M06-05

Question

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

A. 5
B. 10
C. 25
D. 50
E. 100

Bunuel · Accepted Answer

Official Solution:

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

A. 5
B. 10
C. 25
D. 50
E. 100

The x and y intercepts of the graph are (0, 5), (0, -5), (5, 0), and (-5, 0) (we can find these by setting x equal to zero and solving for y, and then setting y equal to zero and solving for x). If we connect these points, we obtain the following region:
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m06-05.gif 
 
The diagonals of the quadrilateral are equal (both have a length of 10) and are also  perpendicular bisectors  of each other (as they lie on the x and y axes). Therefore, the figure must be a square. The area of a square can be calculated as  \frac{	ext{diagonal}^2}{2}=\frac{10^2}{2}=50 .

Answer: D

pate13 · Answer

Another way to find the area of the square, if you don't know how to find it from the diagonal, is to take the area of the four triangles that are made from the x and y axis. You end up with  (\frac{5*5}{2})*4 .
=
Additionally, you can find the difference between any two points with the formula  \sqrt{(X2-X1)^2+(Y2-Y1)^2} . Plug in (5,0) and (0,5) into that equation, then square the result to find the area.

shrive555 · Answer

Thanks B

one more question please

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.

Bunuel · Answer

You can find the area of a square in either way: area_{square}=side^2=\frac{diagonal^2}{2} . The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees. |x+y|+|x-y|=4: https://gmatclub.com/forum/download/file.php?id=47215 |x|+|y| =5: https://gmatclub.com/forum/download/file.php?id=47216 MSP86219e41c890f2h195700001928g4634f4e6h4h.gif MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif

ajit257 · Answer

Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong

Bunuel · Answer

We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure. If x>0 and y>0 we'll have x+y=5; If x>0 and y<0 we'll have x-y=5; If x<0 and y>0 we'll have -x+y=5; If x<0 and y<0 we'll have -x-y=5; So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them.

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

bailz314 · Answer

is there a way to know the sides from the information given? thx

Bunuel · Answer

Yes.

5^2 + 5^2 = side^2 , which gives  side = 5\sqrt{2} .

Or from the area:  area=side^2=50 , which gives  side = 5\sqrt{2} .

susheelh · Answer

I think this is a High quality question and I agree with the explanation.

However, I have a small question. Can someone please help me with an answer?

Are the four points (0, 5), (0, -5), (5, 0), and (-5, 0) the ONLY possible four solutions to this equation? Or are they taken to make the calculation of area simple and direct?

I do understand that even if there are other possible 4 points satisfying the equation in the Coordinate system, the area enclosed will always remain the same. Just curious to know if I am correct in saying this.

Hope, I am able to communicate what I intend to say

TIA!

Bunuel · Answer

|x|+|y|=5 has infinitely may solutions. For any value of x from -5 to 5 there will exist some value of y which will satisfy |x|+|y|=5. Since there are infinitely many real numbers from -5 to 5, then there are infinitely many pairs of (x, y) which satisfy |x|+|y|=5.

(0, 5), (0, -5), (5, 0), and (-5, 0) are x and y-intercepts of |x|+|y|=5. x-intercept is the value of x when y = 0 and y-intercept is the value of y when x = 0. If you plug x = 0, you'll get two value of y and if you plug y = 0, you'll get two value of x. Thus 4 points.

Hope it's clear.

adkikani · Answer

Bunuel   niks18   chetan2u   generis   pushpitkc   KarishmaB   EMPOWERgmatRichC   GMATPrepNow

It is interesting to see jump from absolute modulus to intercepts in this problem.
Do we get a hint about this approach from highlighted text ?

I believe it is very difficult to solve |x| + |y| = 5 on its own. Let me know your thoughts.

KarishmaB · Answer

It is a co-ordinate geometry question (keywords are graph, area of the region). Just that the equations of the lines are given in a more compact format involving absolute values. 
If you know your absolute values, you know the equations you will get are x + y = 5, x - y = 5 etc depending on the signs of x and y.
Also, it is very easy to visualise these lines by putting each variable equal to 0 in turn.
x + y = 5
If x = 0, y = 5 and if y = 0, x = 5 so line will intersect the axis at (0, 5) and (5, 0) and so on.

Darselle · Answer

So when I did this I guessed that it would form a square like that too and chose the correct answer. But can anyone explain to me intuitively why we can connect these four dots and consider that the enclosed area ? Thanks

AndrewN · Answer

Hello,  Darselle . My guess is that if you had already solved the question in the manner outlined above, then you already understand the intuitive reasoning behind the solution. Since the equation places an absolute value around  each  unknown and sets their sum equal to 5, and since absolute value can be thought of as a distance from 0, you are effectively testing extremes by setting one unknown to its highest and lowest values while holding the other at 0. That is, once you set  x  at -5, for instance, you can only set  y  to 0. You can derive all four "extreme" points in this manner. Connecting them with segments graphically shows that for any  valid   x  or  y  value tested within the constraints of the problem, the sum of the distance from 0 does not exceed 5. You can test easier points for reference, such as (-1, 4) or (3, -2), but you cannot break from the figure. In fact, if you tested enough points from the infinite number of possible  x  and  y  values--decimals would just give you more and more points to test--you would create the figure itself. An earlier poster had suggested values of -3 and 8 for the unknowns, but such a combination would violate the absolute value principles at work: -3 is 3 away from 0, which is fine, but 8 is 8 away from 0, and there is no negative absolute value that could pull the sum back to 5. So in short, I would suggest that you attempt to break free of the "box." As long as you observe that the total distance of the sum cannot exceed 5, you can test all you want, and you will only ever create a more and more refined box. It is not an issue of "Why can we?" but more an issue of "How can we not?"

- Andrew

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

sanchitb23 · Answer

A very simple yet tricky question! Correct Figure Construction = correct Answer

vedha0 · Answer

woah, this is a slightly more difficult question that's worth noting!

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