Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Apr 2015, 19:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 09 Jun 2009
Posts: 231
Followers: 1

Kudos [?]: 91 [3] , given: 6

In the x-y plane the area of the region bounded by the [#permalink]  08 Nov 2009, 13:11
3
KUDOS
15
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

60% (02:18) correct 40% (01:27) wrong based on 524 sessions
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
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40359 [8] , given: 5420

Re: graphs_Modulus....Help [#permalink]  08 Nov 2009, 13:34
8
KUDOS
Expert's post
4
This post was
BOOKMARKED
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.
Attachment:

MSP17971c13h40gd024h6g10000466ge1e9df941i96.gif [ 1.86 KiB | Viewed 2897 times ]

_________________
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 3

Kudos [?]: 110 [0], given: 37

Re: graphs_Modulus....Help [#permalink]  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?
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40359 [0], given: 5420

Re: graphs_Modulus....Help [#permalink]  08 Nov 2009, 14:39
Expert's post
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.
_________________
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 3

Kudos [?]: 110 [0], given: 37

Re: graphs_Modulus....Help [#permalink]  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
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40359 [1] , given: 5420

Re: graphs_Modulus....Help [#permalink]  08 Nov 2009, 16:41
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.
_________________
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 312
Followers: 3

Kudos [?]: 110 [0], given: 37

Re: graphs_Modulus....Help [#permalink]  08 Nov 2009, 19:23
Thanks Bunuel , once again wonderful explanation +1 Kudos..

have a good day...
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Senior Manager
Joined: 24 Mar 2011
Posts: 465
Location: Texas
Followers: 4

Kudos [?]: 74 [0], given: 20

Re: graphs_Modulus....Help [#permalink]  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
Attachments

Doc2.docx [11.23 KiB]

Director
Joined: 01 Feb 2011
Posts: 763
Followers: 14

Kudos [?]: 68 [0], given: 42

Re: graphs_Modulus....Help [#permalink]  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
Director
Joined: 01 Feb 2011
Posts: 763
Followers: 14

Kudos [?]: 68 [0], given: 42

Re: graphs_Modulus....Help [#permalink]  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

VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1360
Followers: 13

Kudos [?]: 159 [0], given: 10

Re: graphs_Modulus....Help [#permalink]  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
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

SVP
Joined: 16 Nov 2010
Posts: 1687
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 31

Kudos [?]: 340 [1] , given: 36

Re: graphs_Modulus....Help [#permalink]  21 May 2011, 04:46
1
KUDOS
|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

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 22 May 2011
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: graphs_Modulus....Help [#permalink]  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?

Thanks & Regards,
Vinu
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5429
Location: Pune, India
Followers: 1324

Kudos [?]: 6708 [0], given: 176

Re: graphs_Modulus....Help [#permalink]  22 May 2011, 07:38
Expert's post
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40359 [0], given: 5420

Re: Absolute Values [#permalink]  14 Feb 2012, 23:25
Expert's post
1
This post was
BOOKMARKED
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.
_________________
Current Student
Joined: 25 Aug 2011
Posts: 195
Location: India
GMAT 1: 730 Q49 V40
WE: Operations (Insurance)
Followers: 1

Kudos [?]: 126 [0], given: 11

Re: graphs_Modulus....Help [#permalink]  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.
Math Expert
Joined: 02 Sep 2009
Posts: 27056
Followers: 4184

Kudos [?]: 40359 [1] , given: 5420

Re: graphs_Modulus....Help [#permalink]  27 Feb 2012, 22:42
1
KUDOS
Expert's post
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 7916 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.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 17

Kudos [?]: 255 [0], given: 11

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

Impossible is nothing to God.

Manager
Joined: 24 Mar 2010
Posts: 81
Followers: 0

Kudos [?]: 29 [0], given: 134

Re: graphs_Modulus....Help [#permalink]  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.
_________________

- Stay Hungry, stay Foolish -

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 17

Kudos [?]: 255 [0], given: 11

Re: graphs_Modulus....Help [#permalink]  13 Dec 2012, 03:06
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
_________________

Impossible is nothing to God.

Re: graphs_Modulus....Help   [#permalink] 13 Dec 2012, 03:06

Go to page    1   2   3    Next  [ 43 posts ]

Similar topics Replies Last post
Similar
Topics:
11 Region R is a square in the x-y plane with vertices 8 08 Feb 2013, 10:45
3 In the x-y plane, the square region bound by (0,0), (10, 0) 20 21 Nov 2012, 15:34
8 On the xy-coordinate plane, a quadrilateral is bounded by 15 17 May 2011, 03:07
The area of the region that consists of all points (x,y) 2 16 Aug 2006, 13:54
In xy-plane, what is the area of the region encircled by 1 06 May 2006, 04:23
Display posts from previous: Sort by