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

What you mean by "sufficient"? |x+y| + |x-y| = 4 gives FOUR equations, as explained on page 1.

HI Bunnel,

Have one doubt on this.

Similar kind of question is posted on following link

https://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

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.

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.

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).

Thanks in advance.

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

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!...

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?

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

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.