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# In the x-y plane the area of the region bounded by the

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In the x-y plane the area of the region bounded by the [#permalink]

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08 Nov 2009, 13:11
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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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Feb 2012, 23:29, edited 2 times in total.
Edited the question and added the OA
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08 Nov 2009, 13:34
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papillon86 wrote:
In 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

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

OK, there can be 4 cases:

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

A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2

The area bounded by 4 graphs x=2, x=-2, y=2, y=-2 will be square with the side of 4 so the area will be 4*4=16.
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MSP17971c13h40gd024h6g10000466ge1e9df941i96.gif [ 1.86 KiB | Viewed 10000 times ]

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08 Nov 2009, 14:27
Bunuel wrote:
papillon86 wrote:
In 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

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

I've never seen such kind of question in GMAT before.

OK there can be 4 cases:

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

A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2

The area bounded by 4 graphs x=2, x=-2, y=2, y=-2 will be square with the side of 4 so the area will be 4*4=16.

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?
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08 Nov 2009, 14:39
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2
x=-2
y=2
y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.
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08 Nov 2009, 15:58
Bunuel wrote:
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2
x=-2
y=2
y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.

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
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08 Nov 2009, 16:41
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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.
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08 Nov 2009, 19:23
Thanks Bunuel , once again wonderful explanation +1 Kudos..

have a good day...
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20 May 2011, 11:39
Bunuel wrote:
x=2
x=-2
y=2
y=-2

This lines will make a square with the side 4, hence area 4*4=16.

i am still now able to follow if the area that is formed with these lines is as per fig 1 or fig 2
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20 May 2011, 16:31
Thats same as what you see in fig 2 .

agdimple333 wrote:
Bunuel wrote:
x=2
x=-2
y=2
y=-2

This lines will make a square with the side 4, hence area 4*4=16.

i am still now able to follow if the area that is formed with these lines is as per fig 1 or fig 2
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20 May 2011, 16:33
solving the inequality we have the following as solutions

x=2
y=2
y=-2
x=-2

drawing this in a graph, we can observe that it forms a square with side length of 4.

Hence the area is 4*4 = 16

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20 May 2011, 20:50
so check for ++, +-, -+ and --

giving x=2|-2 and y=2|-2

hence a square with 4*4 area = 16
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21 May 2011, 04:46
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|x-y| = x-y if x-y > 0

|x-y| = -(x-y) if x-y < 0

x+y > 0 => x > -y then x !> y

x+y + x - y = 4

x = 2

-x - y + x - y = 4 (if x < -y, then x !< y)

y = -2

x + y -x + y = 4

=> y = 2

-x-y + x - y = 4

=> y = -2

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22 May 2011, 06:21
Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

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22 May 2011, 07:38
VinuPriyaN wrote:
Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards,
Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations.
One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3.
For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.
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14 Feb 2012, 23:25
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prashantbacchewar wrote:
In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is
(1) 8
(2) 12
(3) 16
(4) 20
(5) 24

Some questions on the same subject to practice:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html
m06-q5-72817.html
if-equation-encloses-a-certain-region-110053.html

Hope it helps.
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27 Feb 2012, 22:33
Hi,
Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:
VinuPriyaN wrote:
Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards,
Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations.
One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3.
For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.
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27 Feb 2012, 22:42
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devinawilliam83 wrote:
Hi,
Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:
VinuPriyaN wrote:
Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards,
Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations.
One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3.
For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.

Yes, it can be done by graphing. |x+y| + |x-y| = 4 can expand in four different wasy:

A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2

So you can draw all these four lines x=2, x=-2, y=2, y=-2 to get a square with the side of 4:
Attachment:

Square.gif [ 1.86 KiB | Viewed 15008 times ]
See more examples here:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html
m06-q5-72817.html
if-equation-encloses-a-certain-region-110053.html

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

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05 Dec 2012, 22:17
(1) derive all equations from |x+y| + |x-y| = 4

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

(3) Notice you have formed a square region bounded by x=2, y=2, y=-2 and x=-2 lines
(4) Area = 4*4 = 16

For more detailed solutions for similar question types:
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12 Dec 2012, 08:06
Quote:

OK there can be 4 cases:

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

A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2

Bunuel,

Would appreciate it, if you could thoroughly explain the above.

Thanks.
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13 Dec 2012, 03:06
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eaakbari wrote:
Quote:

OK there can be 4 cases:

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

A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2

Any absolute values such as |x| = 5 could mean that x = 5 or x = -5.

Derive both (-) and (+) possibilities.

For the problem: |x+y| + |x-y| = 4

We could derive two possibilities for |x+y| could be -(x+y) and (x+y)
We could derive two possibilities for |x-y| could be -(x-y) and (x-y)

This is the reason why we have 4 derived equations.

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

Just simplify those...

If you want more practice on this question type: http://burnoutorbreathe.blogspot.com/2012/12/absolute-values-solving-for-area-of.html
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Re: graphs_Modulus....Help   [#permalink] 13 Dec 2012, 03:06

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