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:
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
3
This post received KUDOS
Expert's post
2
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:
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...