In the x-y plane the area of the region bounded by the : GMAT Problem Solving (PS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 Feb 2017, 06:41

### 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:

### Hide Tags

VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 177

Kudos [?]: 1369 [0], given: 62

### Show Tags

13 Dec 2012, 03:26
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.

Hii Bunuel.
What is the best approach of finding the points of intersection in order to make the square.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37154
Followers: 7276

Kudos [?]: 96857 [0], given: 10808

### Show Tags

13 Dec 2012, 03:30
Marcab 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.

Hii Bunuel.
What is the best approach of finding the points of intersection in order to make the square.

I'd say substituting x=0 and y=0 in the equations of lines and making a drawing.
_________________
VP
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 1420
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75
Followers: 177

Kudos [?]: 1369 [0], given: 62

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

### Show Tags

13 Dec 2012, 03:41
Thanks Bunuel.
But still some confusion.
Can you elaborate a bit further?
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 37154
Followers: 7276

Kudos [?]: 96857 [0], given: 10808

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

### Show Tags

13 Dec 2012, 03:50
Marcab wrote:
Thanks Bunuel.
But still some confusion.
Can you elaborate a bit further?

Can you please tell me what exactly needs further clarification?

Meanwhile check similar questions:
if-equation-x-2-y-2-5-encloses-a-certain-region-126117.html
m06-5-absolute-value-108191.html
_________________
Intern
Joined: 09 Jun 2012
Posts: 31
Followers: 0

Kudos [?]: 19 [0], given: 13

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

### Show Tags

17 Jul 2013, 22:20
This is an interesting combo of absolute values, plotting lines in coordinate system and then finding the resulting figure's area. Thanks all for presenting the approach!
Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 635
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Followers: 71

Kudos [?]: 448 [0], given: 297

### Show Tags

21 Aug 2013, 20:51
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)
_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Math Expert
Joined: 02 Sep 2009
Posts: 37154
Followers: 7276

Kudos [?]: 96857 [0], given: 10808

### Show Tags

22 Aug 2013, 02:21
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 1215 times ]

_________________
Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 635
Location: India
GMAT 1: 710 Q50 V36
GMAT 2: 750 Q51 V41
GMAT 3: 790 Q51 V49
GPA: 3.3
Followers: 71

Kudos [?]: 448 [0], given: 297

### Show Tags

22 Aug 2013, 04:36
Yes I realized it later in the day. Thanks for your help and support.

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

The side of the square IS 4:
Attachment:
MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif
[/quote]
_________________

Like my post Send me a Kudos It is a Good manner.
My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

Current Student
Joined: 21 Oct 2013
Posts: 194
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Followers: 1

Kudos [?]: 37 [0], given: 19

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

### Show Tags

27 Jan 2014, 02:00
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
Joined: 02 Sep 2009
Posts: 37154
Followers: 7276

Kudos [?]: 96857 [0], given: 10808

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

### Show Tags

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.
_________________
Current Student
Joined: 21 Oct 2013
Posts: 194
Location: Germany
GMAT 1: 660 Q45 V36
GPA: 3.51
Followers: 1

Kudos [?]: 37 [0], given: 19

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

### Show Tags

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!
Senior Manager
Joined: 10 Mar 2014
Posts: 250
Followers: 2

Kudos [?]: 85 [0], given: 13

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 37154
Followers: 7276

Kudos [?]: 96857 [0], given: 10808

### Show Tags

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.
_________________
Senior Manager
Joined: 10 Mar 2014
Posts: 250
Followers: 2

Kudos [?]: 85 [0], given: 13

### Show Tags

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
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 887 times ]

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7194
Location: Pune, India
Followers: 2175

Kudos [?]: 14064 [0], given: 222

### Show Tags

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Status: suffer now and live forever as a champion!!! Joined: 01 Sep 2013 Posts: 149 Location: India Dheeraj: Madaraboina GPA: 3.5 WE: Information Technology (Computer Software) Followers: 1 Kudos [?]: 57 [0], given: 75 Re: graphs_Modulus....Help [#permalink] ### Show Tags 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 . Thanks in advance VeritasPrepKarishma wrote: pawankumargadiya 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 Joined: 16 Oct 2010 Posts: 7194 Location: Pune, India Followers: 2175 Kudos [?]: 14064 [0], given: 222 Re: graphs_Modulus....Help [#permalink] ### Show Tags 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 . 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Status: suffer now and live forever as a champion!!!
Joined: 01 Sep 2013
Posts: 149
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 57 [0], given: 75

### Show Tags

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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7194
Location: Pune, India
Followers: 2175

Kudos [?]: 14064 [1] , given: 222

### Show Tags

15 Apr 2014, 04:01
1
KUDOS
Expert's post
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Senior Manager
Joined: 10 Mar 2014
Posts: 250
Followers: 2

Kudos [?]: 85 [0], given: 13

### Show Tags

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
Re: graphs_Modulus....Help   [#permalink] 15 Apr 2014, 08:02

Go to page   Previous    1   2   3    Next  [ 53 posts ]

Similar topics Replies Last post
Similar
Topics:
5 On the xy-coordinate plane, what is the area of a triangle with vertic 4 03 Feb 2017, 03:57
2 In the xy-plane, triangular region S is bounded by the lines x=0,y=0, 4 30 Nov 2016, 02:05
10 In the rectangular coordinate system above, the shaded region is bound 6 04 Dec 2014, 06:46
19 Region R is a square in the x-y plane with vertices 10 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
Display posts from previous: Sort by