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# If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin

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Equation |x/2| + |y/2| = 5 encloses a certain region on [#permalink]

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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?

A. 20
B. 50
C. 100
D. 200
E. 400
[Reveal] Spoiler: OA

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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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19 Oct 2008, 01:48
study wrote:
OA is not C

ANy takers !!!
kindly post the OA
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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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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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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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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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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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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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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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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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If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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21 Mar 2011, 19:16
1
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If equation |x/2| + |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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html

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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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21 Mar 2011, 22:30
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This is the square with vertices (10,0),
(-10,0)
(0,10)
(0,-10)
The side of square is sqrt(2)*10. hence area is 2*10*10=200

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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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21 Mar 2011, 23:01
CMcAboy wrote:
Can someone help me with this question:

If equation |x/2| + |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

The equation can be reduced to intercept form as |x/10| + |y/10| = 1, so these are lines in four quadrants with x and y intercept as 10, so it is a rhombus with diagonals of 20 each and hence area = 1/2*d1*d2 = 1/2*20*20 = 200. Answer D.

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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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22 Mar 2011, 03:39
The equations can be :

x/2 + y/2 = 5 - (1)

x/2 + y/2 = -5 - (2)

x/2 - y/2 = 5 - (3)

y/2 - x/2 = 5 - (4)

These are a set of || and |- lines

(1) and (2) are parallel

and (3) and (4) are also parallel

(1) & (3) -> x = 10, y = 0 = (10,0)

(1) & (4) -> x = 0, y = 10 = (0,10)

(2) & (3) -> x = 0, y = -10 = (0,-10)

(2) & (4) -> x = -10, y = 0 = (-10,0)

So area of square = 1/2 * d1 * d2 = 1/2 * 20 * 20 = 200
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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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15 Jun 2011, 06:09
gmat1220 can u pls explain how you got the square

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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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17 Mar 2016, 02:51
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Re: Equation |x/2| + |y/2| = 5 encloses a certain region on [#permalink]

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17 Mar 2016, 02:52
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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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17 Mar 2016, 02:57
CMcAboy wrote:
If equation |x/2| + |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

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html
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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin [#permalink]

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17 Mar 2016, 02:58
Bunuel wrote:
CMcAboy wrote:
If equation |x/2| + |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

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Similar questions to practice:
the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html
in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html
if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html
what-is-the-area-of-the-region-enclosed-by-lines-y-x-x-y-150487.html
new-set-of-mixed-questions-150204-100.html#p1208441
the-area-bounded-by-the-three-curves-x-y-1-x-1-and-y-186595.html
if-equation-x-2-y-2-5-encloses-a-certain-region-126117.html

Hope this helps.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html
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Re: If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin   [#permalink] 17 Mar 2016, 02:58
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