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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

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Re M25-19  [#permalink]

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New post 16 Sep 2014, 00:23
3
5
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D
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Re: M25-19  [#permalink]

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New post 06 Nov 2014, 17:06
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D



For this equation |x|+|y|=10.
These points also satisfy the equation ( 5 , 5) ( -5 , 5 ) ( 5 ,-5) ( -5, -5) . If i use this then the area would be 100.
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Re: M25-19  [#permalink]

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New post 07 Nov 2014, 03:40
parameswaranprasad wrote:
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D



For this equation |x|+|y|=10.
These points also satisfy the equation ( 5 , 5) ( -5 , 5 ) ( 5 ,-5) ( -5, -5) . If i use this then the area would be 100.


That's not correct.

Though these points lie on the graph of |x|+|y|=10, they are not the vertices of a cube you find the area of.
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Re: M25-19  [#permalink]

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New post 08 Nov 2014, 01:57
Bunuel wrote:
parameswaranprasad wrote:
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D



For this equation |x|+|y|=10.
These points also satisfy the equation ( 5 , 5) ( -5 , 5 ) ( 5 ,-5) ( -5, -5) . If i use this then the area would be 100.


That's not correct.

Though these points lie on the graph of |x|+|y|=10, they are not the vertices of a cube you find the area of.



Sorry, i am not getting it. Why is that we are not suppose to calculate area of this region ? Is the question to calculate max area?
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Re: M25-19  [#permalink]

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New post 09 Nov 2014, 05:04
1
3
parameswaranprasad wrote:
Bunuel wrote:
parameswaranprasad wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D

For this equation |x|+|y|=10.
These points also satisfy the equation ( 5 , 5) ( -5 , 5 ) ( 5 ,-5) ( -5, -5) . If i use this then the area would be 100.


That's not correct.

Though these points lie on the graph of |x|+|y|=10, they are not the vertices of a cube you find the area of.



Sorry, i am not getting it. Why is that we are not suppose to calculate area of this region ? Is the question to calculate max area?


The figure you get when graphing |x|+|y|=10. is below:
Image
Points ( 5 , 5) ( -5 , 5 ) ( 5 ,-5) ( -5, -5) ARE indeed on that graph but they are NOT the vertices of the square (figure), so you cannot say that the side is 10 and the area is 100.

Similar questions to practice:
the-area-bounded-by-the-curves-x-y-1-and-x-y-1-is-93103.html
if-equation-x-2-y-2-5-encloses-a-certain-region-126117.html
in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html
if-equation-x-2-y-2-5-enclose-a-certain-region-101963.html
what-is-the-area-of-the-region-enclosed-by-lines-y-x-x-y-150487.html
new-set-of-mixed-questions-150204-100.html#p1208441

Hope this helps.
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Collection of Questions:
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Re: M25-19  [#permalink]

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New post 01 Dec 2014, 20:03
Diagonal is 5+5 = 10.. area = 100 ????
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New post 01 Dec 2014, 23:41
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Re: M25-19  [#permalink]

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New post 15 Dec 2014, 07:20
After drawing the graph, you can see that each diagonal is equal to 20 units on the coordinate plane. A must memorize formula for the GMAT is \(d=s*\sqrt{2}\) where D=diagonal s=side of square. By plugging in 20 for d in this formula we can find the distance of one side of the square. ---> 20=s*\(\sqrt{2}\)-->\(\frac{20}{\sqrt{2}}=s\)

s=10*\(\sqrt{2}\). So the area equals \(s^2\) or 200.
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New post 04 Nov 2015, 15:00
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I have no idea where 20 comes from and how exactly 10 and -10 is calculated for x and y
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Re M25-19  [#permalink]

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New post 22 Dec 2015, 08:46
I think this is a poor-quality question and I don't agree with the explanation. If we calculate the side of the square formed it comes out to be _/50
Hence, area= _/50*_/50=50
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M25-19  [#permalink]

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New post 01 Oct 2016, 16:29
A much simpler way to solve this is to realize that this should be a rhombus because each vertex is either \((\pm10,0)\) or \((0,\pm10)\), hence distance between vertices should be same (ie, either \(\sqrt{(\pm10)^2 + (0)^2}\) or \(\sqrt{(0)^2 + (\pm10)^2}\))

Hence simply use area of rhombus = \(\frac{1}{2}(d1*d2)\) = \(\frac{1}{2}(20 * 20)\) = 200
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Re: M25-19  [#permalink]

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New post 25 Feb 2018, 06:29
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D


Hi Bunuel,

I answered this question incorreclty because that formula looks like that of a circle with radius of 10.

abs(x)+abs(y)=10

...can this be transferred to

x^2+y^2 =10^2(circle formula)

or is it the case that I can only square the entire left side, which is obviously not a circle?
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Re: M25-19  [#permalink]

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New post 25 Feb 2018, 06:52
Kontaxis wrote:
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D


Hi Bunuel,

I answered this question incorreclty because that formula looks like that of a circle with radius of 10.

abs(x)+abs(y)=10

...can this be transferred to

x^2+y^2 =10^2(circle formula)

or is it the case that I can only square the entire left side, which is obviously not a circle?


\((a + b)^2=a^2+2ab+b^2\), so if you square \(|x|+|y|=10\) you'll get \(x^2 + 2|xy|+y^2 = 100\) not \(x^2+y^2 =100\).
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M25-19  [#permalink]

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New post 26 Feb 2018, 00:34
Bunuel 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

IMO IS D again i will go with plugin values i.e if we try to find most extreme case we know that 5 is possible if we take (10,0),(-10,0),(0,10),(0,-10).
Rest other points will lie under this figure and if we plot it on graph we will know that its a square whose diagonal is 20. so side is diagonal/root 2 which is 20/root2 and area is square of side. so 400/2 which is 200 option D
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New post 10 May 2018, 23:20
Great question Bunuel.

I think it will help to note that the four smaller triangles are 45-45-90 triangle, as the slope is 10/10 = 1 i.e slope of 1. thus processing the last part will be much easier.

So we it can be simplified as 10-10-10 √2, therefore 10 √2 * 10 √2 = 200.
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New post 22 Jun 2018, 07:09
+1 for option D. Draw the graph and check. The figure will lie on all four quadrants. The area will be (1/2*10*10*4)=200. Option D it is !
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New post 28 Jun 2018, 02:17
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D


When I saw this question, i immediately thought that 10 would be the radius of a circle. So I saw x^2 + y^2 = 100. Is this out of the scope of the GMAT?
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M25-19  [#permalink]

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New post 28 Jun 2018, 02:29
yimmyyam wrote:
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D


When I saw this question, i immediately thought that 10 would be the radius of a circle. So I saw x^2 + y^2 = 100. Is this out of the scope of the GMAT?


The solution shows that it's not a circle (not because the equation of a circle is our of scope but because it's NOT a circle) but because it's a square. We don't have x^2 +y^2 = 100, we have \(|x|+|y|=10\). I guess you squared \(|x|+|y|=10\) but if do so you'll get x^2 + 2|xy| + y^2 = 100, not \(|x|+|y|=10\).

Hop it's clear.
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Re: M25-19  [#permalink]

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New post 14 Oct 2018, 06:16
Bunuel wrote:
Official Solution:

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


First of all to simplify the given expression multiply it be 2: \(|x|+|y|=10\).

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 \(x=0\)):

\(y=0\) gives \(|x|=10\), so \(x=10\) and \(x=-10\)

\(x=0\) gives \(|y|=10\), so \(y=10\) and \(y=-10\)

Thus we have 4 points: (10, 0), (-10, 0), (0, 10) and (-10, 0).

When you join them you'll get the region enclosed by \(|x|+|y|=10\):

Image

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.

Since this square has a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(\text{Side} = \sqrt{200}\), so \(\text{area} = \text{side}^2 = 200\).


Answer: D

Another way to calculate the area -

The rectangle, which is also a square, is composed of 4 triangles.
So, the total area of rectangle is = 4 x area of triangles (the vertices of the triangles are 2 vertices of rectangle and origin)
= 4 * 1/2 * 10*10 = 4* 50
= 200

Answer D
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Re: M25-19 &nbs [#permalink] 14 Oct 2018, 06:16
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