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# Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a

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Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a [#permalink]  19 Oct 2008, 00:43
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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.

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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
greatchap wrote:
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 ??

Check this: cmat-club-test-question-m25-101963.html?hilit=encloses#p791302 and graphs-modulus-help-86549.html?hilit=horizontal#p649401

Hope it helps.
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Re: Coordinate Plane   [#permalink] 14 Nov 2010, 06:33
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