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m06 Q5

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m06 Q5 [#permalink] New post 12 Nov 2008, 13:36
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If equation |x| + |y| = 5 encloses a certain region on the graph, what is the area of that region?

(A) 5
(B) 10
(C) 25
(D) 50
(E) 100

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

explaination given
------------------
The end point of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals to and the area of the figure is that expression squared or 50.

How do I know the enclosed region is square ?? Also can anyone please elaborate on the explaination. I did not understand. :cry:

Edited by: GT
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Re: m06 Q5 [#permalink] New post 12 Nov 2008, 14:28
All sides are equal and perpendicular to each other. Therefore it is a square.
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Re: m06 Q5 [#permalink] New post 25 Nov 2008, 08:53
Hi Guys,
The explanation given is correct.
We are forgetting a very small thing here.That is these points lie on the co-ordinate axes.
So that means a polygon of four sides with equal sides, all angles = 90 & diagonals perpendicular.

SO the figure is a square.
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Re: m06 Q5 [#permalink] New post 01 Dec 2008, 09:54
Puneet,

No where it is said that sides are equal to begin with, though they are on coordinate axes.

anyway I am not fretting on this one anymore.. thanks folks

-Mo
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Re: m06 Q5 [#permalink] New post 23 Dec 2008, 06:11
Guys, you need to make sure that questions you post here are copied properly from the Tests. Here's the complete question:

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

(C) 2008 GMAT Club - m06#5

* 5
* 10
* 25
* 50
* 100

The end point of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals to 5*\sqrt{2} and the area of the figure is that expression squared or 50.
The correct answer is D.
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Re: m06 Q5 [#permalink] New post 25 Dec 2008, 22:14
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The trick here is in absolute value signs enclosing x and y.

This equation looks differently in each of the coordinate plane quarters because of the absolute value signs. x and y have these signs in the quarters from I to IV as follows: ++, -+, --, +-. Because of the absolute value signs all negative values are made positive and this is why we have the same equation look differently in coordinate plane quarters.

I'm attaching an image to make it clearer.


Hope this helps.
Attachments

m06-05.png
m06-05.png [ 9.58 KiB | Viewed 12401 times ]


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Re: m06 Q5 [#permalink] New post 16 Sep 2009, 18:37
>How did u know the pts on the basis of which you made the graph?

Please explain this problem from the scratch as my coordinate geometry+graphs knowledge is absolutely nil.
much appreciated.
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Re: m06 Q5 [#permalink] New post 21 Sep 2009, 11:58
In addition to my post above, you have to consider the different equations for 4 quarters of the coordinate plane:

1(++): x+y=5
2(-+): -x+y=5
3(--): -x-y=5
4(+-): x-y=5

So, you have to draw a line in each of the quarters using these equations in a corresponding quarter. That's how I've come up with those values in the graph.

Hope this helps :).
tejal777 wrote:
>How did u know the pts on the basis of which you made the graph?

Please explain this problem from the scratch as my coordinate geometry+graphs knowledge is absolutely nil.
much appreciated.

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Re: m06 Q5 [#permalink] New post 13 Jan 2010, 06:21
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its a tricky one but really easy....points to note in this question are
1) The degree of equation is 1 therefore relation between y and x is 'linear' ( straight line )
2) This question involves |x| and |y| instead of x,y so its obvious that it demands to consider negative values as well

Approach:


1) Going by the equation, we need values of x and y such that their sum is 5.
2) find pair of (x,y) which will total 5 .......(0,5) ...also, as its a |y| and not y....therefore (0,-5) also satisfies the equation ......Mark these points
3) Now, consider x coordinate ....(5,0) and (-5,0) for the reason stated above [ imp note, we are trying to find the restrictions of the region...the boundary, if to put it bluntly]
4) Now you have 4 points, (0,5) (0,-5) (5,0) (-5,0) ......
5) Join these points with a straight line and form the diamond shape figure as shown in the post above
6) The reason why we join the points with a straight line is because the degree of equation is 1 and relation between x, y is linear ( y+x = 5 therefore slope of line is -1 )
7) easier alternative to point6 is by substituting values...example, by plotting other pairs are (1,4) (2,3), (3,2) etc etc ....we get the same shape

even if we consider decimal numbers ( 0.9, 4.1) then also the linear relation does not change.


i hope it helps.....let me know if u have any doubt

by the way, so the answer is 50 .....how?? .....4 times the area of one triangle....[ 4X ( 1/2X5X5) ]
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Re: m06 Q5 [#permalink] New post 13 Jan 2010, 10:39
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given equation |x| +|Y| =5

|y|= -|x|+5

if x=0 then the points are (0,-5) & (0,5)
if y=0 then the points are (5,0) & (-5,0)

note : since we are dealing with absolute values there is 2 possible values


plotting the above 4 points you will get 4 right triangles or a single square


area of 1 right triangle = 1/2*5*5

area of 4 right trianlges = 4*1/2*5*5 = 50

so ans is D
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Re: m06 Q5 [#permalink] New post 13 Jan 2010, 11:41
It is a square. Draw the lines to a graph. Put 5 dashes going all 4 ways. Now connect the "top" dash to each other. (0,5) connects to (5,0) connects to (0,-5) connects to (-5,0) connects to (0,5). It is a square. Area is length times width or length * length since it is a square. We don't know the length or width yet. The square we just drew is separated into 4 triangles for us. With the Pythagorean theorem (spell check says it is spelled correctly) a^2 + b^2 = c^2. So if we find one c^2 we will have a length. We know a and b are 5. So 5^2 + 5^2 = c^2. 25 + 25 = c^2. 50 = c^2. So a length c is sqrt(50). lenth * length = sqrt(50) * sqrt(50) = 50
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Re: m06 Q5 [#permalink] New post 14 Jan 2010, 10:32
I understand how the answer is D. But why are we picking extreme values (5,0), (-5,0) etc in this question? I tried using values (2.5,2.5), (-2.5,2.5) etc as they add up to 5 taking the absolute value into account. This made me think this is an equation of a circle (since the coordinates will always be equidistant from the origin) and I applied the formula for area of circle and got wrong answer (Although the square can cover a circle satisfying the equation in question, the equation is clearly not that of a circle).

My question is whether in questions around absolute values in geometry, it is a thumbrule(at least in some cases) to get the x and y axis intersection points for calculation ?We only need to connect the lines and get to know the exact shape after doing this. The solution does not mention why the extreme values are taken in the first place.

Ideas anyone ?
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Re: m06 Q5 [#permalink] New post 14 Jan 2010, 12:46
You took (2.5,2.5), (-2.5,2.5) etc.....and the fact that these points satisfy equation of a circle indicates that a circle COULD also pass through these points ( only these points )...the points (2.5,2.5), (-2.5,2.5) represents a location on a graph !! that's it.....and if a shape ( with whatever equation, be it circle, arc, parabola, triangle, line ) passes through these points, these points will always satisfy the equation of that shape ( in ur case, it did for circle)

but the question in focus demands us to find the area of |X|+|Y| = 5 ....which is another way to ask " find the solutions of this equation and find the area in which they lie " ......

The approach to any question related with finding area in coordinate geometry is to first define the boundary and that is why we took the extreme points (0,5) (5,0) etc etc..once the boundary is defined, all the points lying inside will satisfy the equation....hence we found the area

i hope this helps



kaptain wrote:
I understand how the answer is D. But why are we picking extreme values (5,0), (-5,0) etc in this question? I tried using values (2.5,2.5), (-2.5,2.5) etc as they add up to 5 taking the absolute value into account. This made me think this is an equation of a circle (since the coordinates will always be equidistant from the origin) and I applied the formula for area of circle and got wrong answer (Although the square can cover a circle satisfying the equation in question, the equation is clearly not that of a circle).

My question is whether in questions around absolute values in geometry, it is a thumbrule(at least in some cases) to get the x and y axis intersection points for calculation ?We only need to connect the lines and get to know the exact shape after doing this. The solution does not mention why the extreme values are taken in the first place.

Ideas anyone ?
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Re: m06 Q5 [#permalink] New post 18 Jan 2011, 19:20
intcan wrote:
Found similar question in OG

Hi Intcan,
please can you make reference to the OG question?
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Re: m06 Q5 [#permalink] New post 11 Jun 2011, 17:58
points satisfying the given expression are

(-5,0) , (5,0) , (0,5) and (0,-5)

and these points form a square with side 5\sqrt{2}

so the area enclosed by these points is the area of the square = 5 \sqrt{2} * 5 \sqrt{2} = 50

Answer is D.
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Re: m06 Q5 [#permalink] New post 21 Jan 2012, 02:42
Can ne one suggest sm gud book for coordinate geometry!!!
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Re: m06 Q5 [#permalink] New post 21 Jan 2012, 02:57
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gmat1982 wrote:
Can ne one suggest sm gud book for coordinate geometry!!!


Check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

As for the original question:
If equation |x| + |y| = 5 encloses a certain region on the graph, what is the area of that region?
(A) 5
(B) 10
(C) 25
(D) 50
(E) 100

Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0):
Attachment:
square.gif
square.gif [ 2.86 KiB | Viewed 9078 times ]
Now, diagonals of the rectangle are equal (10 and 10), and also are perpendicular bisectors of each other (as the are on X and Y axises), so the figure must be a square. Area of a square equals to diagonal^2/2=10^2/2=50.

Answer: D.

This question is also discussed here: m06-5-absolute-value-108191.html

Similar question to practice with detailed solutions:
area-of-region-126117.html
graphs-modulus-help-86549.html

Hope it helps.

Hope it helps.
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Re: m06 Q5 [#permalink] New post 22 Jan 2013, 06:49
Wouldn't the following points satisfy the statement of |x| + |y| = 5?
2, 3
-2, -3
2, -3,
-2, 3

These points won't yield to a square.
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Re: m06 Q5 [#permalink] New post 22 Jan 2013, 07:04
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linglinrtw wrote:
Wouldn't the following points satisfy the statement of |x| + |y| = 5?
2, 3
-2, -3
2, -3,
-2, 3

These points won't yield to a square.


All these points would be on the square. But there are infinitely many points that satisfy the equation (for example, (1, 4), (1.2, 3.8), ...) and all these points will result in square.

Go through the following post for a solution, theory and similar questions: m06-q5-72817.html#p1032170

Hope it helps.
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Re: m06 Q5 [#permalink] New post 20 Apr 2014, 06:40
IMO D

The region that satisfies the requirement is a square whose 4 vertices are (-5,0),(5,0),(0,5),(0,-5). Each side of the square is equal to sqrt(5^2+5^2)=sqrt(50)
-> the area of the region = 50
Re: m06 Q5   [#permalink] 20 Apr 2014, 06:40
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