GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 06 Jul 2020, 23:12 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the x-y plane, the area of the region bounded by the graph of |x +

Author Message
TAGS:

### Hide Tags

Senior Manager  Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 448
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.

The side of the square can't be 4, instead its sqrt(8)
Math Expert V
Joined: 02 Sep 2009
Posts: 64999
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

honchos wrote:
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.

The side of the square can't be 4, instead its sqrt(8)

The side of the square IS 4:
Attachment: MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif [ 1.86 KiB | Viewed 14785 times ]

_________________
Manager  Joined: 21 Oct 2013
Posts: 172
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 64999
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
_________________
Manager  Joined: 21 Oct 2013
Posts: 172
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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!
Manager  B
Joined: 10 Mar 2014
Posts: 177
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 64999
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
_________________
Manager  B
Joined: 10 Mar 2014
Posts: 177
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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
Attachments

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 Query_Diagonal1.jpg [ 12.82 KiB | Viewed 1949 times ]

Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10643
Location: Pune, India
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  Status: suffer now and live forever as a champion!!!
Joined: 01 Sep 2013
Posts: 96
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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 .
Thanks in advance 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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10643
Location: Pune, India
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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 .
Thanks in advance 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.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  Status: suffer now and live forever as a champion!!!
Joined: 01 Sep 2013
Posts: 96
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
Manager  B
Joined: 10 Mar 2014
Posts: 177
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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
Manager  Joined: 29 May 2013
Posts: 88
Location: India
Concentration: Technology, Marketing
WE: Information Technology (Consulting)
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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!...
Senior Manager  Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 393
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
Board of Directors P
Joined: 17 Jul 2014
Posts: 2422
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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.
Director  P
Joined: 14 Dec 2017
Posts: 502
Location: India
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

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
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 15383
Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: In the x-y plane, the area of the region bounded by the graph of |x +   [#permalink] 22 Jul 2019, 02:25

Go to page   Previous    1   2   [ 38 posts ]

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