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Set T consists of all points (x, y) such that x^2+y^2=1. If
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Updated on: 20 Feb 2012, 23:35
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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:
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Originally posted by mirhaque on 15 Apr 2005, 04:45.
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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Set T consists of all points (x, y) such that x^2+y^2=1. If
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20 Feb 2012, 23:36
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. 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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. Attachment:
graph.php.png [ 15.81 KiB  Viewed 13260 times ]
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Re: PLS DRAW & ILLUSTRATE
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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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. Hi Bunnel, Can you pls explain how did u calculate area of segment above the line (pi2/4) for x^2+y^2<1 Anu



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Re: PLS DRAW & ILLUSTRATE
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21 Feb 2012, 11:25
anuu wrote: Hi Bunnel,
Can you pls explain how did u calculate area of segment above the line (pi2/4) 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 (\(\frac{1}{2}*r*r\)): \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi}{4}\frac{1}{2}=\frac{\pi2}{4}\) (since given that \(r=1\)). Hope it's clear.
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Re: PLS DRAW & ILLUSTRATE
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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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. Great explanation! very elegant. thanks Bunuel.



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Re: PLS DRAW & ILLUSTRATE
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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: settconsistsofallpointsxysuchthatx2y21if15626.html#p1047717Hope 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
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28 Dec 2013, 05:22
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: settconsistsofallpointsxysuchthatx2y21if15626.html#p1047717Hope 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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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If
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28 Dec 2013, 06:10
OK sorry now i get it. For a second i forgot that pi = 3.14159 Thanks Cheers J Posted from my mobile device



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Re: PLS DRAW & ILLUSTRATE
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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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). 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



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Re: PLS DRAW & ILLUSTRATE
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09 Jun 2014, 00:50
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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). 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: inthexyplaneatrianglehasvertexes0040and88395.htmlinthecoordinateplanerectangularregionrhasverticesa104869.htmla5meterlongwireiscutintotwopiecesifthelonger106448.htmlatrianglewiththreeequalsidesisinscribedinsidea160874.htmlagameatthestatefairhasacirculartargetwitharadius171756.htmlacylindricaltankhasabasewithacircumferenceof105453.htmlanxycoordinatepairistobechosenatrandomfromthe146005.htmlacylinderhasabasewithacircumferenceof20pimeters132132.htmlacircularracetrackis3milesinlengthandhassignspost106203.htmlinthecoordinateplanerectangularregionrhasverticesa104869.htmlinthexyplanethevertexofasquareare88246.htmlasearchlightontopofthewatchtowermakes3revolutions76069.htmlpointpabisrandomlyselectedintheregionenclosedby160615.htmlHope this helps.
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If
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08 Dec 2014, 13:30
Hi Bunuel, I understand the approach but can you please let me know how did you draw that blue line for f(x)=x+1? Thanks 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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear.



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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If
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08 Dec 2014, 13:34
sunita123 wrote: Hi Bunuel, I understand the approach but can you please let me know how did you draw that blue line for f(x)=x+1? Thanks 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: mathcoordinategeometry87652.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 ycoordinate > xcoordinate + 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 \(\frac{\pi{r^2}}{4}\frac{r^2}{2}=\frac{\pi2}{4}\) so \(P=\frac{area_{segment}}{area_{circle}}=\frac{\frac{\pi2}{4}}{\pi{r^2}}=\frac{\pi2}{4\pi}\). Hope it's clear. The blue line is y = x + 1. The area above it is y > x + 1. So, all the points for which ycoordinate is greater than xcoordinate + 1.
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Re: Set T consists of all points (x, y) such that x^2+y^2=1. If
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22 Jan 2018, 15:34
Hi All, This question is fairly 'highconcept', but you can solve it with a drawing and a bit of logic. When you graph the equation X^2 + Y^2 = 1, you will have a circle with a radius of 1 that is centered around the Origin. There are only four points on that circle that are integer values: (0,1), (1, 0), (0, 1) and (1, 0). All of the other points are positive/negative fractional values. Knowing that, the ONLY way for the inequality Y > X + 1 to occur is when X is NEGATIVE and Y is POSITIVE. That outcome only occurs in the 2nd quadrant of the graph (which is 1/4 of the circle). While you might be unsure about whether every point in that quadrant would 'fit' the inequality or not, the answer choices ARE numbers  and the smallest of them is 1/4. Since there's no smaller possibility, 1/4 must be the answer. Final Answer: GMAT assassins aren't born, they're made, Rich
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Set T consists of all points (x, y) such that x^2+y^2=1. If
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22 Jan 2018, 16:46
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: Its very simple if we proceed in this way... we are given that for any point (a,b) selected from set T \(a^2 + b^2 = 1\). Also, as \(b  a > 1...... i.e. b^2 + a^2  2ab > 1\) or \(ab < 0\). Now, lets consider this sample space Set T = \({(1/\sqrt{2},1/\sqrt{2}) (1/\sqrt{2},1/\sqrt{2}) (1/\sqrt{2},1/\sqrt{2}) (1/\sqrt{2},1/\sqrt{2})}\).. Now as \(ab < 0\), we can select only 2 points from this sample space of \(8.. i.e. 2/8 = 1/4\)... You can try with \(3/\sqrt{5},4/\sqrt{5}\) and write all the possible conditions for this number as above.



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