Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If points A and B are randomly placed on the circumference [#permalink]

Show Tags

19 Jun 2010, 14:14

1

This post received KUDOS

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

48% (01:50) correct
52% (01:52) wrong based on 164 sessions

HideShow timer Statistics

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?

A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 5/6
_________________

Re: Probability of length of chord AB [#permalink]

Show Tags

19 Jun 2010, 15:08

1

This post was BOOKMARKED

IMO D - 2/3

Probability = required length of arc/ circumference

Take chord AB and join A and B with the center. Find the angle subtended by the chord AB at the center

It will be 60 degrees as sin(AOD) = 1/2 => angle AOD = 30 => angle AOB = 60 Where D is mid point of the chord

Now for the AB > 2 the angle subtended by the chord will be > 60 on one half of the circle and 60 degree on other half of the circle. Total angle = 120 = 2π/3 So total excluded arc = angle * radius = 2πr/3 So required arc = 2πr - 2πr/3 = 4πr/3

Probability = (4πr/3) / ( 2πr) = 2/3 hence D
_________________

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?

Place point A on any place on the circumference. Now, if \(\angle{ACB}=60\), then chord \(AB=raius=2\), so if \(\angle{ACB}<60\), \(AB<raius=2\). But point B can be placed either clockwise or anticlockwise from A. Which means that if B falls in 60 degrees in either of direction from A, AB will be less than radius. Total circumference 360 degrees restricted area 60*2=120 degrees so \(P(AB>2)=\frac{360-120}{360}=\frac{2}{3}\).

But answer can also be \(\frac{3}{4}\) or \(\frac{\sqrt{3}}{2}\).

Don't worry about this question.
_________________

Re: Probability of length of chord AB [#permalink]

Show Tags

03 Sep 2012, 11:23

This is the way I did it:

I drew a circle and drew a line from the center to the point, forming the radius. I asked myself, if I were to pick a another point on the circle for which the chord is equal to 2, what would the angle be of the radius to the first point and the second point? Well the radius of the circle is 2, so if I wanted to make a chord of length two, all I would have to do it draw a 60,60,60, equaliteral triangle from the center to the first point and two the first point making a 60 degree angle between them. The same long is used again, If I were to draw a third point, in the other direction to form a chord of length two, the angle between the center and the first and third point would be 60*. Hence for any single point on the circle, the two point that can be possible be drawn to generate a length of 2, from that first point span 60+60=120*. Everything above that degree and we have a chord bigger than 2. Now 1-(120/360*2*pi*r)=1/3. Hope that made sense

Re: Probability of length of chord AB [#permalink]

Show Tags

03 Sep 2012, 11:46

Hussain15 wrote:

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?

A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 5/6

I don't know whether I am correct or not.

Probability of chord greater than 2 = 1- probability of chord less than or equal to 2. Probability of chord less than or equal to 2 = (Length of circumference of shaded circle)/ Length of circumference = [(60/360)2*pi*2] / [2*pi*2] ie. [(4/6)*pi]/[4*pi] ie 1/6. Hence, Probability of chord greater than 2 = 1-1/6 =5/6. Choice E.

Attachments

123.JPG [ 11.15 KiB | Viewed 7514 times ]

_________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Re: Probability of length of chord AB [#permalink]

Show Tags

03 Sep 2012, 18:17

2/3. There are only 2 chords of 2 in that can be drawn..on adjacent equilateral triangles. These triangles take up 120 degrees of the circle...leaving 240 degrees outside the probable area...hence 240/360=2/3.

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?

A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 5/6

I don't know whether I am correct or not.

Probability of chord greater than 2 = 1- probability of chord less than or equal to 2. Probability of chord less than or equal to 2 = (Length of circumference of shaded circle)/ Length of circumference = [(60/360)2*pi*2] / [2*pi*2] ie. [(4/6)*pi]/[4*pi] ie 1/6. Hence, Probability of chord greater than 2 = 1-1/6 =5/6. Choice E.

There is one little point you missed (using the method you have used to find the way the chord is chosen). Say you put the first point anywhere on the circumference. Now, you have found that if you put the other point on 1/6th of the circumference (right next to the first point), the chord length will be less than or equal to 2. But you have to consider the 1/6th of the circle on the other side of the point too. Say, in your diagram, the left vertex of the triangle lying on the circle is A, the first point. Now B can be to the right of A or to the left of A. So you can put B on 1/3rd of the circle and still get a chord less than or equal to 2. So answer will be 2/3.
_________________

Re: Probability of length of chord AB [#permalink]

Show Tags

04 Sep 2012, 01:27

VeritasPrepKarishma wrote:

SOURH7WK wrote:

Hussain15 wrote:

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2?

A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 5/6

I don't know whether I am correct or not.

Probability of chord greater than 2 = 1- probability of chord less than or equal to 2. Probability of chord less than or equal to 2 = (Length of circumference of shaded circle)/ Length of circumference = [(60/360)2*pi*2] / [2*pi*2] ie. [(4/6)*pi]/[4*pi] ie 1/6. Hence, Probability of chord greater than 2 = 1-1/6 =5/6. Choice E.

There is one little point you missed (using the method you have used to find the way the chord is chosen). Say you put the first point anywhere on the circumference. Now, you have found that if you put the other point on 1/6th of the circumference (right next to the first point), the chord length will be less than or equal to 2. But you have to consider the 1/6th of the circle on the other side of the point too. Say, in your diagram, the left vertex of the triangle lying on the circle is A, the first point. Now B can be to the right of A or to the left of A. So you can put B on 1/3rd of the circle and still get a chord less than or equal to 2. So answer will be 2/3.

Thanks Karishma!!!. I have not thought of that possibility. So I have to add another 1/6.
_________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Re: If points A and B are randomly placed on the circumference [#permalink]

Show Tags

11 Nov 2012, 00:18

kapsycumm wrote:

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2? A. \(\frac{1}{4}\) B. \(\frac{1}{3}\) C. \(\frac{1}{2}\) D. \(\frac{2}{3}\) E. \(\frac{3}{4}\)

I think one way of answering this question is:

Let us assume that the center is A and the two ends of the chord are B and C.

Let us first assume that the length of the chord is 2.

If the length of the chord has to be 2 to start with. The triangle created by drawing lines from the two ends of the chord to the center would be an equilateral triangle.

Which means angle BAC would be 60 degrees. If the angle BAC is less than 60 then the length of the chord would be less than 2 and if it is more than 60 it would be greater than 2.

This means that there are 120 possibilities for angle BAC where the length of BC would be greater than 2.

The probability would therefore be 120/180 = > 2/3

If points A and B are randomly placed on the circumference [#permalink]

Show Tags

29 Nov 2013, 23:18

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2? A)1/4 B)1/3 C)1/2 D)2/3 E)3/4

Re: If points A and B are randomly placed on the circumference [#permalink]

Show Tags

29 Nov 2013, 23:26

If the length of the chord is 2, then the radii joining the ends of the chord to the centre form an equilateral triangle. i.e the angle between A & B from the centre has to be 60.

If A is a random point on the circumference then B can be any point further than 60 degrees of A on either side of A. i.e 60 degrees on either side of A is out of bounds. i.e 120 degress of the circumference is out of bounds. So probability = 360-120/360 = 240/360 = 2/3

Answer is D
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

If points A and B are randomly placed on the circumference of a circle with radius 2, what is the probability that the length of chord AB is greater than 2? A)1/4 B)1/3 C)1/2 D)2/3 E)3/4

Merging similar topics. Please refer to the solutions above.
_________________

Military MBA Acceptance Rate Analysis Transitioning from the military to MBA is a fairly popular path to follow. A little over 4% of MBA applications come from military veterans...

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...