In the figure above, the five circles have points in common as shown

let radius of largest circle be 8
Radius of largest = diameter of Medium circle = 8 ; radius of medium = 4
Radius of smallest will be 2
ratio = 2 x π x 2^2 / π x 8^2 = 1/8

Let the larger circle have radius R.

So, QP = R/2 which is the diameter of the smaller circle.

So radius of smaller circle = 0.5*R/2 = R/4

Hence area of 2 smaller circles = 2* (pi*\(R^2\)/16) = pi*\(R^2\)/8

Ratio = (pi*\(R^2\)/8)/pi*\(R^2\) = 1/8

Hence Option B

pandeyashwin then whats wrong with my solution ?

okay I got it, missed medium sized circles and assigned inconvenient radius number initially because if radius of large circle is 6, medium sized 3 and then smallest is 1.5

the region of the largest circle is πr^2
the region of the shaded circles is: 2 * π(1/4r)^2
=> 1/8 (B)

Let the radius of bigger circle = 8
radius of smallest circle = 4
radius of shaded circle = 2

Area of a largest circle = 64
Area of shaded circles = 8

8/64

1/8

B

Let the radius of the bugger circle= 2R
Hence, radius of the medium circle = R
and radius of the smaller circle = R/2
Required Ratio = {π(R/2)^2 + π(R/2)^2}/ π(2R)^2 = 1/8

Hi Brent,

I have learnt a lot from your explanations. You make Maths very easy to understand. Just wanted to say thanks to you for the great work you are doing. God Bless

Hi Farina,

Thanks for the kind words!

Cheers,
Brent

don't really understand how we're assuming values for radius. nowhere is it mentioned that the points are equidistant from each other, they just happen to be at the centre.

let radius of q circle is q and r circle is r, diameter for large circle=\( q+q+r+r= 2q+2r= 2(q+r)\)

therefore, radius of P circle is diameter/2 that is: \(q+r\). Area of P circle= \( pi. r^2= pi * (q+r)^2\)

radius for shaded region is= \(pi* q^2 + pi* r^2\) (area of circle q and r respectively)

so, factor is:\(pi* (q^2 + r^2)\) /\( pi* (q+r)^2\)

\(pi* (q^2 + r^2)\) /\( pi* q^2+r^2 + 2qr \)


\((q^2 + r^2)\) /\(q^2+r^2 + 2qr \)

would apricate if someone can help here.

Hi all!
Since most experts have answered this question in the best ways possible we would attempt to present the solution in a way that helps you organize your GMAT essential concepts from the topic of circles and apply the same with effective strategies.

Key GMAT Concepts necessary for this question:
To solve a question like this you need to remind yourself of the area of a circle. Area is the space occupied whereas perimeter or circumference in the case of a circle is the length of the boundary for the circle. Any "shaded region" is a trigger to bring in the concept of area.
Area of any circle with radius "r" is π\(r^2\).
Diameter of any circle with radius "r" is 2*r =2r

Key GMAT Focus point - In GMAT questions on circle DO NOT jump into assuming you have a radius because a point "looks like" the centre! Classic GMAT traps!
However, in this question we are being told that P,Q,R are the "centres" and they are in a straight line. So its safe to proceed now.

Coming back to our question, YES we can use variables for radiuses to solve this but would you want to do that under a time pressure of 2 min and particularly for this question of an intermediate level that should be sorted in under 90 seconds?
Not really!
I would suggest using the plugin approach that is explained by BrentGMATPrepNow beautifully too!

GMAT Track of thought
Make it a habit to go through the answer choices and question their validity based on logic. Here the two small shaded circles cannot be occupying 50%(1/2) or 25%(1/4) of the area. Eliminate D,E.
It can also not be so small that it occupies 1/16(6.25%) of the area. Eliminate A.

1. On GMAT always take a simplest and easy value to plugin.
A radius cannot be negative or zero. So what's the simplest value you can think of for the radius of the smallest circle? Its "1".
This means we have an area of π*\(1^2\) for the smallest circle = π square units
Thus the area of the two shaded circles = π + π = 2π square units
That also makes the diameter of the smallest circle to be 2*r =2*1 = 2 units.

2. The second circle(centre Q) has a radius which is equal to the diameter of the smallest circle and hence the radius of the second circle with
(centre Q) is = 2 units

3. The third circle and the largest of all with centre P has a radius which is equal to the diameter of the second circle(centre Q)= 2 * 2= 4 units
The area of this circle = π* 4^2 = π*16 = 16 π

Fraction of the largest circle shaded = \(\frac{Area of the two shaded circles}{ Area of the largest circle}\\
\)
=2π square units/ 16 π square units

= \(\frac{2}{16 }\)
=\(\frac{1}{8}\)

(option b)

Devmitra Sen
GMAT mentor

The area of the largest circle is pi*r^2, let r=1 for this circle. So relative the largest circle, the area of the smallest circle is (1/4)^2 because it's radius is a quarter of the largest circle's radius. Simplifies to 1/16

But because there are two small circles you multiply by 2 to get 1/8, answer B

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