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# m20 #12

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24 Feb 2009, 09:23
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Would someone please explain this problem to me? Thank you.

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 $$b \gt a + 1$$ ?

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

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Answer states that b> a+1 is in the upper left quartile because b=a + 1 (y=X+1) covers ponts -1,0 and 0,1, therefore b> a + 1 needs to be above that.

I understand it slightly but I'm not understanding the whole picture. The line b=a+1 that crosses pts -1,0 and 0,1 also is in lower left quartile and upper right quartile so why can't b>a+1 also be in those quartiles above the line?

Thank you.
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24 Feb 2009, 12:43
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take both
y=x+1 and the circle x^2+y^2=1
line is touching the circle in IInd quadrant only
while passing from III rd to Ist Quadrant.

dczuchta wrote:
Would someone please explain this problem to me? Thank you.

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 ?

answer: 1/4. Answer states that b> a+1 is in the upper left quartile because b=a + 1 (y=X+1) covers ponts -1,0 and 0,1, therefore b> a + 1 needs to be above that.

I understand it slightly but I'm not understanding the whole picture. The line b=a+1 that crosses pts -1,0 and 0,1 also is in lower left quartile and upper right quartile so why can't b>a+1 also be in those quartiles above the line?

Thank you.

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24 Feb 2009, 20:13
Thanks nitya34.

Now I understand after reading your answer then looking back at the test answer. The line crosses over the circle and connects with it at two points, cutting off the top left side of the circle which includes all the points of T in which b>a+1.
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28 Dec 2009, 08:36
From b>1+a we get b-a>1. then (b-a)^2>1, b^2+a^2-2ab>1: since b^2+a^2=1 we get 1-2ab>1 or ab<0. Therefore b and a must be of different sign and because b cannot be negative all such points of the circumference belong to the second quadrant (excluding x-and-y intercepts). Answer is A.
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01 Jan 2010, 20:53
Why b can't be negative? Can u pls explain?
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02 Jan 2010, 03:26
|a| is less or equal to 1, so the least value of a is -1, then b>a+1>=0.
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04 Jan 2010, 18:34
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If |a| = 1, a can be equal to 1 then b = 2 and a^2+b^2 > 1
-1<=a<=0, since then b^2+a^2 will be equal to 1 and would still lie in the circle.

The proposed solution of ab<0 where b cannot be negative is most appropriate.

Why b cannot be negative is can be illustrated with an example,
if b is -0.1 then a would be -1.1 since a+1=b => a = b-1 and a^2+b^2 does not lie inside the circle.
I do not understand the explanation where y=x+1 is treated as a line since the line is cutting the circle in the 2nd quadrant and it doesn't clearly tell about the points below or above the line as the points of interest.
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04 Jan 2010, 18:36
dczuchta wrote:
Thanks nitya34.

Now I understand after reading your answer then looking back at the test answer. The line crosses over the circle and connects with it at two points, cutting off the top left side of the circle which includes all the points of T in which b>a+1.

These do not account for 1\4 of the points of the circle. Could you elaborate?
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04 Jan 2011, 03:19
what's the answer guys? my calculation says that answer should be ((pi/4) - (1/2)) / (pi) which makes 1/12 if pi = 3
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04 Jan 2011, 03:40
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Expert's post
dczuchta wrote:
Would someone please explain this problem to me? Thank you.

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 $$b \gt a + 1$$ ?

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

[Reveal] Spoiler: OA
A

Look at the diagram below.
Attachment:

graph.php.png [ 15.81 KiB | Viewed 6138 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.

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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04 Jan 2011, 03:49
thanks just realized the = sign...
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04 Jan 2011, 11:21
nice. what level is this question? anyone know?

thanks.
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04 Jan 2011, 21:08
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Its 1/4.

total points lie inside the circle with radius =1. when divided into 4 quadrants, the points (a,b) would lie in 3rd quadrant, as given the condition that b-a>1, for this value of b needs to be positive and value of a should be negative , accordingly this set lies in quad 3rd, so all the points in this quadrant satisfies the condition and total sample is 4 quadrants
so the probability = 1/4 ( Ans)
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04 Jan 2011, 22:36
it has been well explained above.
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13 Jan 2011, 11:25
Yes, well explained, but really hard
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14 Apr 2011, 13:21
Its difficult !! but I guessed - A and it was right :D
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06 Jan 2012, 08:24
Bunuel, excellent explanation!

Btw, which application did you use to plot the graph?
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07 Jan 2012, 06:12
Great problem. I got it wrong as 1/2. I deduced that a is negative which left me with two quadrants worth of points. But failed to consider the line theory.
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07 Jan 2012, 17:10
I did an estimation. X and Y both have the same numbers that can go into either. If you plug in 1 for x (or y), the other variable is 0. Therefore, adding 1 to any variable will make the probability that the other number is above 1 + variable very unlikely.
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08 Jan 2013, 05:19
Wonderful explaination Bunuel....I took area into consideration but I was wrong, I should have taken circumference into consideration...
+1 Kudos..
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Re: m20 #12   [#permalink] 08 Jan 2013, 05:19

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# m20 #12

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