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# M25 Q19

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13 Mar 2009, 11:31
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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain this? Thanks!

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16 Mar 2009, 08:55
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jenyang5268 wrote:
Equation |x| + |y| = 10 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

Can someone please explain this? Thanks!

Explanation:
|x| + |y| = 10 means there are 4 combination.

x + y = 10
-x - y = 10
x - y = 10
-x + y = 10

Now if you draw these four lines you will get square with diagonal size (20, and half diagonal size = 10) and it will make 4 small triangles with sides are 10, 10 and so
its hypotaneous becomes 10 root 2.

Square area = 10 root 2 * 10 root 2 = 200.

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07 Jan 2010, 19:19
patedhav wrote:
jenyang5268 wrote:
Equation |x| + |y| = 10 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

Can someone please explain this? Thanks!

Explanation:
|x| + |y| = 10 means there are 4 combination.

x + y = 10
-x - y = 10
x - y = 10
-x + y = 10

Now if you draw these four lines you will get square with diagonal size (20, and half diagonal size = 10) and it will make 4 small triangles with sides are 10, 10 and so
its hypotaneous becomes 10 root 2.

Square area = 10 root 2 * 10 root 2 = 200.

Since all the sides are the same and you determine that the figure they are referring to is a square, then why wouldn't you just multiply 10*10 to get the area? Please clarify so I may have a better understanding. Thank you!

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08 Jan 2010, 08:37
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@ hatethegmat

you are missing a trick here.

consider |x| = 2. this can be broken into x= 2 and x = -2.
Similary |x| + |y|=10 has to be broken so that all possible values of x,y are taken into consideration.
ie. x+y=10 , x-y=10, -x+y=10 , -x-y=10.
What you are considering is only the 1st scenario ( x+y=10), but there are 3 more scenarios.
Now if you draw all the above 4 graphs in X-Y axis , :
x+y=10 ... slope 135 degree in 1st quadrant
x-y=10 ... slope 225 degree in 4th quadrant
-x+y=10 ... slope 45 degree in 2nd quadrant
-x-y=10 ... slope 315 degree in 3rd quadrant .

All of them intersect to make a rhombus of side 10*(2^(1/2)) [Sqrroot( 10^2 + 10^2 )]
or if you are looking 45 degree anticlockwise , a square with side 10*(2^(1/2)) . Thus the total area becomes {10*(2^(1/2)) }^2 = 200

I hope I have not made things even more complicated.
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09 Feb 2010, 11:59
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Here is a graph of the 4 equations. The reason that you multiply 10*sqrt(2) is easier to understand after viewing graph. good luck.
Attachments

File comment: Graph of equations

m25 q19.JPG [ 63.3 KiB | Viewed 10092 times ]

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28 Jun 2010, 16:36
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|x/2| + |y/2| = 5
=> |x| + |y| = 10...graph of the form y = mx + c [m is the slope]
straight line graphs with x and y intercepts at +/- 10
four possibilities: y(0): x: {10, -10} and x(0): y: {10, -10}
The shape is a square (two paired parallels at 90 degrees and same size)
We only know diagonal = 20 units (from +10 to -10)
length of a side: 2a^2 = 400 (a is one side of the square)
a^2 = 200
Ans = D
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08 Jul 2010, 09:18
Guys,

need some help on this one.

What if we consider

|x|=8 and |y|=2

which would on addtion be equal to 10.

also, now the diagnols would be 16 and 4 which means each side of the parallelogram is

sqrt of 68 and thus area would be 68.

SO once again why are we only considering X and Y as 10???

Thanks,
Prash

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13 Jul 2010, 12:51
think of it as a "diamond"

and picture 4 equal triangles.

top right quandrant triangle looks like this:

| .\
| .....\
| .........\
| ............ \

(horribly drawn, but you get my point. this is a 45-45 triangle, with the sides 10 length). the area is just 10x10 / 2 = 50

50x4 (since we have 4 of these triangles) = 200.

i did it by the same method.

when x=0, y= +/- 10
x=1, y=+/-9
x=2,y=+/-8
..

similarly
y=0, x=+/- 10

and go on

when you plot on x-y plane, you will 4 45-45 right triangles each with height = base=10

so 4*1/2*10*10 = 200

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10 Oct 2010, 02:42
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jenyang5268 wrote:
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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Can someone please explain this? Thanks!

Orange08 wrote:
Any better explanation for this please?

In quadrant 1, this is just the line x+y=10. Which is a line with X-int=10 and Y-int=10. So area enclosed with axes in Q1, is just 0.5*10*10 = 50 (a right angled triangle)

Now notice, if you replace x by -x the equation does not change --> So figure must be symmetric about the Y-axis
So Q2, must also have a right angled triangle, of area 50 with the axis

Finally, if you replac y by -y, the equation is still the same --> So the figure must be symmetric about the X-axis
So Q3 and Q4 have a mirror image of the above two triangles

So total area = 50*4 = 200

And the figure that this equation form is a diamond shape with (0,0) at the centre of it

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05 Jul 2011, 12:07
Question - can we not generalize that the equation of a square in the form of a diamond around the origin will be -> |x| + |y| = A, where A is the side of the square?

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03 Feb 2012, 23:34
in the explanation it says you can move the denominator of 2 out to the other side of the equation. even though they are in separate abs value... is it okay to apply this to all instances where there are 2 different fractions but with the same denominator each with it's own abs value?
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04 Feb 2012, 09:14
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calreg11 wrote:
in the explanation it says you can move the denominator of 2 out to the other side of the equation. even though they are in separate abs value... is it okay to apply this to all instances where there are 2 different fractions but with the same denominator each with it's own abs value?

Generally: $$|xy|=|x|*|y|$$, so if you have instead of $$y$$ some specific number then you can factor its absolute value from the modulus: $$|-3x|=|3x|=3*|x|$$ or $$|\frac{x}{-10}|=|\frac{x}{10}|=\frac{|x|}{10}$$.

Hope it's clear.

As for the question itself:
If equation |x/2|+|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

First of all to simplify the given expression: 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 $$|x|+|y|=10$$:
Attachment:

Enclosed region.gif [ 2.04 KiB | Viewed 12501 times ]
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 $$Side= \sqrt{200}$$ --> $$area=side^2=200$$.

This question is also discussed here: area-of-region-126117.html and here: cmat-club-test-question-m25-101963.html

Similar questions:
m06-5-absolute-value-108191.html
graphs-modulus-help-86549.html
m06-q5-72817.html
if-equation-encloses-a-certain-region-110053.html

Hope it helps.
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26 May 2012, 11:28
@schumi252 i think the point 8,2 will lie within the region graphed by the diamond fig. hence while 8,2 is a true region, the area enclosed by it doesnot include all the possible values. hence the 10's and 0's work the best to graph the total area and hence all the possible solution points.

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26 May 2012, 11:31
@calreg11 yes you can multiply inequality by 2. in general the rule is that you can multiply or divide both sides of an inequality by a positive no. without changing the inequality sign. if you multiply or divide by a negative no then the sign of the inequality changes.

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04 Jul 2013, 06:34
If you quickly sketch out the outer boundaries of this equation you quickliy notice that you have a square bounded by (5,5), (-5,5), (-5,-5) and (5,-5). Additionally you also have the points (10,0), (0,10), (-10,0) and (0, -10). These points form 4 equal triangles outside the square formed by the previous coordinates. Therefore, area of square is 10x10 and area of each triangle is (10x5)/(2). Since there are 4 equal triangles we multiply 25x4.
Total area is 100+100 = 200 (D)

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06 Jul 2013, 00:38
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what I feel whenever we get this type equatn
|x| + |y| = A, will always be a SQUARE
2*A will be diagonal of that square and side of square will be A*(sqrt of 2)
and area will be 2*A*A

so we have |x/2| +|y/2| = 5 we can write it as |x|+|y| = 10
and area will be 2*10*10 = 200

if u plot the graph once it will be easier for u understand

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15 Jul 2013, 13:33
This one got me at first. Once I realized you could multiply the 2 over to the other side of the equation, it was only a matter of plugging in numbers for x and y to find the area. Answer is D!

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Re: M25 Q19   [#permalink] 15 Jul 2013, 13:33
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# M25 Q19

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