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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] New post 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
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Re: In the xy- plane, [#permalink] New post 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.

P=8/10=4/5

Answer: E.
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Re: In the xy- plane, [#permalink] New post 25 Dec 2009, 05:27
Bunuel

great solution!!!

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Re: In the xy- plane, [#permalink] New post 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

P = 8/10 = 4/5

Answer: E

Detailed Solution: Line x=y as a Boundary
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Re: In the xy- plane, [#permalink] New post 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. :-D
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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink] New post 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] New post 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] New post 26 Mar 2013, 07:55
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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] New post 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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COLLECTION OF QUESTIONS:
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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink] New post 27 Mar 2013, 08:38
Ok, Thank you very much!

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Re: In the xy- plane, a triangle has vertexes (0,0), (4,0) and [#permalink] New post 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] New post 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] New post 06 Mar 2014, 01:32
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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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