Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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 15190 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]

Show Tags

10 Sep 2012, 11:25

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.
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

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

Show Tags

15 Oct 2014, 18:17

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]

Show Tags

15 May 2015, 01:37

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]

Show Tags

10 Jun 2015, 08:21

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
_________________

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]

Show Tags

12 Jun 2015, 13:43

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

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

Show Tags

20 Jun 2015, 00:15

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, 00:15

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/22/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :3qa61fk6]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

[rss2posts title=The MBA Manual title_url=https://mbamanual.com/2016/11/21/mba-vs-mim-guest-post/ sub_title=MBA vs. MiM :2kn54ay1]Hey, guys! We have a great guest post by Abhyank Srinet of MiM-Essay . In a quick post and an...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...