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 thoughtMake 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 _________________