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:
Graphic approach to problems with inequalities [#permalink]
29 Jul 2008, 03:51
53
This post received KUDOS
Expert's post
74
This post was BOOKMARKED
Hi all! My friend Tarek PMed me and asked to show how to use the graphic approach to problems with inequalities. I really love the approach because it isn't only fast one after training, but also reliable. So, I try to illustrate how to use it.
There are two distinct ways to solve math problems: logical and visual. The approach described in this post is for those who use the visual approach. If you like imaging math problems in the visual scratchpad in your mind, you will enjoy this method. At the same time, don't be discourage if it doesn't work for you, try the logical approach instead (described by Bunuel).
1) If \((x/y)>2\), is \(3x+2y<18?\)
(1) \(x-y\) is less than \(2\) (2) \(y-x\) is less than \(2\)
1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.
2. Next, we draw our main inequality: 3x+2y<18. 3x+2y=18 - is a boundary. (see figure 2).
3. Now, we should combine our main inequality with the restriction, x/y>2. (see figure 3). Eventually, we defined two areas (sets) were the main inequality is TRUE and were it is FALSE. Two lines intersect in point P with coordinates: (4.5;2.25).
4. Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region. Therefore, the first statement is sufficient to answer the question. We should be careful and check where line x-y=2 passes point P, through left side or right side. We can put y=2.25 into x-y=2 and find that x=4.25<4.5 (left side). In other words, line x-y=2 passes y=2.25 (y-coordinate of P) early and goes above P.
5. Finally, let's check last condition: y-x<2. y-x=2 is a boundary. (see figure 5). As we can see all y,x that satisfies the second condition lie in both "green-TRUE" and "red-FALSE" regions. Thus, the second condition is insufficient.
So, answer is A
This approach took less than 2 minutes.
Tips:
1) How fast can we draw a line, for example 3x+2y=18? Simple approach: we need two points to draw line, let's choose intersections with x- and y- axes. x=0 (intersection with y-axis) --> y=9; y=0 (intersection with x-axis) --> x=6.
2) Let's suppose we have a linear inequality, such as 38y-11x>121, suppose we've already drawn the line. How can we find what side is "true" and what side is "false"? The fastest method is just use y=0,x=-infinity. In our case, 0-(-infinity)=infinity>121 - true. Therefore, we take a left side.
Re: Graphic approach to problems with inequalities [#permalink]
30 Jul 2008, 07:44
walker wrote:
Hi all! My friend, Tarek, PM me and asked me to show how to use the graphic approach to problem with inequalities. I really love such approach because it is not only fast one after training, but also reliable. So, I try to illustrate how to use it.
1) If \((x/y)>2\), is \(3x+2y<18?\)
(1) \(x-y\) is less than \(2\) (2) \(y-x\) is less than \(2\)
1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.
2. Next, we draw our main inequality: 3x+2y<18. 3x+2y=18 - is a boundary. (see figure 2).
3. Now, we should combine our main inequality with the restriction, x/y>2. (see figure 3). Eventually, we defined two areas (sets) were the main inequality is TRUE and were it is FALSE. Two lines intersect in point P with coordinates: (4.5;2.25).
4. Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region. Therefore, the first statement is sufficient to answer the question. We should be careful and check where line x-y=2 passes point P, through left side or right side. We can put x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).
5. Finally, let's check last condition: y-x<2. y-x=2 is a boundary. (see figure 5). As we can see all y,x that satisfies the second condition lie in both "green-TRUE" and "red-FALSE" regions. Thus, the second condition is insufficient.
So, answer is A
This approach took less than 2 minutes.
Additional tip: Let's suppose we have a linear inequality, such as 38y-11x>121, suppose we've already drawn the line. How can we find what side is "true" and what side is "false"? The fastest method is just use y=0,x=-infinity. In our case, 0-(-infinity)=infinity>121 - true. Therefore, we take a left side.That's all
Re: Graphic approach to problems with inequalities [#permalink]
03 Aug 2008, 09:27
3
This post received KUDOS
Expert's post
bigfernhead wrote:
Hi - can someone help me explain Tip #2? I don't really understand what it is saying. Thanks.
See fig.2 - we have line 3x+2y=18. Where is 3x+2y<18? left or right side? we put x=-infinity and y=0 --> -infinity<18. Is is correct? Yes. Therefore, left side corresponds to 3x+2y<18 (orange color in fig.2) _________________
Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 04:40
walker wrote:
Hi all! My friend, Tarek, PM me and asked me to show how to use the graphic approach to problem with inequalities. I really love such approach because it is not only fast one after training, but also reliable. So, I try to illustrate how to use it.
1) If \((x/y)>2\), is \(3x+2y<18?\)
(1) \(x-y\) is less than \(2\) (2) \(y-x\) is less than \(2\)
1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.
How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks
Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 08:24
Expert's post
kbulse wrote:
How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks
One of approaches is to check any point. Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true. We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false. In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast). _________________
Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 09:07
walker wrote:
kbulse wrote:
How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks
One of approaches is to check any point. Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true. We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false. In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast).
Hi walker,
how about this way: \(x/y>2\) => \(y>x/2\) then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?
Re: Graphic approach to problems with inequalities [#permalink]
07 Mar 2009, 10:05
Expert's post
kbulse wrote:
how about this way: \(x/y>2\) => \(y>x/2\) then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?
1) you should be careful with first operation because it is an inequality and you multiply both sides by y that can have a different sign and change a sign of the inequality.
The correct answer will be: \(x/y>2\) => \(x/y * y>2 *y\) at y>=0 and \(x/y * y<2 *y\) at y<0 ==> \(y<x/2\) at y>=0 and \(y>x/2\) at y<0 _________________
Re: Graphic approach to problems with inequalities [#permalink]
10 Mar 2009, 00:30
1
This post received KUDOS
Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region. Therefore, the first statement is sufficient to answer the question. We should be careful and check where line x-y=2 passes point P, through left side or right side. We can put x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).
Hi ! Can you please clarify the above part for me. According to me, to check for point p, if we put x=4.5 and y = 2.25, we get 4.5-2.25=2.25, which does not satisfy the inequality x-y<2 and thus lie in the false region. I am a little confused here, trying my hands at this for the first time, please correct me if I am wrong.
Re: Graphic approach to problems with inequalities [#permalink]
10 Mar 2009, 02:34
1
This post received KUDOS
Expert's post
Thanks, there is a typo here: Instead of
walker wrote:
... We can put x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).
should be: ... We can put x=4.5 into x-y=2 and find that y=2.5>2.25 (left side, line x-y=2 goes above P). or even better: ... We can put y=2.25 into x-y=2 and find that x=4.25<4.5 (left side). In other words, line x-y=2 passes y=2.25 (y-coordinate of P) early and goes above P.
I've fixed it in original post. +1 _________________
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...