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:

Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 14:39

Expert's post

srini123 wrote:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2 x=-2 y=2 y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4. _________________

Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 15:58

Bunuel wrote:

srini123 wrote:

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2 x=-2 y=2 y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.

Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was

what is the area bounded by graph|x/2| + |y/2| = 5?

I got hunderd since x=10 x=-10 y=10 y=-10

isnt the area 400 ? the answer given was 200, please explain _________________

Thanks, Sri ------------------------------- keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Re: graphs_Modulus....Help [#permalink]
08 Nov 2009, 16:41

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

srini123 wrote:

Thanks Bunuel, I used similar method for a similar question and I got wrong answer the question was

what is the area bounded by graph|x/2| + |y/2| = 5?

I got hunderd since x=10 x=-10 y=10 y=-10

isnt the area 400 ? the answer given was 200, please explain

I think this one is different.

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

After solving you'll get equation of four lines:

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

These four lines will also make a square, BUT in this case the diagonal will be 20 so the Area=\frac{20*20}{2}=200. Or the Side= \sqrt{200}, area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).

Re: graphs_Modulus....Help [#permalink]
22 May 2011, 07:38

Expert's post

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them. _________________

Re: graphs_Modulus....Help [#permalink]
27 Feb 2012, 22:33

Hi, Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.

Re: graphs_Modulus....Help [#permalink]
27 Feb 2012, 22:42

1

This post received KUDOS

Expert's post

devinawilliam83 wrote:

Hi, Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:

VinuPriyaN wrote:

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards, Vinu

Look at the solution given by Bunuel above. When you solve it, you get four equations. One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.

Yes, it can be done by graphing. |x+y| + |x-y| = 4 can expand in four different wasy:

A. x+y+x-y = 4 --> x=2 B. x+y-x+y = 4 --> y=2 C. -x-y +x-y= 4 --> y=-2 D. -x-y-x+y=4 --> x=-2

So you can draw all these four lines x=2, x=-2, y=2, y=-2 to get a square with the side of 4:

Today, 1st year Rotman students had a great simulation event hosted by Scotiabank, one of Canada’s best and largest banks. Attended by entire Rotman 1st year students, the...

Nope. I never learned finance ever in my life until I came to Rotman. This is why I got really frustrated when this term started because I was certain...