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

Question

https://gmatclub.com/forum/download/file.php?id=45188 In the figure above, the five circles have points in common as shown. P is the center of the largest circle, Q and R are centers of the medium-sized circles, and Q, P, and R are points on a straight line. What fraction of the largest circular region is shaded? (A) \frac{1}{16} (B) \frac{1}{8} (C) \frac{3}{16} (D) \frac{1}{4} (E) \frac{1}{2} Circles.JPG

BrentGMATPrepNow · Accepted Answer

APPROACH #1 : Apply geometric properties Let's start by assigning a "nice" value to the radius of the smallest circle. Let's say the blue circle has radius 1 https://i.imgur.com/7yB5K1D.png If the blue circle has radius 1 , then the DIAMETER of the blue circle must be 2. This means the radius of the green circle must be 2 https://i.imgur.com/XAJ0YT1.png If the radius of the green circle is 2 , then the DIAMETER of the green circle must be 4. This means the radius of the red circle must be 4 https://i.imgur.com/XDczKng.png Area of circle = \pi r^2 So, the area of the red circle = \pi (4^2) = 16\pi And the area of ONE blue circle = \pi (1^2) = \pi So, the area of TWO blue circles = 2\pi What fraction of the largest circular region is shaded? Answer = \frac{2\pi}{16\pi} = \frac{1}{8} Answer: B ------------------------------------------------------------------------------ APPROACH #2 : Visual estimation IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise. So, if we don't know how to answer a particular question, we can always use this helpful property to eliminate some answer choices by simply "eyeballing" the diagram. We can see that the 2 shaded circles occupy less than 1/4 of the largest circle. This allows us to eliminate answer choices D and E Likewise, it looks like the 2 shaded circles occupy more than 1/16 of the largest circle. So, we can also eliminate answer choice A. So, without performing any calculations whatsoever, we're left with answers choices B and C. Those are pretty good odds if you weren't sure how to answer the question in the first place

Answer: B

dave13 · Answer

have no idea if my solution is correct

how is it possble to find fraction if i dont know radius

anyway i assigned some random values let diamertre of large circle be 12, so radius is 6 hence diametre of smaller circle be 6 and radius is 3 Area of large circle 36 Area of two smaller circles is 18 so \frac{18}{36} i.e. = \frac{1}{2} IMO: E

pandeyashwin · Answer

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

nkin · Answer

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

dave13 · Answer

pandeyashwin  then whats wrong with my solution ?

dave13 · Answer

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

tuyetngantran1993 · Answer

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

KanishkM · Answer

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

anuj02sharma · Answer

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

Farina · Answer

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

BrentGMATPrepNow · Answer

Hi Farina,

Thanks for the kind words!

Cheers,
Brent

ueh55406 · Answer

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.

CrackverbalGMAT · Answer

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

criticall · Answer

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

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

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

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

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