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In the diagram above, points A, O, and B all lie on a dia
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Updated on: 15 Dec 2017, 05:13
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In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ? A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 Attachment:
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Originally posted by saintforlife on 15 Oct 2013, 07:09.
Last edited by Bunuel on 15 Dec 2017, 05:13, edited 2 times in total.
Edited the question.




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Re: In the diagram above, points A, O, and B all lie on a dia
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15 Oct 2013, 07:43
In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\). The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\). Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\); \(D^2+d^2=\frac{5}{8}(D+d)^2\); \(3D^2+3d^210Dd=0\); Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\); Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d). Answer: D.
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Re: In the diagram above, points A, O, and B all lie on a dia
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06 Nov 2014, 20:29
russ9 wrote: Bunuel wrote: Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\);
\(D^2+d^2=\frac{5}{8}(D+d)^2\);
\(3D^2+3d^210Dd=0\);
Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\);
Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).
Answer: D. Hi Bunuel, Can you expand on this a little? If I substitute x for D/D, I get x^2  (10/3)x + 1 = 0. Something that add's up to 10/3 and multiplies to get 1? When you substitute \frac{(D}{d)} = X
Your equation becomes: \(3X^2 + 3  10 X = 0\)
Rearranging terms
Your equation becomes:\(3X^2  10 X + 3 = 0\)
You only have to substitute X in place of \frac{(D}{d)} and then solve the quadratic equationNow, \(3X^2  10 X + 3 = 0\) is of the form \(AX^2 + BX + C = 0\) i.e. A=3 B=10 C=3
When we come across quadratic equations where coefficient of \(X^2\) is other than 1, use the following method
1. Write B ( coefficient of X) as the sum of two quantities whose product is equal to A*C
In this case 10 has to be written as the sum of two quantities whose product is 9. We can write 10 as (9) + (1), so that the product of (9) and (1) is equal to 9
2. Rewrite the equation with the BX term split.
In this case the equation can be written as \(3X^2  9X  X +3 = 0\)
Now solve quadratic the usual way
\(3X (X  3)  1 (X  3) = 0\)
\((X3) (3X  1) = 0\)
\(X=3\) or \(X = \frac{1}{3}\)
Therefore X = 3 is the required solution since X= \frac{1}{3} makes D<d which is an invalid caseSince X = 3, therefore \frac{(D}{d)} = \frac{(3}{1)}
Hence answer is D




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Re: In the diagram above, points A, O, and B all lie on a dia
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15 Oct 2013, 10:43
Bunuel wrote: Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\);
Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).
Answer: D. Can you expand this a bit? I reached till this point pretty fast, but I got stuck on this 'nonstandard' quadratic equation. Is there an easy way to solve these by inspection?



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Re: In the diagram above, points A, O, and B all lie on a dia
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15 Oct 2013, 13:08
saintforlife wrote: Bunuel wrote: Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\);
Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).
Answer: D. Can you expand this a bit? I reached till this point pretty fast, but I got stuck on this 'nonstandard' quadratic equation. Is there an easy way to solve these by inspection? Put D/d = x The equation becomes 3x^2  10x + 3 = 0
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Re: In the diagram above, points A, O, and B all lie on a dia
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22 Oct 2013, 21:10
Bunuel wrote: In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\). The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\). Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\); \(D^2+d^2=\frac{5}{8}(D+d)^2\); \(3D^2+3d^210Dd=0\); Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\); Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d). Answer: D. I have taken area as (pie*dia), Is there any wrong in doing this because i am not getting the answer....My equation becomes (D+d)=5/8(D+d). Kindly tell bunuel



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Re: In the diagram above, points A, O, and B all lie on a dia
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22 Oct 2013, 22:06
anu1706 wrote: Bunuel wrote: In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\). The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\). Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\); \(D^2+d^2=\frac{5}{8}(D+d)^2\); \(3D^2+3d^210Dd=0\); Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\); Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d). Answer: D. I have taken area as (pie*dia), Is there any wrong in doing this because i am not getting the answer....My equation becomes (D+d)=5/8(D+d). Kindly tell bunuel Hi Anu1706, Area of Circle is Pi* (Radius)^2 or Pi*(Diameter/2)^2 What you have considered as Area is actually Circumference of the circle (Pi*Dia or Pi*2*radius) hope it helps
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Re: In the diagram above, points A, O, and B all lie on a dia
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25 Mar 2014, 04:39
Sometimes the best sol in geometry is estimation just look closely D/d is asked right
just see D+d is the full diameter of the large circle
D is close to 75% of D+d
d is close to 25% of D+d
D/D+d divided by d/D+d
D+d's cancel out and we are left with 75/25=3 ie 3:1



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Re: In the diagram above, points A, O, and B all lie on a dia
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06 Nov 2014, 12:13
Bunuel wrote: Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\);
\(D^2+d^2=\frac{5}{8}(D+d)^2\);
\(3D^2+3d^210Dd=0\);
Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\);
Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).
Answer: D. Hi Bunuel, Can you expand on this a little? If I substitute x for D/D, I get x^2  (10/3)x + 1 = 0. Something that add's up to 10/3 and multiplies to get 1?



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Re: In the diagram above, points A, O, and B all lie on a dia
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18 Sep 2015, 00:05
here is another approach which might help. I got the eqn to the form 10Dd = 3(D^2 + d^2)...
from here I substituted D and d in the eqn, with values from the answer options (ratios D:d). only 3:1 works.



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Re: In the diagram above, points A, O, and B all lie on a dia
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18 Nov 2015, 02:19
Bunuel wrote: In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\). The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\). Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\); \(D^2+d^2=\frac{5}{8}(D+d)^2\); \(3D^2+3d^210Dd=0\); Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\); Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d). Answer: D. Took me nearly 4.5 minutes to solve this question. Does the real GMAT test such lengthy equations ?



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Re: In the diagram above, points A, O, and B all lie on a dia
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15 Apr 2016, 13:35
Bunuel wrote: In the diagram above, points A, O, and B all lie on a diameter of the large circle, and D and d are the diameters of the two smaller circles. If the sum of the areas of the two smaller circles is 5/8 of the large circle’s area, what is the ratio of D to d ?A. 3:2 B. 5:3 C. 2:1 D. 3:1 E. 4:1 The radius of the larger circle is (D+d)/2. Its area is \(\pi{(\frac{D+d}{2})^2}\). The sum of the areas of the two smaller circles is \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}\). Given that: \(\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}\); \(D^2+d^2=\frac{5}{8}(D+d)^2\); \(3D^2+3d^210Dd=0\); Divide by d^2: \(3(\frac{D}{d})^2+310\frac{D}{d}=0\); Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d). Answer: D. Hello. I was wondering if the following is the right approach: Large circle area: pie (D/2 + d/2)^2 Small cricles: pie (D/2)^2 + pie(d/2)^2 (ignore pies since they cancel out) Expand large circle using the quadratic template: (D/2)^2 + (d/2)^2 + (2)(D/2)(d/2) Since small circles are 5/8 of the big circle. We can say that [Large circle]  [sum of two small circles area] = 1  5/8 or 3/8 Therefore: Large: (D/2)^2 + (d/2)^2 + (2)(D/2)(d/2) (subtract) Small: (D/2)^2 + (d/2)^2 => (2)(D/2)(d/2) = 3/8 Now solve from here onwards. Is this correct approach? Or am i forcing myself to the right answer by using an incorrect approach? Thanks.



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Re: In the diagram above, points A, O, and B all lie on a dia
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15 May 2017, 05:35
any other way of solving that doesn't take an insane amount of computing time?
looks like all these approaches take 5 min to solve unless you've seen this problem before and are solving it based on memory. i'm sure there has to be a shorter way of doing it



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Re: In the diagram above, points A, O, and B all lie on a dia
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21 Jul 2017, 12:22
Hi I used back solving. After after creating the equation 3(Dd)2+3−10Dd=0 Substituted the values D= 3d



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Re: In the diagram above, points A, O, and B all lie on a dia
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25 Feb 2018, 20:32
Given that : 5(D+d)^2 =8(D^2 + d^2) (Cancelling pie and 4) We can directly substitute the options assuming them to be the same units such as 3:1 can be taken as 3cm and 1cm, to get to the answer. A faster way to solve this question under 2 minutes. Another hint I believe is the sum of ratio of D and d has to be divisible by 8. Feel free to comment



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In the diagram above, points A, O, and B all lie on a dia
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14 Jul 2018, 13:58
vinayakaggrawal wrote: Given that : 5(D+d)^2 =8(D^2 + d^2) (Cancelling pie and 4) We can directly substitute the options assuming them to be the same units such as 3:1 can be taken as 3cm and 1cm, to get to the answer. A faster way to solve this question under 2 minutes. Another hint I believe is the sum of ratio of D and d has to be divisible by 8. Feel free to comment The ratio is \(\frac{Area (D) + Area (d)}{Area (large circle)}\) = \(\frac{5}{8}\). Sum area D and d must be divisible by 5. Area of the larger circle must be divisible by 8. In answer choice D (3:1) you have diameter D=6 and diameter d=2. The sum of these two diameters will provide the diameter of the large circle=8. You will then get the ratio \(\frac{9π+π}{16π}\) which reduces to \(\frac{5}{8}\). Cheers!




In the diagram above, points A, O, and B all lie on a dia &nbs
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