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Equation |x/2| + |y/2| = 5 encloses a certain region on [#permalink ]
18 Oct 2008, 23:43

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Question Stats:

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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?

A. 20

B. 50

C. 100

D. 200

E. 400

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Re: Coordinate Plane [#permalink ]
19 Oct 2008, 00:09

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, 00:48

study wrote:

OA is not C

ANy takers !!!

kindly post the OA

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Re: Coordinate Plane [#permalink ]
19 Oct 2008, 01:17

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, 10:20

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, 22:34

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, 01:06

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, 01:14

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, 01:52

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, 05:29

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, 05:33

Re: Coordinate Plane
[#permalink ]
14 Nov 2010, 05:33