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

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Manager
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In the diagram above, points A, O, and B all lie on a dia  [#permalink]

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Updated on: 15 Dec 2017, 06:13
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Question Stats:

65% (02:25) correct 35% (01:42) wrong based on 469 sessions

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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:

3092.png [ 5.8 KiB | Viewed 14790 times ]

Originally posted by saintforlife on 15 Oct 2013, 08:09.
Last edited by Bunuel on 15 Dec 2017, 06:13, 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  [#permalink]

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15 Oct 2013, 08:43
9
3

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:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).

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Re: In the diagram above, points A, O, and B all lie on a dia  [#permalink]

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06 Nov 2014, 21:29
4
1
russ9 wrote:
Bunuel wrote:
Given that:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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 equation

Now, $$3X^2 - 10 X + 3 = 0$$ is of the form $$AX^2 + BX + C = 0$$
i.e.
A=3
B=-10
C=3

When we come across quadratic equations where co-efficient of $$X^2$$ is other than 1, use the following method

1. Write B ( co-efficient 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

$$3X (X - 3) - 1 (X - 3) = 0$$

$$(X-3) (3X - 1) = 0$$

$$X=3$$ or $$X = \frac{1}{3}$$

Therefore X = 3 is the required solution since X= \frac{1}{3} makes D<d which is an invalid case

Since X = 3, therefore \frac{(D}{d)} = \frac{(3}{1)}

##### General Discussion
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Re: In the diagram above, points A, O, and B all lie on a dia  [#permalink]

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15 Oct 2013, 11:43
Bunuel wrote:
Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).

Can you expand this a bit? I reached till this point pretty fast, but I got stuck on this 'non-standard' 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  [#permalink]

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15 Oct 2013, 14:08
1
saintforlife wrote:
Bunuel wrote:
Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>d).

Can you expand this a bit? I reached till this point pretty fast, but I got stuck on this 'non-standard' 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  [#permalink]

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22 Oct 2013, 22:10
1
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:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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  [#permalink]

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22 Oct 2013, 23:06
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:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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  [#permalink]

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25 Mar 2014, 05:39
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  [#permalink]

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06 Nov 2014, 13:13
Bunuel wrote:
Given that:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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  [#permalink]

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18 Sep 2015, 01:05
2
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  [#permalink]

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18 Nov 2015, 03:19
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:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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  [#permalink]

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15 Apr 2016, 14:35
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:
$$\pi{(\frac{D}{2})^2}+\pi{(\frac{d}{2})^2}=\frac{5}{8}\pi{(\frac{D+d}{2})^2}$$;

$$D^2+d^2=\frac{5}{8}(D+d)^2$$;

$$3D^2+3d^2-10Dd=0$$;

Divide by d^2: $$3(\frac{D}{d})^2+3-10\frac{D}{d}=0$$;

Solve for D/d: D/d=3 or D/d=1/3 (discard as D>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  [#permalink]

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15 May 2017, 06:35
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  [#permalink]

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21 Jul 2017, 13:22
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  [#permalink]

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25 Feb 2018, 21:32
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  [#permalink]

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14 Jul 2018, 14:58
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 [#permalink] 14 Jul 2018, 14:58
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