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In the x-y plane, the area of the region bounded by the graph of |x +

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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27 Jan 2014, 02:04
unceldolan wrote:
Bunuel,

wouldn't it be sufficient to look at only two cases?

(x+y) + (x-y) = 4 ==> x=2
(x+y) - (x-y) = 4 ==> y=2

Which would give us 2*2 * 4 = 16?

What you mean by "sufficient"? |x+y| + |x-y| = 4 gives FOUR equations, as explained on page 1.
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27 Jan 2014, 04:07
Bunuel wrote:
unceldolan wrote:
Bunuel,

wouldn't it be sufficient to look at only two cases?

(x+y) + (x-y) = 4 ==> x=2
(x+y) - (x-y) = 4 ==> y=2

Which would give us 2*2 * 4 = 16?

What you mean by "sufficient"? |x+y| + |x-y| = 4 gives FOUR equations, as explained on page 1.

Hey Bunuel,

I think I confused some things here. Just this morning, I did Chapter 9 of MGMAT Strategy Guide 2. Here it's stated that, if I have an equation with 2 absolute value expressions and one variable, I only need to set up 2 cases. Case A, when the absolute values have the same sign, Case B when the absolute values have different signs. But since here we got 2 variables and two absolute value expressions, I think I have to set up 4 equations. I just confused what I read a bit...sorry!
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 03:09
HI Bunnel,

Have one doubt on this.

Similar kind of question is posted on following link

http://gmatclub.com/forum/m25-q19-76535.html

why here we are doing different then defined on above link.

on the above link we have square of side 20 then why we are not getting ans as 20*20 = 400
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 03:40
HI Bunnel,

Have one doubt on this.

Similar kind of question is posted on following link

http://gmatclub.com/forum/m25-q19-76535.html

why here we are doing different then defined on above link.

on the above link we have square of side 20 then why we are not getting ans as 20*20 = 400

In that link the square does NOT have the side of 10, it has the side of $$10\sqrt{2}$$ and the diagonal of 20:

The figure from original question is different:

Check this: in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html#p649401

Hope it helps.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 07:50
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8
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File comment: If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 19:47
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8

The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect.
You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 20:48
Hi karishma,

Could you please elaborate more on this one .
I am really having a hard time figuring this one out .
I understood the first question i.e
In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16.
But in Second Question
what is the area bounded by graph|x/2| + |y/2| = 5?
Why was the graph drawn different from that of the previous Question.
Why was diagonal considered for second question and Side of a square considered for first question.

Help is appreciated .
VeritasPrepKarishma wrote:
If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8

The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect.
You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 21:43
dheeraj24 wrote:
Hi karishma,

Could you please elaborate more on this one .
I am really having a hard time figuring this one out .
I understood the first question i.e
In x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 . The area is 16.
But in Second Question
what is the area bounded by graph|x/2| + |y/2| = 5?
Why was the graph drawn different from that of the previous Question.
Why was diagonal considered for second question and Side of a square considered for first question.

Help is appreciated .

The graph was drawn differently because the equations given to you are different.

|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?

y = 10+x
y = 10 -x
y = x - 10
y = -x - 10

When you draw them out, you get slanting lines and the figure which looks like a kite.
The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.

In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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14 Apr 2014, 22:18
Yeah karishma,

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

VeritasPrepKarishma wrote:
dheeraj24 wrote:
Hi karishma,

The graph was drawn differently because the equations given to you are different.

|x/2| + |y/2| = 5 gives us a set of 4 equations. What are they?

y = 10+x
y = 10 -x
y = x - 10
y = -x - 10

When you draw them out, you get slanting lines and the figure which looks like a kite.
The coordinates of the square will be (10, 0), (0, 10), (-10, 0) and (0, -10). Here, 20 is the length of the diagonal.

In our original question, when we draw out the 4 equations, we get horizontal/vertical lines and hence a regular looking square. In that case, 4 is the length of the side.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Apr 2014, 04:01
1
dheeraj24 wrote:
Yeah karishma,

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

Because you are asked the area of the region bounded by |x+y| + |x-y| = 4.
This equation gives you ONLY horizontal/vertical lines passing through points (2,0), (-2,0), (0,2) and (0,-2) such as x = 2, y = 2, x = -2, y = -2.

Note that x = 2 is the equation of a line (it is not a coordinate) which passes through point (2, 0) and is parallel to the y axis. Similarly, y = 2 is the equation of a line which is parallel to x axis and passes through the point (0, 2) and so on. I think you are taking x = 2 as a coordinate but that is not the case. A coordinate has a value for y too. x =2 is the equation of a line. It implies that x coordinate is always 2 and y can be anything. So all points lying on a line passing through x = 2 and parallel to y axis satisfy this criteria.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Apr 2014, 08:02
Hi Karishma,

I am still not clear

in question |x/2| + |y/2| = 5 we are getting following cordinates.

x=10
x=-10
y=10
y=-10

and in question |x+y| + |x-y| = 4. we are having following cordinates
x=2
x=-2
y=2
y=-2

why we are drawing graph differently?

Thanks
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Apr 2014, 08:16
1
PathFinder007 wrote:
Hi Karishma,

I am still not clear

in question |x/2| + |y/2| = 5 we are getting following cordinates.

x=10
x=-10
y=10
y=-10

and in question |x+y| + |x-y| = 4. we are having following cordinates
x=2
x=-2
y=2
y=-2

why we are drawing graph differently?

Thanks

For $$|x+y| + |x-y| = 4$$ the equations are:

$$x = 2$$;
$$y = 2$$;
$$y = -2$$;
$$x = -2$$.

For $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ the equations are:

$$y=-10-x$$;
$$y=10+x$$;
$$y=10-x$$;
$$y=x-10$$.

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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20 Jun 2015, 01:58
Bunuel wrote:
srini123 wrote:
Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph$$|x/2| + |y/2| = 5$$?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

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

After solving you'll get equation of four lines:

$$y=-10-x$$
$$y=10+x$$
$$y=10-x$$
$$y=x-10$$

These four lines will also make a square, BUT in this case the diagonal will be 20 so the $$Area=\frac{20*20}{2}=200$$. Or the $$Side= \sqrt{200}$$, area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).

Hope it's clear.

Hi bunnel,

How did u rhombus for this one and a square for the other one?...I got the limits for both the questions, but could not figure out they turn out to be a square and rhombus!...
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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20 Jun 2015, 02:53
1
jayanthjanardhan wrote:
Bunuel wrote:
srini123 wrote:
Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph$$|x/2| + |y/2| = 5$$?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

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

After solving you'll get equation of four lines:

$$y=-10-x$$
$$y=10+x$$
$$y=10-x$$
$$y=x-10$$

These four lines will also make a square, BUT in this case the diagonal will be 20 so the $$Area=\frac{20*20}{2}=200$$. Or the $$Side= \sqrt{200}$$, area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).

Hope it's clear.

Hi bunnel,

How did u rhombus for this one and a square for the other one?...I got the limits for both the questions, but could not figure out they turn out to be a square and rhombus!...

Even that is a square but never forget that A Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that figure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won't be wrong either but you are right about the figure being a Square.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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26 Jun 2015, 03:23
Well in this case I didn't even consider the moduli.

Since we are looking for areas, I thought that the area would be of either a square, a triangle or a rectangle. And probaby not a triangle cause then you also have to find the height which makes it that much more complicated and sort of out of the scope of the question.

I tried to see how we could create number 4 and it can be done in 3 ways:
0 + 4 or the opposite
1 + 3 or the opposite
2+2 the opposite is the same.

Out of these numbers, the area resulting in any of the given answer choices could only be 16, and it would be a square of size 4. So, I went for 16.

If I had to do it, I would also use Bunuel's way.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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09 Apr 2016, 17:57
papillon86 wrote:
In the x-y plane, the area of the region bounded by the graph of |x+y| + |x-y| = 4 is

A. 8
B 12
C. 16
D. 20
E. 24

Need help in solving equations involving Mod......
help?

4 options
+ + -> line x=2
+ - -> line y=2
- + -> line x=-2
- - -> line y=-2

basically we have a square with s=4. area is 16.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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16 Jun 2018, 10:40
papillon86 wrote:
In the x-y plane, the area of the region bounded by the graph of |x + y| + |x - y| = 4 is

A. 8
B. 12
C. 16
D. 20
E. 24

Given |x + y| + |x - y| = 4

Case 1 - (x + y) > 0 & (x - y) > 0, we get

x + y + x - y = 4

x = 2......(i)

Case 2 - (x + y) < 0 & (x - y) < 0, we get

- x - y - x + y = 4

x = - 2......(ii)

Case 3 - (x + y) < 0 & (x - y) > 0, we get

- x - y + x - y = 4

y = - 2......(iii)

Case 4 - (x + y) > 0 & (x - y) < 0, we get

x + y - x + y = 4

y = 2......(iv)

So, the region |x + y| + |x - y| = 4, is bounded by lines x = 2, x = -2, y = 2 & y = -2

Giving us a square of length 4 units,

Hence area of the region = 16 units.

Thanks,
GyM
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