Equation |\frac{x}{2}| + |\frac{y}{2}| = 5 encloses a : Quant Question Archive [LOCKED]
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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]

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03 Oct 2008, 21:47
5
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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03 Oct 2008, 22:11
you can ignore the absolute value sign since its distance (can't be -)

x/2+y/2 = 5

x+y = 10

if x=0 then y=10

if y=0 then x=10

xy = 10*10 = 100

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04 Oct 2008, 01:59
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x can be 5 and y 5
x can be 7 and y 3
there are too many combinations of x and y.
there has to be a logical reason why you would choose 0 as one and 10 for the other.

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04 Oct 2008, 03:48
ast wrote:
x can be 5 and y 5
x can be 7 and y 3
there are too many combinations of x and y.
there has to be a logical reason why you would choose 0 as one and 10 for the other.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

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04 Oct 2008, 04:38
Greenberg wrote:
ast wrote:
x can be 5 and y 5
x can be 7 and y 3
there are too many combinations of x and y.
there has to be a logical reason why you would choose 0 as one and 10 for the other.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

i think you all are missing something the figure formed is symmetrical about the x and y axis
so we get 4 triangles in each quadrant that are congurent
area of each is 1/2(base * height)
=1/2(10*10)
=50
as we get 4 triangles total area=4(50)
=200
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04 Oct 2008, 07:54
rohit929 wrote:
Greenberg wrote:
ast wrote:
x can be 5 and y 5
x can be 7 and y 3
there are too many combinations of x and y.
there has to be a logical reason why you would choose 0 as one and 10 for the other.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

i think you all are missing something the figure formed is symmetrical about the x and y axis
so we get 4 triangles in each quadrant that are congurent
area of each is 1/2(base * height)
=1/2(10*10)
=50
as we get 4 triangles total area=4(50)
=200

I think you are correct

But the answer should be (E)

{10,-10} {-10,10} {-10,-10} {10,10}

10*10*4 = 400

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04 Oct 2008, 07:58
the region formed is a square with sides 10*sqrt(2)...therefore area=200
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07 Oct 2008, 00:57
1
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hope this will help
Attachments

triangle.JPG [ 15.38 KiB | Viewed 679 times ]

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07 Oct 2008, 01:15
rohit929 wrote:
hope this will help

Wow ! thanks

Yes you are all correct - 200

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07 Oct 2008, 01:32
Greenberg wrote:
rohit929 wrote:
hope this will help

Wow ! thanks

Yes you are all correct - 200

i think you went wrong because you ignored the absolute value.....
to be honest i think members like you makes this forum more interesting, thanks
Re: Math Question   [#permalink] 07 Oct 2008, 01:32
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