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

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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 ?

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.

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

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.

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.

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

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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]

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02 May 2016, 00:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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