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Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a [#permalink]
 19 Oct 2008, 00:43
Question Stats:
0% (00:00) correct
100% (00:20) wrong based on 1 sessions
Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a certain region on the coordinate plane. What is the area of this region?
* 20
* 50
* 100
* 200
* 400
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Re: Coordinate Plane [#permalink]
19 Oct 2008, 01:09
study wrote: Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a certain region on the coordinate plane. What is the area of this region?
* 20
* 50
* 100
* 200
* 400 Represents a squarte with side 10 hence Area = 100 x/10 +y/10 =10 -x/10 +y/10 =10 -x/10 -y/10 =10 x/10 -y/10 =10 hence IMO C
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Re: Coordinate Plane [#permalink]
19 Oct 2008, 01:48
study wrote: OA is not C ANy takers !!! kindly post the OA
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Re: Coordinate Plane [#permalink]
19 Oct 2008, 02:17
200 (10,0),(0,10),(-10,0),(0,-10) This are the four points which acn have max area. so diagonal of this square is of lengh 20.. 20^2 = r^2 +r^2 400 = 2r^2 r^2 = area = 200.. 
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Re: Coordinate Plane [#permalink]
19 Oct 2008, 11:20
vishalgc wrote: 200 (10,0),(0,10),(-10,0),(0,-10) This are the four points which acn have max area. so diagonal of this square is of lengh 20.. 20^2 = r^2 +r^2 400 = 2r^2 r^2 = area = 200.. That's exactly what I get. IMO, the best way to solve this kind of problems is to calculate the points where the area cuts the axis. I think this problem has already come up in a recent post. Cheers,
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Re: Coordinate Plane [#permalink]
19 Oct 2008, 23:34
study wrote: OA is C What do you mean? Earlier, you have mentioned that OA is not C. And, OA cannot be C. It has to be D.
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Re: Coordinate Plane [#permalink]
22 Oct 2008, 02:06
While your answer is correct (it's a co-incidence) the method is not correct.
'Coz the (x,y) values you have chosen (0,-10) or(-10,0) do not satisfy the question stem info: |\frac{x}{2}| + |\frac{y}{2}| = 5
|\frac{-10}{2}| + |\frac{0}{2}| = -5 does NOT equal to 5.
Any takers for this one?
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Re: Coordinate Plane [#permalink]
22 Oct 2008, 02:14
study wrote: While your answer is correct (it's a co-incidence) the method is not correct.
'Coz the (x,y) values you have chosen (0,-10) or(-10,0) do not satisfy the question stem info: |\frac{x}{2}| + |\frac{y}{2}| = 5
|\frac{-10}{2}| + |\frac{0}{2}| = -5 does NOT equal to 5.
Any takers for this one? When we have numbers/variables in modulus then the result is always positive |-10| = 10 so |\frac{-10}{2}| + |\frac{0}{2}| = 5 + 0 = 5
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Re: Coordinate Plane [#permalink]
22 Oct 2008, 02:52
oh man...i think i have too much math in my head to overlook a basic feature!! Thanks
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Re: Coordinate Plane [#permalink]
14 Nov 2010, 06:29
I am unable to understand how to solve this question.
If |X| + |Y| = 10 then we get sample no like (5,5) (-5,5) (5,-5) (-5,-5). so after plotting them I get each side as 10 and thus area as 100.
What am I missing here ??
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Re: Coordinate Plane [#permalink]
14 Nov 2010, 06:33
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Re: Coordinate Plane
[#permalink]
14 Nov 2010, 06:33
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