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

Question

https://gmatclub.com/forum/download/file.php?id=21364 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 3092.png

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=21364 
 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).

Answer: D.

Ashishmathew01081987 · Answer

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)}

Hence answer is D

saintforlife · Answer

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?

imsahlahno · Answer

Put D/d = x
The equation becomes 3x^2 - 10x + 3 = 0

anu1706 · Answer

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

WoundedTiger · Answer

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

prasi009 · Answer

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

russ9 · Answer

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?

Success2015 · Answer

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.

nishantsharma87 · Answer

Took me nearly 4.5 minutes to solve this question. Does the real GMAT test such lengthy equations ? :?:

kaaaaku · Answer

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   -   = 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.

LakerFan24 · Answer

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

12.nehasingh · Answer

Hi I used back solving. After after creating the equation 3(Dd)2+3−10Dd=0
Substituted the values D= 3d

vinayakaggrawal · Answer

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

Frosty · Answer

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!

Fdambro294 · Answer

probably took the long way to do this, but...

Let the Entire Large Circle's Radius = R

Let the Radius with Center A = a

Let the Radius with Center B = b

Let the Distance from Center O to the Edge where the 2 Smaller Circles "TOUCH" = x

then:

The Large Circle's Radius R from Center O to the Right Side of the Circle = x + 2b

The Large Circle's Radius R from Center O to the Left Side of the Circle = 2a - x

Setting R Equal to itself:

x + 2b = 2a - x

2x + 2b = 2a

x + b = a

Radius of Circle with Center A = (b + x)

Radius of Circle with Center B = b

Entire Large Circle's Radius with Center 0 = (2b + x)

Setting up the Ratio of Areas of Circles:

SUM of Smaller Circles' Areas / Entire Circles = 5/8

/   = 5/8

----Canceling (pi) on the L.H.S. and FOIL'ing out the Quadratics ---

/   = 5/8

---Cross-Multiplying---

16b^2 + 16xb + 8x^2 = 20b^2 + 20xb + 5x^2

4b^2 + 4xb - 3x^2 = 0

---Factoring the Quadratic with Co-efficient a =4 and b =4 and c = -3-----

4b^2 - 2xb + 6xb - 3x^2 = 0

2b * (2b - x) + 3x * (2b - x) = 0

(2b - x) * (2b + 3x) = 0

Rule: According to the Zero Product Rule, at least 1 of the Factors (or both) must equal 0

1st)

2b + 3x = 0
2b = (-)3x

Since X is a (+)Positive Length Value, as well as b -----this can NOT be a Valid solution because we would end up with (-)Negative Lengths

2nd)

2b - x = 0

2b = x******

Radius of Circle with Center A = (b + x) = b + 2b = 3b
Diameter of Circle with Center A = 2 * 3b = 6b

Radius of Circle with Center B = b
Diameter of Circle with Center B = 2b

Ratio of Diameters:

D/d = 6b/2b = 3/1

-answer d-

draftpunk · Answer

30 second method: pick easy numbers

5/8 is equal to 10/16
(D^2 + d^2 = 10) / 4^2
(3^2 + 1^2 = 10) / 4^2

Hence, D:d is 3:1

dave13 · Answer

let r = radius of smaller circle and R = radius of bigger one r^2+R^2 = \frac{5}{8}(R+r)^2 3r^2-10rR+3R^2 3r^2 - (rR-9)(rR-1)+3R^2 hi BrentGMATPrepNow can you pls explain how to proceed further ?

I saw posts above but couldnt understand ..

BrentGMATPrepNow · Answer

r^2+R^2 = \frac{5}{8}(R+r)^2

3r^2-10rR+3R^2 = 0

(3r - R)(r - 3R)= 0

So, either  3r - R= 0  or  r - 3R= 0

If  3r - R= 0 , then  3r = R  then  \frac{R}{r}= \frac{3}{1} , which also means  R : r = 3 : 1 , which also means  D : r = 3 : 1

If  r - 3R= 0 , then  r = 3R  then  \frac{R}{r}= \frac{1}{3} , which also means  R : r = 1 : 3 , which also means  D : r = 1 : 3

Since r = radius of  smaller  circle and R = radius of  bigger  one, the correct answer is D

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

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