Question: If two points, A and B, are randomly placed on the circumference of a circle with circumference 12\(\pi\)inches, what is the probability that the length of chord AB is atleast 6 inches?

A. \(1\over2\pi\)
B. \(1\over\pi\)
C. \(1\over3\)
D. \(2\over\pi\)
E. \(2\over3\)

Answer:
Approach Solution (1):
Let’s first determine the details of this circle
For any circle, circumference = (diameter) (\(\pi\))
The circumference of the given circle is 12\(\pi\) inches, so we can write: 12\(\pi\)inches = (diameter)(\(\pi\))
This tells us that the diameter of the circle = 12\(\pi\)
It also tells us that the radius of the circle = 6 inches
Okay, now let’s solve the question
We will begin by arbitrarily placing point A somewhere on the circumference

So, we want to know the probability that a randomly- placed point B will yield a chord AB that is at least 6 inches long
So, let’s first find a location for point B that creates a chord that is exactly 6 inches

There’s also another location for point B that creates another chord that is exactly 6 inches

Important: For chord AB to be greater than or equal to 6 inches, point B must be placed somewhere along the red portion of the circle’s circumference

So, the question really boils down to, “What is the probability that point B is randomly placed somewhere on the red line?”
To determine this probability, notice that the 6-inch chords are the same length as the circle’s radius (6 inches)

Since these 2 triangles have sides of equal length, they are equilateral triangles, which means each interior angle is 60 degrees

The 2 central angles (from the equilateral triangles) add to 120 degrees
This means the remaining central angle must be 240 degrees

This tells us that the red portion of the circle represents 240/360 of the entire circle
So, P (point B is randomly placed somewhere on the red line) = 240/360 = 2/3
Correct option: E

Approach Solution (2):
Radius is 6\((2\pi*r=12\pi=2\pi*6)\)
Fix point A in the circumference (let’s choose the North Pole)
Fro A, draw the two possible chords of the length 6 (‘left’ and ‘right’)
Ending points are X and Y
Join A, X, and Y to the centre O
We have two equilateral triangles (each side is 6), so angle XOY is 120 degrees
Now, point B of our ‘chord greater than 6’ can be chosen anywhere outside arch YX. As inside is 120, outside is 240 degrees
So the probability is 240 (degrees / 360 (degrees) = 2/3
Correct option: E

Approach Solution (3):
Probability questions involving circles are best solved using area or angle subtended on the centre
As 12is the area, diameter of the circle is 12 inches. Taking the limiting case of 6 inches chord- A chord of 6 inches will subtend 60 degrees on the centre
Any chord more than 6 inches will subtend more than 60 degrees on the centre
So the probability of the chord being in that sector is\({240\over360}={2\over3}\)
Correct option: E

“If two points, A and B, are randomly placed on the circumference of a circle with circumference 12inches, what is the probability that the length of chord AB is atleast 6 inches?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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