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# M20-12

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Intern
Joined: 10 Mar 2017
Posts: 9

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22 Sep 2017, 02:49
Bunuel wrote:
Subham10 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "You can see that portion of the circle which is above the line is 1/4 of the circumference of the circle"--

How did u conclude this? Please show calculation

This should be easy. I'll give you the hints and let you figure out the rest.

1. The circle is centred at the origin, so each quadrant has 1/4 of the circumference.
2. y = x + 1 cuts the x and y-axis at (-1, 0) and (0, 1) respectively at the same exact points the circle cuts the axis.

Got it. Yes, it was easy!

Thanks & Regards
Intern
Joined: 10 Sep 2015
Posts: 44
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GPA: 4

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23 Nov 2017, 13:25
Hi Bunuel,

A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (-1,0) or the point (0,1) and none of them satisfies the condition.
So shouldn't exact answer be a little less than 1/4.
Math Expert
Joined: 02 Sep 2009
Posts: 43363

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23 Nov 2017, 20:25
asthagupta wrote:
Hi Bunuel,

A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (-1,0) or the point (0,1) and none of them satisfies the condition.
So shouldn't exact answer be a little less than 1/4.

A point has no dimension, so it does not matter. The same way a line has only one dimension, length, but no area.
_________________
Intern
Joined: 10 Sep 2015
Posts: 44
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GPA: 4

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30 Nov 2017, 11:06
Bunuel wrote:
asthagupta wrote:
Hi Bunuel,

A small doubt. Though i agree with solution and got the same answer but in case if we are selecting 1/4 of the circumference, we are including either the point (-1,0) or the point (0,1) and none of them satisfies the condition.
So shouldn't exact answer be a little less than 1/4.

A point has no dimension, so it does not matter. The same way a line has only one dimension, length, but no area.

Thanks
Manager
Joined: 14 Oct 2015
Posts: 227

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23 Dec 2017, 00:34
awal_786@hotmail.com wrote:
Bunuel wrote:
Official Solution:

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. $$\frac{2}{3}$$

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 $$r=\sqrt{1}=1$$.

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 $$b \gt a+1$$? All points $$(a,b)$$ which satisfy this condition (belong to $$T$$ and have y-coordinate &gt; 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 $$\frac{1}{4}$$ of the whole circumference, hence $$P=\frac{1}{4}$$.

Is there any other simple solution??

We have a point a,b such that ... $$b > a + 1$$, we can simplify it as.

$$b - a > 1$$

We know b can max out at $$1$$ since $$x^2 + y^2 = 1$$ is a circle of radius $$1$$. For $$b - a$$ to be greater than $$1$$, $$b$$ has to be positive and $$a$$ has to negative. This only happens in the 2nd quadrant which is contains 1 fourth of the total circle. So probability should be $$1/4$$.
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Re: M20-12   [#permalink] 23 Dec 2017, 00:34

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# M20-12

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