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 350,000 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:

The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

The correct answer is D.

How do we know that the area is square and not something else ? _________________

The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

The correct answer is D.

How do we know that the area is square and not something else ?

Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0):

Attachment:

MSP254019e3df2hii74417g00000g7c2817hhbe0a3b.gif [ 2.86 KiB | Viewed 2087 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 it must be a square. Area of a square equals to diagonal^2/2=10^2/2=50.

Re: M06#5 Absolute value [#permalink]
25 Jan 2011, 09:01

1

This post received KUDOS

Expert's post

shrive555 wrote:

Thanks B

one more question please

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.

You can find the area of a square in either way: \(area_{square}=side^2=\frac{diagonal^2}{2}\).

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees. |x+y|+|x-y|=4:

Attachment:

MSP86219e41c890f2h195700001928g4634f4e6h4h.gif [ 1.79 KiB | Viewed 2048 times ]

|x|+|y| =5:

Attachment:

MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 2047 times ]

Re: M06#5 Absolute value [#permalink]
26 Feb 2011, 16:42

1

This post received KUDOS

Expert's post

ajit257 wrote:

Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong

We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure.

If x>0 and y>0 we'll have x+y=5; If x>0 and y<0 we'll have x-y=5; If x<0 and y>0 we'll have -x+y=5; If x<0 and y<0 we'll have -x-y=5;

So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them.

Follow the links in my previous posts for more. _________________

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

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

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3 x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?

Merging similar topics. Please ask if anything remains unclear.

Yes, why can't it be a circle?

Well it cannot be a circle because it's a square. The reason why it's a square is given in previous posts (you can also see a diagram there).

Next, in an xy plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((x-a)^2+(y-b)^2=r^2\). There is no resemblance whatsoever between this equation and the equation given in the question.

Re: M06#5 Absolute value [#permalink]
14 Apr 2014, 21:04

Hi Bunuel,

I understood your concept of equations and plotting points. But, i have a small doubt.

Why can't the graph in the second question (|x|+|y| =5 )be a horizontal square than a square turned by 45 degrees in the first question (|x+y|+|x-y|=4)

My question is there any reason why in the second question we have plotted a graph which is turned by 45 degrees rather than a horizontal square.

Help is appreciated.

Thanks in advance

Bunuel wrote:

shrive555 wrote:

Thanks B

one more question please

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.

You can find the area of a square in either way: \(area_{square}=side^2=\frac{diagonal^2}{2}\).

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees. |x+y|+|x-y|=4:

Attachment:

MSP86219e41c890f2h195700001928g4634f4e6h4h.gif

|x|+|y| =5:

Attachment:

MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif

gmatclubot

Re: M06#5 Absolute value
[#permalink]
14 Apr 2014, 21:04