pi\) \(A_1 = \pi, r_1=1\) \(A_2=A_1+3\pi=4\pi, r_2=2\) \(A_3=A_2+5\pi=9\pi, r_3=3\) \(A_4=A_3+7\pi=16\pi, r_4=4\) 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
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16 Oct 2010, 20:59
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
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16 Oct 2010, 21:20
Thanks  yeah you guys are right... but thats exactly how the question was worded in gmat club teststhere 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 rewording... thanks.



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Re: Four concentric circles share the same center. The smallest circle has
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17 Oct 2010, 07:29
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
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17 Oct 2010, 09:02
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
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18 Oct 2010, 19:56
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_n1 + (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 \(A_n=A_{n1} + (2n1)\pi\) \(A_1 = \pi, r_1=1\) \(A_2=A_1+3\pi=4\pi, r_2=2\) \(A_3=A_2+5\pi=9\pi, r_3=3\) \(A_4=A_3+7\pi=16\pi, r_4=4\) 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
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18 Oct 2010, 20:55
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_n1 + (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 \(A_n=A_{n1} + (2n1)\pi\) \(A_1 = \pi, r_1=1\) \(A_2=A_1+3\pi=4\pi, r_2=2\) \(A_3=A_2+5\pi=9\pi, r_3=3\) \(A_4=A_3+7\pi=16\pi, r_4=4\) 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
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19 Oct 2010, 04:42
thanks for your explanation .....



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Re: Four concentric circles share the same center. The smallest circle has
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12 May 2011, 03:08
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
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12 May 2011, 03:24
ratio = 30pi/20pi
1.5



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Re: Four concentric circles share the same center. The smallest circle has
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29 Dec 2011, 10:10
I hate this question, while doing the MGMAT challenge, I wasted 45 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
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24 Oct 2014, 01:41
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 \(A_n = A_{n1} + (2n  1)\pi\). 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
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28 Nov 2018, 23:25
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