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# In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)

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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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Updated on: 05 May 2017, 07:48
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In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1), and (-1, 1). If a point falls into the square region, what is the probability that the ordinates of the point (x, y) satisfy that $$x^2+y^2>1$$?

(A) $$1-\frac{\pi}{4}$$

(B) $$\frac{\pi}{2}$$

(C) $$4-\pi$$

(D) $$2-\pi$$

(E) $$\pi-2$$

Originally posted by delta09 on 21 Dec 2009, 03:33.
Last edited by Bunuel on 05 May 2017, 07:48, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 56376
In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Dec 2009, 04:40
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6
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution

First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

$$x^2+y^2=1$$ is an equation of a circle also centered at the origin, with radius 1 and the $$area=\pi{r^2}=\pi$$.

We are told that the point is IN the square and want to calculate the probability that it's outside the circle ($$x^2+y^2>1$$ means that the point is outside the given circle):

P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=$$4-\pi$$
Total number of possible outcomes is the area of the square (as given that the point is in the square) =$$4$$

$$P=\frac{4-\pi}{4}=1-\frac{\pi}{4}$$

Hope it's clear.

Attachment:

Untitled.png [ 9.91 KiB | Viewed 4274 times ]

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Joined: 13 Oct 2009
Posts: 19
Schools: ISB, UCLA,Darden
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Dec 2009, 04:37
1
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution

$$x^2 + y^2 = R^2$$ is the equation of a circle with centre (0,0) and radius R

If we draw the square and the circle, then we will see that the circle is inscribed in the square i.e. the diameter of the circle is equal to the length of the side of square.

Area of the circle= pi (1)^2=pi
Area of the square=(2)^2 = 4

The proabibility that the point lies within the square and outside the circle is

(Area of the sqaure - area of the circle)/area of the circle
= 4-pi/4
=1 - pi/4

Plz. let me know if the OA is A.

Thanks
Math Expert
Joined: 02 Aug 2009
Posts: 7763
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Dec 2009, 05:46
hi .. i think ans will be D...
2 is the diag and not side and each side is 2^(1/2)... area is 2...
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Posts: 56376
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Dec 2009, 06:05
chetan2u wrote:
hi .. i think ans will be D...
2 is the diag and not side and each side is 2^(1/2)... area is 2...

Not so.

In our case we have "horizontal" square: side=2, area=4.

We would have the square with diagonal 2 if the vertices were: (0,1), (1,0), (0,-1), (-1,0). In this case if the point is IN the square it can not be outside the circle, as the square, in this case, is inscribed in the circle. Hence the probability would be 0.
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Posts: 7763
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Dec 2009, 06:18
u r correct , i should have marked it on graph before ans .....
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Feb 2018, 04:13
Bunuel wrote:
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution

First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

$$x^2+y^2=1$$ is an equation of a circle also centered at the origin, with radius 1 and the $$area=\pi{r^2}=\pi$$.

We are told that the point is IN the square and want to calculate the probability that it's outside the circle ($$x^2+y^2>1$$ means that the point is outside the given circle):
Attachment:
Untitled.png

P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=$$4-\pi$$
Total number of possible outcomes is the area of the square (as given that the point is in the square) =$$4$$

$$P=\frac{4-\pi}{4}=1-\frac{\pi}{4}$$

Hope it's clear.

hey bunuel, first of all thanks for your effort !!

I would like to know why it is clear that x^2 + y^2 = 1 is a circle ? Or how did you know.

Math Expert
Joined: 02 Sep 2009
Posts: 56376
Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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21 Feb 2018, 05:52
gmatmo wrote:
Bunuel wrote:
delta09 wrote:
In the xy-plane, the vertex of a square are (1, 1), (1,-1), (-1, -1), and (-1,1). If a point falls into the square region, what is the probability that the ordinates of the point (x,y) satisfy that x^2+y^2>1?
(A) 1-pi/4
(B) pi/2
(C) 4-pi
(D) 2-pi
(E) Pi-2

kindly help me to understand the q and provide a simple , step by step digesteble solution

First note that the square we have is centered at the origin, has the length of the sides equal to 2 and the area equal to 4.

$$x^2+y^2=1$$ is an equation of a circle also centered at the origin, with radius 1 and the $$area=\pi{r^2}=\pi$$.

We are told that the point is IN the square and want to calculate the probability that it's outside the circle ($$x^2+y^2>1$$ means that the point is outside the given circle):
Attachment:
Untitled.png

P = (Favorable outcome)/(Total number of possible outcomes).

Favorable outcome is the area between the circle and the square=$$4-\pi$$
Total number of possible outcomes is the area of the square (as given that the point is in the square) =$$4$$

$$P=\frac{4-\pi}{4}=1-\frac{\pi}{4}$$

Hope it's clear.

hey bunuel, first of all thanks for your effort !!

I would like to know why it is clear that x^2 + y^2 = 1 is a circle ? Or how did you know.

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
$$(x-a)^2+(y-b)^2=r^2$$

This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to:
$$x^2+y^2=r^2$$.

According to the above, $$x^2+y^2=1$$ is an equation of a circle centered at the origin, with radius 1.

Check more here: http://gmatclub.com/forum/math-coordina ... 87652.html
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)  [#permalink]

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14 Mar 2019, 05:19
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Re: In the xy-plane, the vertex of a square are (1, 1), (1, -1), (-1, -1)   [#permalink] 14 Mar 2019, 05:19
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