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# In the xy- plane, a triangle has vertexes (0,0), (4,0) and

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In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  24 Dec 2009, 07:09
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In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5
[Reveal] Spoiler: OA
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Re: In the xy- plane, [#permalink]  24 Dec 2009, 07:35
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sagarsabnis wrote:
In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: Area of this region/Area of the triangle.

Favorable region is also right triangle with vertexes at (0,0) (4,0) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

P=8/10=4/5

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Re: In the xy- plane, [#permalink]  25 Dec 2009, 05:27
Bunuel

great solution!!!

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Re: In the xy- plane, [#permalink]  03 Jan 2013, 00:31
sagarsabnis wrote:
In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

My approach is to use the x=y line which is the boundary that divides x>y and x<y coordinate system. Below the x=y line is the region of points with x > y. Above the x=y line is the region of points with x < y.

Get the area of the given triangle: (4)(5) / 2 = 10
Get the area of the smaller triangle drawn below the line x=y which has (0,0), (4,0) and (4,4) for its coordinates: (4)(4)/2 = 8

P = 8/10 = 4/5

Detailed Solution: Line x=y as a Boundary
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Re: In the xy- plane, [#permalink]  12 Mar 2013, 06:03
Bunuel wrote:
sagarsabnis wrote:
In the xy- plane, a triangle has vertexes (0,0), (4,0) and (4,5). If a point (x,y) is selected at random from the triangular region, What is the probability that x-y>0 ?

A. 1/5
B. 1/3
c. 1/2
D. 2/3
E. 4/5

We have right triangle with the area=4*5/2=10. Consider the line y<x. All the points which satisfy this equation (are below the line y=x) and lie in the triangular region obviously will have x more than y, which is exactly what we want (as x>y --> x-y>0).

The probability that the point will be from this region is: Area of this region/Area of the triangle.

Favorable region is also right triangle with vertexes at (0,0) (0,4) and (4,4). As y=x intersects the side of our original triangle at the point (4,4). You''ll see it easily if you draw it. So favorable are=4*4/2=8.

P=8/10=4/5

I did not understand the concept of the line y<x. Could u explain it to me in detail maybe with the help of a diagram or something. That would be appreciated. Thanks.
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  15 Mar 2013, 13:55
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See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E
Attachments

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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  17 Mar 2013, 01:07
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

Please correct me if i am wrong.. What if, y=3 and x=2?
That will still remain in the shaded part of the triangle and won't satisfy the question..
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  26 Mar 2013, 07:55
1
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johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  27 Mar 2013, 07:47
Expert's post
perilm41 wrote:
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?

A point by definition has no length, area or any other dimension, thus this won't affect the answer.

Check this discussion for more: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  27 Mar 2013, 08:38
Ok, Thank you very much!

Posted from my mobile device
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  01 Apr 2013, 11:36
navigator123 wrote:
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

Please correct me if i am wrong.. What if, y=3 and x=2?
That will still remain in the shaded part of the triangle and won't satisfy the question..

y=3; x=2 is not in the shaded region
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  01 Apr 2013, 11:43
perilm41 wrote:
johnwesley wrote:
See attached drawing.

The probability will be the area of the dark triangle divided by the area of the original triangle:

Area dark triangle (where X>Y is true): $$\frac{1}{2}*4*4$$

Area original triangle: $$\frac{1}{2}*4*5$$

$$Probability=\frac{(0.5*4*4)}{(0.5*4*5)}=\frac{4}{5}$$

Solution E

What about (4,4)? that is in the shaded region, but is not greater than 0 (or algebraically x=4 is not greater than y=4)?

The point (4,4) is effectively not in the domain of the equation X>Y... but it is located in the borderline... and the area of the triangle will not change - a point does not have dimension. Take this example: you don't take 4, but you take 3,99999999999999999999999999 (and put more nines)... now calculate the area and you will see the result is exactly the same.
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  06 Mar 2014, 01:32
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Bumping for review and further discussion.

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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink]  17 Apr 2015, 23:30
Hello from the GMAT Club BumpBot!

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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and   [#permalink] 17 Apr 2015, 23:30
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