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

 It is currently 27 Nov 2015, 11:31

### 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

# If equation |x/2| + |y/2| = 5 enclose a certain region

Author Message
TAGS:
Intern
Joined: 19 Sep 2010
Posts: 26
Followers: 0

Kudos [?]: 55 [4] , given: 0

If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  30 Sep 2010, 03:10
4
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

46% (02:03) correct 54% (01:13) wrong based on 622 sessions
If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400

M25-19

[Reveal] Spoiler:
ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 30377
Followers: 5091

Kudos [?]: 57319 [5] , given: 8811

Re: CMAT Club Test Question - m25 [#permalink]  30 Sep 2010, 03:22
5
KUDOS
Expert's post
3
This post was
BOOKMARKED
Barkatis wrote:
Hello,
Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Check similar problem at: graphs-modulus-help-86549.html?hilit=horizontal#p649401 it might help to get this one better.

Hope it helps.
_________________
Intern
Joined: 19 Sep 2010
Posts: 26
Followers: 0

Kudos [?]: 55 [0], given: 0

Re: CMAT Club Test Question - m25 [#permalink]  30 Sep 2010, 03:44
Thanks, the 20 was actually for the diagonals not the sides !
Manager
Joined: 22 Aug 2008
Posts: 186
Followers: 5

Kudos [?]: 75 [2] , given: 11

Re: CMAT Club Test Question - m25 [#permalink]  30 Sep 2010, 16:41
2
KUDOS
1
This post was
BOOKMARKED
|X/2| + |Y/2| = 5
so when x = 0, y= |10|
when y=0 , x=|10|

so the sides of the enclosed area touches (0,10),(10,0),(0,-10) and (-10,0).
so its a square having the diagonal =20unit
So the area of the region = (20/1.414)^2 = 200
Manager
Joined: 07 Feb 2010
Posts: 159
Followers: 2

Kudos [?]: 302 [3] , given: 101

Re: CMAT Club Test Question - m25 [#permalink]  16 Nov 2010, 08:45
3
KUDOS
|x|+||y|=10

put x=0

you get |y|=10 .....y=+-10

when you y=0

you get |x| = 10 ..... x=+-10

plot this on the co-ordinate plane.

you will get a rhombus
area of rhombus = 1/2 (d1 x d2)= 20X20/2= 200
Manager
Joined: 03 Oct 2009
Posts: 62
Followers: 0

Kudos [?]: 63 [2] , given: 8

If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Jan 2012, 08:42
2
KUDOS
6
This post was
BOOKMARKED
If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400
Math Expert
Joined: 02 Sep 2009
Posts: 30377
Followers: 5091

Kudos [?]: 57319 [2] , given: 8811

Re: Area of region [#permalink]  15 Jan 2012, 09:26
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
Apex231 wrote:

If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A 20
B 50
C 100
D 200
E 400

First of all to simplify the given expression a little bit let's multiply it be 2: $$|\frac{x}{2}|+|\frac{y}{2}|=5$$ --> $$|x|+|y|=10$$.

Now, find x and y intercepts of the region (x-intercept is a value(s) of x for y=0 and similarly y-intercept is a value(s) of y for x=0):
$$y=0$$ --> $$|x|=10$$ --> $$x=10$$ and $$x=-10$$;
$$x=0$$ --> $$|y|=10$$ --> $$y=10$$ and $$y=-10$$.

So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by $$|x|+|y|=10$$:
Attachment:

Enclosed region.gif [ 2.04 KiB | Viewed 10343 times ]
You can see that it's a square. Why a square? Because diagonals of the rectangle are equal (20 and 20), and also are perpendicular bisectors of each other (as they are on X and Y axis), so it must be a square. As this square has a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Similar questions:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html

Hope it's clear.
_________________
Manager
Joined: 03 Oct 2009
Posts: 62
Followers: 0

Kudos [?]: 63 [0], given: 8

Re: Area of region [#permalink]  15 Jan 2012, 09:58
I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?
Math Expert
Joined: 02 Sep 2009
Posts: 30377
Followers: 5091

Kudos [?]: 57319 [2] , given: 8811

Re: Area of region [#permalink]  15 Jan 2012, 10:07
2
KUDOS
Expert's post
Apex231 wrote:
I had solved till this point - So we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

But instead of joining these points i did this - 4 * (10 * 10) = 400 , which is wrong of course.

So when we join these points, how |x|+|y| = 10 stays satisfied , what's the maths behind it?

Given: $$|x|+|y|=20$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-x-y=10$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-x+y=10$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$x-y=10$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$x+y=10$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll get the figure shown in my previous post.

Hope it's clear.
_________________
Director
Joined: 24 Aug 2009
Posts: 505
Schools: Harvard, Columbia, Stern, Booth, LSB,
Followers: 10

Kudos [?]: 525 [0], given: 241

If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  10 Sep 2012, 11:25
CMcAboy wrote:
Can someone help me with this question:

If equation |x/2| + |y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A) 20
B) 50
C) 100
D) 200
E) 400

I believe this is the simplest & the quickest solution
|x/2| + |y/2| = 5
Put x = 0 in the above equation we get |y/2| = 5, which means y= 10, - 10
Put y = 0 in the above equation we get |y/2| = 5, which means x= 10, - 10

If you see plot these four points you get a square with two equal diagonals of length 20 units
Thus area = 1/2 * (Diagonal)^2 -----> 1/2 * 400 = 200

I hope this will help many.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS.
Kudos always maximizes GMATCLUB worth
-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 18

Kudos [?]: 309 [0], given: 11

Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  05 Dec 2012, 22:30
(1) derive all equations
x+y = 10
x-y = 10
x+y=-10
-x+y=10

(2) Get your x and y intercepts

(0,10), (10,0)
(0,-10),(10,0)
(0,-10),(-10,0)
(0,10),(-10,0)

(3) You will have a square with a diagonal of 20
(4) Calculate area = $$(10 * \sqrt{2})\sqrt{^2}$$ = 200

_________________

Impossible is nothing to God.

Intern
Joined: 12 Jul 2014
Posts: 10
Location: India
Concentration: Operations, Technology
GMAT Date: 11-06-2014
GRE 1: 303 Q159 V144
Followers: 0

Kudos [?]: 7 [0], given: 40

If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Oct 2014, 18:17
Apex231 wrote:
If equation |x/2|+|y/2| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

A. 20
B. 50
C. 100
D. 200
E. 400

Hello There,
Equation of a straight line whose x and y intercepts are a and b resp. is (x/a) + (y/b) = 1 i.e., coordinates of two ends of the line are (a,0) and (0,b).
Now, from the given question,
|x/2|+|y/2| = 5, reducing this to intercept form we get,
|x/10|+|y/10| = 1
Considering the equation without modulus, coordinates are (10,0) and (0,10). Since there is modulus, other two coordinates are (-10,0) and (0,-10).
Now coordinates (10,0), (0,10), (-10,0) and (0,-10) form a square with diagonal length = 20.
Here diagonal length can be obtained by calculating the distance between (10,0) and (-10,0) or (0,10) and (0,-10).
In a square,
Diagonal = Side * sqrt(2)
Side = 10 * sqrt(2)
Area = Side * Side = 200.

Ans : D

Hope this helps!
Thanks!
_________________

Regards,
Bharat Bhushan Sunkara.

"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"

Math Expert
Joined: 02 Sep 2009
Posts: 30377
Followers: 5091

Kudos [?]: 57319 [0], given: 8811

Re: If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]  15 Oct 2014, 23:48
Expert's post
Manager
Joined: 23 Nov 2014
Posts: 51
Location: India
GMAT 1: 700 Q47 V40
GMAT 2: 690 Q48 V36
GMAT 3: 730 Q49 V40
GPA: 3.11
WE: Sales (Consumer Products)
Followers: 1

Kudos [?]: 13 [0], given: 62

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  15 May 2015, 01:37
Bunuel wrote:
Barkatis wrote:
Hello,
Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Check similar problem at: graphs-modulus-help-86549.html?hilit=horizontal#p649401 it might help to get this one better.

Hope it helps.

Hi Bunuel,

Could you point me to any resources on how to draw the equations of the four lines without necessarily calculating the intersection points?

TIA.
Intern
Joined: 25 Jan 2014
Posts: 17
Concentration: Technology, General Management
Followers: 0

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

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  10 Jun 2015, 08:21
Bunuel wrote:
Barkatis wrote:
Hello,
Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Check similar problem at: graphs-modulus-help-86549.html?hilit=horizontal#p649401 it might help to get this one better.

Hope it helps.

hey Bunuel,

I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.

We can re-write the question as below

$$x^2/4 +y^2/4 = 5$$ (since $$|x| = x^2$$)

$$x^2 + y^2 = 20$$

This is the equation is a circle having the centre at (0,0) (general form is $$x^2 + y^2= r^2$$)

area =$$3.14 * R^2$$ = $$3.14 * 20$$ = 62.8

What am i assuming wrong here?? Thanks!
VP
Joined: 08 Jul 2010
Posts: 1007
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 31

Kudos [?]: 677 [0], given: 39

If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  10 Jun 2015, 08:50
Expert's post
arshu27 wrote:
Bunuel wrote:

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.

We can re-write the question as below

$$x^2/4 +y^2/4 = 5$$ (since $$|x| = x^2$$)

$$x^2 + y^2 = 20$$

This is the equation is a circle having the centre at (0,0) (general form is $$x^2 + y^2= r^2$$)

area =$$3.14 * R^2$$ = $$3.14 * 20$$ = 62.8

What am i assuming wrong here?? Thanks!

The part that I have highlighted above is WRONG which the first step in your solution

|x| is NOT equal to x^2 for all values of x[/highlight]

The Function "Modulus" only keeps the final sign Positive but that doesn't mean what you mentioned in the quoted Highlighted section.

Alternatively you can solve this question in this way

Step 1: Substitute y=0, $$|\frac{x}{2}| + |\frac{0}{2}| = 5$$ i.e. $$|\frac{x}{2}| = 5$$ i.e. $$|x| = 10$$ i.e. $$x = +10$$

So on the X-Y plane you get two Point (+10,0) and (-10,0)

Step 2:Substitute x=0, $$|\frac{0}{2}| + |\frac{y}{2}| = 5$$ i.e. $$|\frac{y}{2}| = 5$$ i.e. $$|y| = 10$$ i.e. $$y = +10$$

So on the X-Y plane you get two lines parallel to X-Axis passing through Y=+10 and Y=-10

So on the X-Y plane you get two Point (0, +10) and (0, -10)

Join all the four points, It's a Square with Side $$10\sqrt{2}$$

i.e. Area =$$(10\sqrt{2})^2$$ = 200
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal FREE Demo Class

______________________________________________________
Please press the if you appreciate this post !!

VP
Joined: 08 Jul 2010
Posts: 1007
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 31

Kudos [?]: 677 [0], given: 39

If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  10 Jun 2015, 09:02
Expert's post
arshu27 wrote:
Bunuel wrote:
Barkatis wrote:
Hello,
Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Check similar problem at: graphs-modulus-help-86549.html?hilit=horizontal#p649401 it might help to get this one better.

Hope it helps.

hey Bunuel,

I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.

We can re-write the question as below

$$x^2/4 +y^2/4 = 5$$ (since $$|x| = x^2$$)

$$x^2 + y^2 = 20$$

This is the equation is a circle having the centre at (0,0) (general form is $$x^2 + y^2= r^2$$)

area =$$3.14 * R^2$$ = $$3.14 * 20$$ = 62.8

What am i assuming wrong here?? Thanks!

One More Clarification

( $$|x| is NOT equal to x^2$$)

Instead, $$|x| = \sqrt{(x^2)}$$
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal FREE Demo Class

______________________________________________________
Please press the if you appreciate this post !!

Intern
Joined: 20 Apr 2014
Posts: 16
Followers: 0

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

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  12 Jun 2015, 13:43
why should I suppose that x or y equals +- 10 & zeros ? what about +- 5 as following :
|+ 5 |+ |-5| = 10
|-5|+|+5| = 10
|-5|+|-5|= 10
|+5|+|+5|=10
S0 we have Square with side of 10 length
Its area is 100
VP
Joined: 08 Jul 2010
Posts: 1007
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 31

Kudos [?]: 677 [0], given: 39

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  12 Jun 2015, 22:02
Expert's post
hatemnag wrote:
why should I suppose that x or y equals +- 10 & zeros ? what about +- 5 as following :
|+ 5 |+ |-5| = 10
|-5|+|+5| = 10
|-5|+|-5|= 10
|+5|+|+5|=10
S0 we have Square with side of 10 length
Its area is 100

Hi Hatemnag,

The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane

One Linear equation when x is +ve and y is +ve i.e. X+Y = 10
Second Linear equation when x is +ve and y is -ve i.e. X-Y = 10
Third Linear equation when x is -ve and y is +ve i.e. -X+Y = 10
Forth Linear equation when x is -ve and y is -ve i.e. -X-Y = 10

So you need to plot these equation and then take the area of Quadrilateral formed

Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (-10,0), (0,10) and (0,-10)

Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the Square

I hope this clears your doubt!
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha

e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772

http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal FREE Demo Class

______________________________________________________
Please press the if you appreciate this post !!

Manager
Joined: 29 May 2013
Posts: 96
Location: India
Concentration: Technology, Strategy
WE: Information Technology (Consulting)
Followers: 0

Kudos [?]: 13 [0], given: 42

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]  20 Jun 2015, 00:15
Bunuel wrote:
Barkatis wrote:
Hello,
Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ encloses a certain region on the coordinate plane, what is the area of this region?
20
50
100
200
400

OA: 200

ME: well, since $$|x| + |y| = 10$$ ; X can range from (-10) to (10) (when Y is 0) and the same for Y
So the length of the side of the square should be 20.

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

$$|\frac{x}{2}| + |\frac{y}{2}| = 5$$

You will have 4 case:

$$x<0$$ and $$y<0$$ --> $$-\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=-10-x$$;

$$x<0$$ and $$y\geq{0}$$ --> $$-\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10+x$$;

$$x\geq{0}$$ and $$y<0$$ --> $$\frac{x}{2}-\frac{y}{2}=5$$ --> $$y=x-10$$;

$$x\geq{0}$$ and $$y\geq{0}$$ --> $$\frac{x}{2}+\frac{y}{2}=5$$ --> $$y=10-x$$;

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the $$Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200$$.

Or the $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

Check similar problem at: graphs-modulus-help-86549.html?hilit=horizontal#p649401 it might help to get this one better.

Hope it helps.

Sorry, i dont know what i am missing, how do i get the diagonal to be 20?...from the square i got, i have all the sides equal to 20, hence the area=400.
Re: If equation |x/2| + |y/2| = 5 enclose a certain region   [#permalink] 20 Jun 2015, 00:15

Go to page    1   2    Next  [ 25 posts ]

Similar topics Replies Last post
Similar
Topics:
13 The equation x = 2y^2 + 5y - 17, describes a parabola in the 6 17 Oct 2012, 11:18
If equation |x/2|+|y/2| = 5 encloses a certain region 0 15 Oct 2014, 23:48
If equation |x| + |y| = 5 encloses a certain region on the coordinate 1 26 Feb 2011, 09:51
If equation |x/2| + |y/2| + 5 encloses a certain region on the coordi 1 03 Sep 2010, 13:45
Equation |x/2| + |y/2| = 5 encloses a certain region on 10 18 Oct 2008, 23:43
Display posts from previous: Sort by