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Updated on: 26 May 2017, 13:50
Originally posted by BrentGMATPrepNow on 16 Feb 2017, 11:59.
Last edited by BrentGMATPrepNow on 26 May 2017, 13:50, edited 1 time in total.
Last edited by BrentGMATPrepNow on 26 May 2017, 13:50, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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55%
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If two points, A and B, are randomly placed on the circumference of a circle with circumference 12pi inches, what is the probability that the length of chord AB is at least 6 inches?
(A) 1/(2pi)
(B) 1/pi
(C) 1/3
(D) 2/pi
(E) 2/3
*kudos for all correct solutions
(A) 1/(2pi)
(B) 1/pi
(C) 1/3
(D) 2/pi
(E) 2/3
*kudos for all correct solutions
16 Feb 2017, 16:20
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GMATPrepNow
pitroncoso's solution is perfect.
Below is a very similar solution (with a few diagrams to help students visualize the solution)
Let's first determine the details of this circle.
For any circle, circumference = (diameter)(pi)
The circumference of the given circle is 12pi inches, so we can write: 12pi inches= (diameter)(pi)
This tells us that the diameter of the circle = 12 inches
It also tells us that the radius of the circle = 6 inches
Okay, now let's solve the question.
We'll 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
Answer: E
pitroncoso
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Updated on: 17 Feb 2017, 05:05
Originally posted by pitroncoso on 16 Feb 2017, 12:44.
Last edited by pitroncoso on 17 Feb 2017, 05:05, edited 1 time in total.
Last edited by pitroncoso on 17 Feb 2017, 05:05, edited 1 time in total.
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(Another method)
Raduis is 6 (2pi*r=12pi=2pi*6)
Fix point A in the circumference (let's choose the north pole)
From A, draw the two possible chords of length 6 ('left' and 'right')
Ending points are X and Y.
Join A, X and Y to the center O.
We have two equilateral triangles (each side is 6), so angle XOY is 120 degrees.
Now, point B of our 'chord grater than 6' can be chose anywhere outside arch YX. As inside is 120, outside is 240 degrees.
So the probability is 240 (degrees)/ 360 (degrees) = 2/3 letter E
Raduis is 6 (2pi*r=12pi=2pi*6)
Fix point A in the circumference (let's choose the north pole)
From A, draw the two possible chords of length 6 ('left' and 'right')
Ending points are X and Y.
Join A, X and Y to the center O.
We have two equilateral triangles (each side is 6), so angle XOY is 120 degrees.
Now, point B of our 'chord grater than 6' can be chose anywhere outside arch YX. As inside is 120, outside is 240 degrees.
So the probability is 240 (degrees)/ 360 (degrees) = 2/3 letter E
