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

Oldest

Best Reply

Author

Message

TAGS:

Manager

Joined: 09 Jun 2009

Posts: 231

Followers: 1

Kudos [?]: 30 [0], given: 6

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

08 Nov 2009, 14:11

00:00

Question Stats:

69% (02:30) correct

30% (00:58) wrong

based on 68 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 15 Feb 2012, 00:29, edited 2 times in total.

Edited the question and added the OA

GMAT Club team member

Joined: 02 Sep 2009

Posts: 12096

Followers: 1876

Kudos [?]: 10096 [0], given: 959

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

08 Nov 2009, 14:34

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.

Answer: C
_________________

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!, 11 New DS set. NEW!!!

What are GMAT Club Tests?
25 extra-hard Quant Tests

Find out what's new at GMAT Club - latest features and updates

Senior Manager

Affiliations: PMP

Joined: 13 Oct 2009

Posts: 319

Followers: 2

Kudos [?]: 78 [0], given: 37

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

08 Nov 2009, 15:27

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.

Answer: C

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

GMAT Club team member

Joined: 02 Sep 2009

Posts: 12096

Followers: 1876

Kudos [?]: 10096 [0], given: 959

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

08 Nov 2009, 15:39

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.
_________________

Senior Manager

Affiliations: PMP

Joined: 13 Oct 2009

Posts: 319

Followers: 2

Kudos [?]: 78 [0], given: 37

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

08 Nov 2009, 16:58

Bunuel wrote:

srini123 wrote:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

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

GMAT Club team member

Joined: 02 Sep 2009

Posts: 12096

Followers: 1876

Kudos [?]: 10096 [1] , given: 959

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

08 Nov 2009, 17:41

This post received
KUDOS

srini123 wrote:

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: 319

Followers: 2

Kudos [?]: 78 [0], given: 37

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

08 Nov 2009, 20:23

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: 478

Location: Texas

Followers: 4

Kudos [?]: 33 [0], given: 20

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

20 May 2011, 12:39

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]
Downloaded 41 times

To download please login or register as a user

Director

Joined: 01 Feb 2011

Posts: 788

Followers: 11

Kudos [?]: 64 [0], given: 42

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

20 May 2011, 17:31

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: 788

Followers: 11

Kudos [?]: 64 [0], given: 42

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

20 May 2011, 17:33

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

Answer is C.

Status: There is always something new !!

Affiliations: PMI,QAI Global,eXampleCG

Joined: 08 May 2009

Posts: 1395

Followers: 8

Kudos [?]: 86 [0], given: 10

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

20 May 2011, 21:50

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: 1719

Location: United States (IN)

Concentration: Strategy, Technology

Followers: 26

Kudos [?]: 229 [0], given: 35

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

21 May 2011, 05:46

|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

Answer - C
_________________

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

Find out what's new at GMAT Club - latest features and updates

Intern

Joined: 22 May 2011

Posts: 1

Followers: 0

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

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

22 May 2011, 07:21

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: 3165

Location: Pune, India

Followers: 596

Kudos [?]: 2119 [0], given: 97

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

22 May 2011, 08:38

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

GMAT Club team member

Joined: 02 Sep 2009

Posts: 12096

Followers: 1876

Kudos [?]: 10096 [0], given: 959

Re: Absolute Values [#permalink]

15 Feb 2012, 00:25

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

Merging similar topics. Please ask if anything remains unclear.

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.
_________________

Manager

Joined: 25 Aug 2011

Posts: 200

Location: India

GMAT 1: 730 Q49 V40

WE: Operations (Insurance)

Followers: 1

Kudos [?]: 22 [0], given: 11

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

27 Feb 2012, 23:33

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

GMAT Club team member

Joined: 02 Sep 2009

Posts: 12096

Followers: 1876

Kudos [?]: 10096 [0], given: 959

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

27 Feb 2012, 23:42

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

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 2260 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: 468

Followers: 13

Kudos [?]: 88 [0], given: 11

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

05 Dec 2012, 23:17

(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

(2) Plot your four lines
(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

Answer: C

For more detailed solutions for similar question types:

Manager

Joined: 24 Mar 2010

Posts: 82

Followers: 0

Kudos [?]: 8 [0], given: 133

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

12 Dec 2012, 09:06

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: 468

Followers: 13

Kudos [?]: 88 [0], given: 11

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

13 Dec 2012, 04:06

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

Re: graphs_Modulus....Help [#permalink] 13 Dec 2012, 04:06

		Similar topics	Author	Replies	Last post
Similar Topics:		In xy-plane, what is the area of the region encircled by	getzgetzu	1	06 May 2006, 05:23
		The area of the region that consists of all points (x,y)	kevincan	2	16 Aug 2006, 14:54
	3	On the xy-coordinate plane, a quadrilateral is bounded by	fluke	13	17 May 2011, 04:07
	1	In the x-y plane, the square region bound by (0,0), (10, 0)	mikemcgarry	8	21 Nov 2012, 16:34
	5	Region R is a square in the x-y plane with vertices	mikemcgarry	7	08 Feb 2013, 11:45

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

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.