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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0).

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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). If these 4 points makes a rectangle, what is the probability that x+y<4?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 2/5

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[Reveal] Spoiler: OA

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 13 Jun 2016, 11:05
Answer is 1/3.

Solved using the areas of the triangle rectangle.

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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As you can see from the diagram below, the correct answer is 1/3. Hence, B is our answer choice.
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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 22 Jun 2016, 08:02
MathRevolution wrote:
As you can see from the diagram below, the correct answer is 1/3. Hence, B is our answer choice.


Can you please share the complete solution, but in the question it is mentioned as x+y < 4.

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 22 Jun 2016, 08:04
Senthil1981 wrote:
Answer is 1/3.

Solved using the areas of the triangle rectangle.


Please share us the method that you used to solve this question, please share us in detail.

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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msk0657 wrote:
Senthil1981 wrote:
Answer is 1/3.

Solved using the areas of the triangle rectangle.


Please share us the method that you used to solve this question, please share us in detail.


as shown in above figure the line x+y<4 shows a triangle.
the area of triangle =1/2*4*4=8

are od rectangle given =4*6=24

probability any point lies x+y<4 in rectangle given is =8/24=1/3

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 07 Oct 2016, 06:58
MathRevolution wrote:
As you can see from the diagram below, the correct answer is 1/3. Hence, B is our answer choice.


Can you please explain why do we take (0,4) and (4,0) to form a triangle and not (0,3) and (3,0) ?
We want x+y to be <4.
At the point (0,4) , x=0 and y=4 , hence x+y are not <4.
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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 12 Jun 2017, 04:20
This question isn't very clear in what it is asking. I thought it wanted the probability of selecting an x+y < 4 value from the endpoints given. It's just too ambiguous

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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 01 Oct 2017, 05:23
Bunuel karishma Could you please explain why do we take (0,4) and (4,0) to form a triangle and not (0,3) and (3,0) ?
We want x+y to be <4.
At the point (0,4) , x=0 and y=4 , hence x+y are not <4. What am I missing here?

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 18 Oct 2017, 04:03
MathRevolution wrote:
As you can see from the diagram below, the correct answer is 1/3. Hence, B is our answer choice.

How is X+Y<4?
If you have posted a question can you also provide a proper solution? Not everyone is brilliant in maths

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0). [#permalink]

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New post 18 Oct 2017, 08:45
I am too, having a problem comprehending this question.
If x+y < 4, why the area included (4, 0) & (0, 4) ?
I'm quite sure you meant x+y <= 4, otherwise, we cannot calculate this using areas.

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Re: In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0).   [#permalink] 18 Oct 2017, 08:45
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In the x- y plane, there are 4 points (0,0), (0,4), (6,4), and (6,0).

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