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If equation |x| + |y| = 5 encloses a certain region on the

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If equation |x| + |y| = 5 encloses a certain region on the [#permalink]

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08 Mar 2008, 12:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

* 5
* 10
* 25
* 50
* 100
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08 Mar 2008, 12:37
Mode X + Mode Y = 5 represents a square with side 10.

So Answer should be 10*10 = 100

Explanation
Mode X can be any value from and between -5 to 5.
Similarly if above statement is true then Mode Y can be any value from and between 5 to -5.
Valye of Mode X and Mode Y will be such that their sum is 5.

Now extreme condrinates can be (-5,0), (0,-5), (5,0), and (0,5)

Last edited by abhijit_sen on 08 Mar 2008, 16:10, edited 1 time in total.
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08 Mar 2008, 16:07
D for me.

It must be a square enclosed with coordinates - (-5,0), (0,-5), (5,0), and (0,5)

Each side is sqrt(5^2 + 5^2) = 5*sqrt(2)

Area of sqaure = 5*sqrt(2) * 5*sqrt(2) = 50
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08 Mar 2008, 16:13
I agree with sreehari, I miscalculated it. Thanks mate.
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Re: Maths: what is the area of the region? [#permalink]

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08 Mar 2008, 18:59
suntaurian wrote:
If equation $$|x| + |y| = 5$$ encloses a certain region on the graph, what is the area of that region?

* 5
* 10
* 25
* 50
* 100

Points are (0,5) (5,0) (0,-5) and (-5,0) Thus diameter is 10. xsqrt2=10 --> x=5sqrt2 x^2=50

D
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08 Mar 2008, 20:26
Yes D is the correct answer.

I actualy drew the graph with that region and I spent a lot of time. What is an easier and quicker way to do it?
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Re: Maths: what is the area of the region? [#permalink]

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09 Mar 2008, 01:16
Not sure if I am the odd man out...but why can't the values of x and y also be (1, 4) or (2, 3) or (3, 2) or (4, 1) which will give +ve and -ve co-ordinates in the four quadrants.

It does not say anywhere in the equation that either X or Y is 0. How did all of you guys assume that the values are only 5 & 0.
Also it does not say that its the biggest region encompassed...

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09 Mar 2008, 07:29
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suntaurian wrote:
Not sure if I am the odd man out...but why can't the values of x and y also be (1, 4) or (2, 3) or (3, 2) or (4, 1) which will give +ve and -ve co-ordinates in the four quadrants.

It does not say anywhere in the equation that either X or Y is 0. How did all of you guys assume that the values are only 5 & 0.
Also it does not say that its the biggest region encompassed...

Here is the deal - you need to find the points of intersection to figure out the area enlclosed by the given equations. The above points lie on the lines, but they are not where the lines intersect.

Let us take an example - two out four equations from the equation in question are x+y=5 and -x+y=5.

I use the form x/a + y/b = 1 to find out the x and y intercept (a is the x intercept and b is the y intercept). Or you can substitute x=0 and y=0 to get the same result - it's your choice.

For x+y=5 - (5,0) and (0,5) are the points where it intersects x and y axis.

For -x+y=5 - (-5,0) and (0,5) are the points where it intersects x and y axis.

So clearly they intersect at (0,5). You don't need to complete the whole process here. Based on the analysis so far you can see that the two above equations are pependicular to each other and intersect on y axis. So by logic of symmetry you can fill in the blanks to complete the picture and can see it's a square with vertices - (-5,0), (0,-5), (5,0), and (0,5).
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Re: Maths: what is the area of the region? [#permalink]

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09 Mar 2008, 07:38
sreehari wrote:
suntaurian wrote:
Not sure if I am the odd man out...but why can't the values of x and y also be (1, 4) or (2, 3) or (3, 2) or (4, 1) which will give +ve and -ve co-ordinates in the four quadrants.

It does not say anywhere in the equation that either X or Y is 0. How did all of you guys assume that the values are only 5 & 0.
Also it does not say that its the biggest region encompassed...

Here is the deal - you need to find the points of intersection to figure out the area enlclosed by the given equations. The above points lie on the lines, but they are not where the lines intersect.

Let us take an example - two out four equations from the equation in question are x+y=5 and -x+y=5.

I use the form x/a + y/b = 1 to find out the x and y intercept (a is the x intercept and b is the y intercept). Or you can substitute x=0 and y=0 to get the same result - it's your choice.

For x+y=5 - (5,0) and (0,5) are the points where it intersects x and y axis.

For -x+y=5 - (-5,0) and (0,5) are the points where it intersects x and y axis.

So clearly they intersect at (0,5). You don't need to complete the whole process here. Based on the analysis so far you can see that the two above equations are pependicular to each other and intersect on y axis. So by logic of symmetry you can fill in the blanks to complete the picture and can see it's a square with vertices - (-5,0), (0,-5), (5,0), and (0,5).

Good explanation. That's what I wanted to understand. +1
Re: Maths: what is the area of the region?   [#permalink] 09 Mar 2008, 07:38
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