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Set T consists of all points (x, y) such that x^2+y^2=1. If

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Set T consists of all points (x, y) such that x^2+y^2=1. If [#permalink] New post 15 Apr 2005, 04:45
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Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:
[Reveal] Spoiler: OA

Attachments

0012.jpg
0012.jpg [ 9.03 KiB | Viewed 3570 times ]


Last edited by Bunuel on 20 Feb 2012, 23:35, edited 2 times in total.
Edited the question added the answer choices with OA
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 20 Feb 2012, 23:36
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mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png
graph.php.png [ 15.81 KiB | Viewed 3505 times ]

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

Hope it's clear.
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 21 Feb 2012, 11:01
Bunuel wrote:
mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

Hope it's clear.


Hi Bunnel,

Can you pls explain how did u calculate area of segment above the line (pi-2/4) for x^2+y^2<1

Anu
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 21 Feb 2012, 11:25
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anuu wrote:
Hi Bunnel,

Can you pls explain how did u calculate area of segment above the line (pi-2/4) for x^2+y^2<1

Anu


Sure. Look at the diagram. The area above the line equals to 1/4th of the area of the circle (\frac{\pi{r^2}}{4}) minus the area of the isosceles right triangle made by radii (\frac{1}{2}*r*r): \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4} (since given that r=1).

Hope it's clear.
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 24 Mar 2012, 10:39
anuu wrote:
Bunuel wrote:
mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.



Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 24 Mar 2012, 13:22
Expert's post
pkonduri wrote:
Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?


Check this: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html#p1047717

Hope it helps.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If [#permalink] New post 20 Oct 2013, 12:16
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 26 Nov 2013, 04:48
Bunuel wrote:
mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

Hope it's clear.


Great explanation! very elegant. thanks Bunuel.
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 28 Dec 2013, 04:34
Bunuel wrote:
pkonduri wrote:
Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?


Check this: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html#p1047717

Hope it helps.


Hi Bunuel,

Still don't get how did you arrive at the conclusion that the area above the line within the circle is 1/4th of the circle for the original question

Would you kindly elaborate on this

Thanks
Cheers!
J :)
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 28 Dec 2013, 05:22
Expert's post
jlgdr wrote:
Bunuel wrote:
pkonduri wrote:
Hi Bunuel,

Can you please take some time and clarify my doubt?

How did you arrive at P = 1/4, The portion of the circle that is above the line is (pi*r^2/4) - 1/* r^2 correct?


Check this: set-t-consists-of-all-points-x-y-such-that-x-2-y-2-1-if-15626.html#p1047717

Hope it helps.


Hi Bunuel,

Still don't get how did you arrive at the conclusion that the area above the line within the circle is 1/4th of the circle for the original question

Would you kindly elaborate on this

Thanks
Cheers!
J :)


Original question does not ask about the area, it asks about the portion of the circumference
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If [#permalink] New post 28 Dec 2013, 06:10
OK sorry now i get it. For a second i forgot that pi = 3.14159

Thanks
Cheers
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If [#permalink] New post 25 May 2014, 02:58
Is that really a question which can come up in the GMAT? I believe the mathematics involved go well beyond! The MGMAT for example do not cover circles in xy planes at all.
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 08 Jun 2014, 20:42
Bunuel wrote:
mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

Hope it's clear.


Hi Bunuel,

Do we have more variations of such questions. If yes, could you share it i want to practice.

R/
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Re: PLS DRAW & ILLUSTRATE [#permalink] New post 09 Jun 2014, 00:50
Expert's post
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BOOKMARKED
ammuseeru wrote:
Bunuel wrote:
mirhaque wrote:
Set T consists of all points (x, y) such that x^2+y^2=1. If point (a, b) is selected from set T at random, what is the probability that b>a+1?

(A) \frac{1}{4}
(B) \frac{1}{3}
(C) \frac{1}{2}
(D) \frac{3}{5}
(E) \frac{2}{3}


PLS DRAW & ILLUSTRATE:


Look at the diagram below.
Attachment:
graph.php.png

The circle represented by the equation x^2+y^2 = 1 is centered at the origin and has the radius of r=\sqrt{1}=1 (for more on this check Coordinate Geometry chapter of math book: math-coordinate-geometry-87652.html ).

So, set T is the circle itself (red curve).

Question is: if point (a,b) is selected from set T at random, what is the probability that b>a+1? All points (a,b) which satisfy this condition (belong to T and have y-coordinate > x-coordinate + 1) lie above the line y=x+1 (blue line). You can see that portion of the circle which is above the line is 1/4 of the whole circumference, hence P=1/4.

Answer: A.

If it were: set T consists of all points (x,y) such that x^2+y^2<1 (so set T consists of all points inside the circle). If point (a,b) is selected from set T at random, what is the probability that b>a+1?

Then as the area of the segment of the circle which is above the line is \frac{\pi{r^2}}{4}-\frac{r^2}{2}=\frac{\pi-2}{4} so P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi-2}{4}}{\pi{r^2}}=\frac{\pi-2}{4\pi}.

Hope it's clear.


Hi Bunuel,

Do we have more variations of such questions. If yes, could you share it i want to practice.

R/
Ammu


I searched our questions banks and was able to find the following questions.

PROBABILITY AND GEOMETRY:


in-the-xy-plane-a-triangle-has-vertexes-0-0-4-0-and-88395.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
a-triangle-with-three-equal-sides-is-inscribed-inside-a-160874.html
a-game-at-the-state-fair-has-a-circular-target-with-a-radius-171756.html
a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.html
an-x-y-coordinate-pair-is-to-be-chosen-at-random-from-the-146005.html
a-cylinder-has-a-base-with-a-circumference-of-20pi-meters-132132.html
a-circular-racetrack-is-3-miles-in-length-and-has-signs-post-106203.html
in-the-coordinate-plane-rectangular-region-r-has-vertices-a-104869.html
in-the-xy-plane-the-vertex-of-a-square-are-88246.html
a-searchlight-on-top-of-the-watch-tower-makes-3-revolutions-76069.html
point-p-a-b-is-randomly-selected-in-the-region-enclosed-by-160615.html

Hope this helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: PLS DRAW & ILLUSTRATE   [#permalink] 09 Jun 2014, 00:50
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Set T consists of all points (x, y) such that x^2+y^2=1. If

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