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nick1816
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chetan2u
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chetan2u
nick1816
What is the probability that a point satisfying the conditions |x| ≤ 2 and |y| ≤ 2 also satisfies |x + y| ≤ 1?

A. 1/16
B. 1/4
C. 7/16
D. 9/16
E. 3/4


Please check the question it may not be correct.
|x|≤ 2 means x can be -2,-1,0,1,2 and |y| ≤ 2 means y can be -2,-1,0,1,2
So total ways (x,y) can be chosen is 5*5=25 but none of the choices has 25 or 5 in the denominator

Source please
He didn't say x is an integer, why did you assume all integer values

Posted from my mobile device
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chetan2u
nick1816
What is the probability that a point satisfying the conditions |x| ≤ 2 and |y| ≤ 2 also satisfies |x + y| ≤ 1?

A. 1/16
B. 1/4
C. 7/16
D. 9/16
E. 3/4


Please check the question it may not be correct.
|x|≤ 2 means x can be -2,-1,0,1,2 and |y| ≤ 2 means y can be -2,-1,0,1,2
So total ways (x,y) can be chosen is 5*5=25 but none of the choices has 25 or 5 in the denominator

Source please
He didn't say x is an integer, why did you assume all integer values

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Yes you are correct and that is what I mentioned above. Although I realised that even while I wrote the post but felt that the post actually meant them as integers since most of the GMAT question would be in that language.
I would delete my post as it would unnecessarily create confusion.
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chetan2u
I will take care of that from next time. Thanks for letting me know. :
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nick1816
x and y can be any real number in the range.

|x| ≤ 2 and |y| ≤ 2
Area bounded by these 2 curves= 4*4=16

Area bounded by |x| ≤ 2, |y| ≤ 2 and |x+y| ≤ 1
16-(1/2*3*3)-(1/2*3*3)=7

Probability= 7/16

Source: Career Launcher


I am unable to follow this, could please explain one more time?
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nick1816
x and y can be any real number in the range.

|x| ≤ 2 and |y| ≤ 2
Area bounded by these 2 curves= 4*4=16

Area bounded by |x| ≤ 2, |y| ≤ 2 and |x+y| ≤ 1
16-(1/2*3*3)-(1/2*3*3)=7

Probability= 7/16

Source: Career Launcher


I am unable to follow this, could please explain one more time?
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chetan2u I got 7/25 (I assumed that it mentions integer) Without the assumption the answer would be Infinity/Infinity. :-D . However I clicked on C because that was the only option with 7 in the numerator. :lol:

Good question, poorly written.
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Hi Rajshek,

the total area filled by the conditions for x,y is 2*2 per quadrant. We have 4 quandrants, so its 4^2=16.
Now, how much of that area is occupied with Ix+yI<=1?
For that, check the picture that nick posted.
Now, if you draw the square (4x4 square) into the coordinate system, and also draw the stuff that nick posted, you can see that two equal triangles are created. Each has two sides of length 3 (a,b, hypotenuse c leave open, u dont need it now).

So in total, the two triangles take 3*3=9 of the area.
This means that 9/16 is free -> 7/16 is occupied.
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@Experts
Can someone please explain me this question?
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chetan2u
nick1816
What is the probability that a point satisfying the conditions |x| ≤ 2 and |y| ≤ 2 also satisfies |x + y| ≤ 1?

A. 1/16
B. 1/4
C. 7/16
D. 9/16
E. 3/4


Please check the question it may not be correct.
|x|≤ 2 means x can be -2,-1,0,1,2 and |y| ≤ 2 means y can be -2,-1,0,1,2
So total ways (x,y) can be chosen is 5*5=25 but none of the choices has 25 or 5 in the denominator

Source please
[/quote]

Yes, I missed out that you have not mentioned that x and y are integers, but a GMAT question would be in a format which will either tell you that x,y are integers or it will mention area bounded by |x|<2 and |y|<2. This is more of a language of a CAT question and Career launcher also has more to do with CAT preparation.
Please mention source along with the question.[/quote]

Chetan2u may you kindly solve the question?
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iam unable to understand why 0 is excluded?
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hadimadi
Hi Rajshek,

the total area filled by the conditions for x,y is 2*2 per quadrant. We have 4 quandrants, so its 4^2=16.
Now, how much of that area is occupied with Ix+yI<=1?
For that, check the picture that nick posted.
Now, if you draw the square (4x4 square) into the coordinate system, and also draw the stuff that nick posted, you can see that two equal triangles are created. Each has two sides of length 3 (a,b, hypotenuse c leave open, u dont need it now).

So in total, the two triangles take 3*3=9 of the area.
This means that 9/16 is free -> 7/16 is occupied.

i didn't get how do u take 2*2 per quadrant?please explain ...I'm unable to understand it
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