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ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-\frac{x}{2}-\frac{y}{2}=5\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(\frac{x}{2}-\frac{y}{2}=5\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

|X/2| + |Y/2| = 5 so when x = 0, y= |10| when y=0 , x=|10|

so the sides of the enclosed area touches (0,10),(10,0),(0,-10) and (-10,0). so its a square having the diagonal =20unit So the area of the region = (20/1.414)^2 = 200

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 a little bit let's 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 14463 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 \(Side= \sqrt{200}\) --> \(area=side^2=200\).

If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]

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10 Sep 2012, 12:25

CMcAboy wrote:

Can someone help me with this question:

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

I believe this is the simplest & the quickest solution |x/2| + |y/2| = 5 Put x = 0 in the above equation we get |y/2| = 5, which means y= 10, - 10 Put y = 0 in the above equation we get |y/2| = 5, which means x= 10, - 10

If you see plot these four points you get a square with two equal diagonals of length 20 units Thus area = 1/2 * (Diagonal)^2 -----> 1/2 * 400 = 200

I hope this will help many.
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If equation |x/2|+|y/2| = 5 encloses a certain region [#permalink]

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15 Oct 2014, 19:17

Apex231 wrote:

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

Hello There, Equation of a straight line whose x and y intercepts are a and b resp. is (x/a) + (y/b) = 1 i.e., coordinates of two ends of the line are (a,0) and (0,b). Now, from the given question, |x/2|+|y/2| = 5, reducing this to intercept form we get, |x/10|+|y/10| = 1 Considering the equation without modulus, coordinates are (10,0) and (0,10). Since there is modulus, other two coordinates are (-10,0) and (0,-10). Now coordinates (10,0), (0,10), (-10,0) and (0,-10) form a square with diagonal length = 20. Here diagonal length can be obtained by calculating the distance between (10,0) and (-10,0) or (0,10) and (0,-10). In a square, Diagonal = Side * sqrt(2) Side = 10 * sqrt(2) Area = Side * Side = 200.

Ans : D

Hope this helps! Thanks!
_________________

Regards, Bharat Bhushan Sunkara.

"You need to sacrifice what you are TODAY, for what you want to be TOMORROW!!"

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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15 May 2015, 02:37

Bunuel wrote:

Barkatis wrote:

Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-\frac{x}{2}-\frac{y}{2}=5\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(\frac{x}{2}-\frac{y}{2}=5\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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10 Jun 2015, 09:21

Bunuel wrote:

Barkatis wrote:

Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-\frac{x}{2}-\frac{y}{2}=5\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(\frac{x}{2}-\frac{y}{2}=5\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

I had another way of solving. The answer is wrong but i wanted to know what is wrong in the method.

We can re-write the question as below

\(x^2/4 +y^2/4 = 5\) (since \(|x| = x^2\))

\(x^2 + y^2 = 20\)

This is the equation is a circle having the centre at (0,0) (general form is \(x^2 + y^2= r^2\))

area =\(3.14 * R^2\) = \(3.14 * 20\) = 62.8

What am i assuming wrong here?? Thanks!

The part that I have highlighted above is WRONG which the first step in your solution

|x| is NOT equal to x^2 for all values of x[/highlight]

The Function "Modulus" only keeps the final sign Positive but that doesn't mean what you mentioned in the quoted Highlighted section.

Alternatively you can solve this question in this way

Step 1: Substitute y=0, \(|\frac{x}{2}| + |\frac{0}{2}| = 5\) i.e. \(|\frac{x}{2}| = 5\) i.e. \(|x| = 10\) i.e. \(x = +10\)

So on the X-Y plane you get two Point (+10,0) and (-10,0)

Step 2:Substitute x=0, \(|\frac{0}{2}| + |\frac{y}{2}| = 5\) i.e. \(|\frac{y}{2}| = 5\) i.e. \(|y| = 10\) i.e. \(y = +10\)

So on the X-Y plane you get two lines parallel to X-Axis passing through Y=+10 and Y=-10

So on the X-Y plane you get two Point (0, +10) and (0, -10)

Join all the four points, It's a Square with Side \(10\sqrt{2}\)

i.e. Area =\((10\sqrt{2})^2\) = 200
_________________

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-\frac{x}{2}-\frac{y}{2}=5\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(\frac{x}{2}-\frac{y}{2}=5\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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12 Jun 2015, 14:43

why should I suppose that x or y equals +- 10 & zeros ? what about +- 5 as following : |+ 5 |+ |-5| = 10 |-5|+|+5| = 10 |-5|+|-5|= 10 |+5|+|+5|=10 S0 we have Square with side of 10 length Its area is 100

why should I suppose that x or y equals +- 10 & zeros ? what about +- 5 as following : |+ 5 |+ |-5| = 10 |-5|+|+5| = 10 |-5|+|-5|= 10 |+5|+|+5|=10 S0 we have Square with side of 10 length Its area is 100

Hi Hatemnag,

The given equation is basically representing FOUR linear equations which are representing 4 lines on the plane

One Linear equation when x is +ve and y is +ve i.e. X+Y = 10 Second Linear equation when x is +ve and y is -ve i.e. X-Y = 10 Third Linear equation when x is -ve and y is +ve i.e. -X+Y = 10 Forth Linear equation when x is -ve and y is -ve i.e. -X-Y = 10

So you need to plot these equation and then take the area of Quadrilateral formed

Also, Please Note that Four Vertices of Quadrilateral are obtained where two lines Intersect, and The intersections of the lines are obtained at points (10,0), (-10,0), (0,10) and (0,-10)

Whereas, what you have done is taking any FOUR RANDOM POINTS on those four lines as per your convenience and then have assumed that these points form the Square

READ:http://gmatclub.com/forum/620-to-760-getting-reborn-161230.html Classroom Centre Address: GMATinsight 107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075 ______________________________________________________ Please press the if you appreciate this post !!

Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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20 Jun 2015, 01:15

Bunuel wrote:

Barkatis wrote:

Hello, Am new here. I just took the m25 GMAT CLub Test and I don't get the solution of a question. (Q19)

If equation \(|\frac{x}{2}| + |\frac{y}{2}| = 5\) encloses a certain region on the coordinate plane, what is the area of this region? 20 50 100 200 400

OA: 200

ME: well, since \(|x| + |y| = 10\) ; X can range from (-10) to (10) (when Y is 0) and the same for Y So the length of the side of the square should be 20. My Answer : 400

I think I am making a silly mistake some where but I just can't figure it out.

Thanks

Hi and welcome to the Gmat Club. Below is the solution for your problem. Hope it's clear.

\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)

You will have 4 case:

\(x<0\) and \(y<0\) --> \(-\frac{x}{2}-\frac{y}{2}=5\) --> \(y=-10-x\);

\(x<0\) and \(y\geq{0}\) --> \(-\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10+x\);

\(x\geq{0}\) and \(y<0\) --> \(\frac{x}{2}-\frac{y}{2}=5\) --> \(y=x-10\);

\(x\geq{0}\) and \(y\geq{0}\) --> \(\frac{x}{2}+\frac{y}{2}=5\) --> \(y=10-x\);

So we have equations of 4 lines. If you draw these four lines you'll see that the figure which is bounded by them is square which is turned by 90 degrees and has a center at the origin. This square will have a diagonal equal to 20, so the \(Area_{square}=\frac{d^2}{2}=\frac{20*20}{2}=200\).

Or the \(Side= \sqrt{200}\) --> \(area=side^2=200\).

Sorry, i dont know what i am missing, how do i get the diagonal to be 20?...from the square i got, i have all the sides equal to 20, hence the area=400.

gmatclubot

Re: If equation |x/2| + |y/2| = 5 enclose a certain region
[#permalink]
20 Jun 2015, 01:15

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