this one is pretty straight forward once you draw the figure on paper

X(max) = 5( i.e max value of x is 5)
Y(max) = 5

so, plot both X and Y on co-ordinate plane and join X and Y. You will get a straight line of length 5*sqrt(2)
Over here were took
x =+
y =+

now similarly, we have get 4 sets of straight lines( rather a square) of length 5*sqrt(2)


so, area = [5*sqrt(2)]^2
= 25*2 = 50

Hope this Helps

Excellent answer!

Let me see if I can reinforce this for myself and add to this:

1) If x(max)=5
and
y(max)=5
x= (5,0)
y= (0,5)

The distance between (0,0) and X(5,0) is 5.
The distance between (0,0) and Y(0,5) is 5.

Drawing a straight line between X and Y results in the distance being 5sqrt 2.

2) Next make X negative which results in (-5,0) but maintain Y as positive such that it is (0,5).
Once again the distance between X and Y is 5sqrt 2.

3) Next Make X and Y both negative such that X is (-5,0) and Y is (0,-5) resulting in a distance of 5sqrt2.

4) Lastly make X positive but Y negative such that X is (5,0) and Y is (0,-5). The distance is 5sqrt2.

Since all sides are equal, (5sqrt2), it forms a square. The area would be (5sqrt2)^2 = (5)^2 * (sqr2)^2 = 25 * 2 = 50.

gmat740 wrote:this one is pretty straight forward once you draw the figure on paper

X(max) = 5( i.e max value of x is 5)
Y(max) = 5

so, plot both X and Y on co-ordinate plane and join X and Y. You will get a straight line of length 5*sqrt(2)
Over here were took
x =+
y =+

now similarly, we have get 4 sets of straight lines( rather a square) of length 5*sqrt(2)


so, area = [5*sqrt(2)]^2
= 25*2 = 50

Hope this Helps

Hi Karan,

I am very poor at co-ordinate geometry. Can you guide me with any good resource/material to learn about co-ordinate geometry essential for GMAT?

Regards,
Viju

Hi, I would suggest you to refer to your High-School text books. they can really be helpful in developing the coordinate fundamentals and then you can move on to MGMAT math guides.

Hope this Helps

Thanks Karan.

Regards,
Viju

modX + mody = 5 represents the following line equations:

x + y = 5 => (x/5) + (y/5) = 1;
x - y = 5 => (x/5) + (y/-5) = 1;
-x + y = 5 => (x/-5) + (y/5) = 1;
-x -y = 5 => (x/-5) + (y/-5) = 1;

A line equation of the type : (x/a) + (y/b) = 1 will intersect the X axis at (a,0) and the Y axis at (0,b) respectively.

So, the 4 equations will correspond to (5,0) (0,5) (-5,0) and (0, -5). This area represents a rhombus in the XY co-ordinate space

The area of a rhombus = product of the diagonals/2

both the diagonals of the region (a rhombus) are 10.

Therefore the area of the region = (10 X 10)/2 = 50.