If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin

Question

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

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. 
My Answer : 400

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

Thanks

Bunuel · Accepted Answer

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

geturdream · Answer

|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

anilnandyala · Answer

|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

Bunuel · Answer

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 : https://gmatclub.com/forum/download/file.php?id=16332 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 . Answer: D. Similar questions: https://gmatclub.com/forum/m06-5-absolut ... 08191.html https://gmatclub.com/forum/graphs-modulu ... 86549.html Hope it's clear. Enclosed region.gif

Apex231 · Answer

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?

Bunuel · Answer

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.

Bunuel · Answer

Similar questions to practice: the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html what-is-the-area-of-the-region-enclosed-by-lines-y-x-x-y-150487.html new-set-of-mixed-questions-150204-100.html#p1208441 the-area-bounded-by-the-three-curves-x-y-1-x-1-and-y-186595.html if-equation-x-2-y-2-5-encloses-a-certain-region-126117.html Hope this helps.

arshu27 · Answer

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!

GMATinsight · Answer

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

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

GMATinsight · Answer

One More Clarification

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

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

hatemnag · Answer

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

GMATinsight · Answer

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!

jayanthjanardhan · Answer

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.

GMATinsight · Answer

Hi Jayanthjanardan,

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

NOTE: PLEASE PLOT THE LINES TO UNDERSTAND THE FIGURE (REFER THE FIGURE) and see that Diagonal of Square is 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!

jayanthjanardhan · Answer

Hi GMATInight,

thanks a lot for the expalnantion. I get the logic now.

However, kindly refer to this link below.

in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549-40.html#p1540033

Its a similar problem, but the diagram we end getting is a square and not a rhombus...what am i missing here?

GMATinsight · Answer

Even that is a square but never forget that   a Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that fugure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees" . So calling it a Rhombus won;t be wrong either but you are right about the figure being a Square.

rahul16singh28 · Answer

|x| + |y| = 10.
Take the first quadrant => x + y = 10
Draw a line with (10,0) and (0,10) .. Area of this quadrant will be = 0.5 * 10 * 10 = 50
As this is mod it will be same in all quadrant = 50 * 4 = 200

sananoor · Answer

Hi Bunuel and other experts,
Can someone help me draw the diagram because with diagram i am getting 40 not 20.

Bunuel · Answer

Here it is:

https://gmatclub.com/forum/download/file.php?id=16332

If equation |x/2| + |y/2| = 5 encloses a certain region on the coordin

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