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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 ?

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

please help with this one.

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.

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

please help with this one.

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

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

please help with this one.

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

Answer: E.

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.

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

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. _________________

More:"All I wish someone had told me about GMAT beforehand" There are many things you want to know before doing the GMAT exam (how is exam day, what to expect, how to think, to do's...), and you have them in this blog, in a simple way

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

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