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

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

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

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

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?

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.

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

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.

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

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

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!

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

[ pi * (b + x)^2 + pi * (b)^2] / [ pi * (2b + x)^2 ] = 5/8

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

[2b^2 + 2xb + x^2] / [4b^2 + 4xb + x^2] = 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-

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

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 ..

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

\(3r^2-10rR+3R^2 = 0\)

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

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