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:

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.

\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive.

In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)).

So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)?

(1) \(x-y<2\).

Subtract inequalities \(x>2y\) and \(x-y<2\) (we can do this as signs are in opposite direction) --> \(x-(x-y)>2y-2\) --> \(y<2\).

Now add inequalities \(x-y<2\) and \(y<2\) (we can do this as signs are in the same direction) --> \(x-y+y<2+2\) --> \(x<4\).

We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\).

Sufficient.

(2) \(y-x<2\) and \(x>2y\): \(x=3\) and \(y=1\) --> \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) --> \(3x+2y=43<18\) false.

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]

Show Tags

20 Mar 2009, 13:11

3

This post received KUDOS

1

This post was BOOKMARKED

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 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]

Show Tags

10 Mar 2009, 01:30

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.

... 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
_________________

... But at x=4.5, actual value of y is 2.25, which is, indeed, not greater than 2.5!! What should I look for from this statement?

I guess I am missing something..... .

Figure 3. We have "true" and "false" regions. What happens in remaining area we don't care as the area doesn't satisfies problem's conditions. Figure 4. Part of "true/false" regions is gray because it doesn't satisfy x-y<2 condition. Only in "color" part x-y<2 is true.

But in Figure4 we could have doubt about point P: where line x-y<2 passes P, left-above or bottom-right. In first case we will have only "true" region and in second case - "true" and "false" regions. As you correctly pointed out at x=4.5 y=2.5>2.25. So, point P is not included in "color" region (P does not satisfy x-y<2 condition) and x-y<2 passes left-above. So that, we can conclude we have only "true" region.
_________________

Why condition 1 is sufficient and condition 2 is not.As to me both of them have some points on true region and some on false region.

Let's say you have y>2x+1. You draw line y=2x+1 and above region is TRUE, below is FALSE. If you have some doubts about that, you may check any point from regions. For example, point (1,100) is above and 100>2+1 is TRUE.

No, only second condition has points on true and false regions. All points for first condition are on true region. (see figures)
_________________

i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y .....

Your problem is a typical one for inequalities and modulus questions. You just forget to consider the case when y is negative. Here is what you should do:

1. x/y >2 2. x >2y (y>0) & x <2y (y<0) 3. solve both inequalities BUT don't forget to apply conditions (y>0 and y<0). For example, in your plot y can't be negative.

By the way, try to check out whether the answer makes sense. Moreover, sometimes it's useful think a bit about expression at the beginning. For example, x/y > 2 only if x and y have the same sign. So, your last graph fails to pass this test.

Actually, that is why I used graphic approach as it allows to avoid such kind of mistakes.
_________________

\(\frac{x}{y}>2\) tells us that \(x\) and \(y\) are either both positive or both negative, which means that all points \((x,y)\) satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality \(3x+2y<18\) is always true, so we should check only for I quadrant, or when both \(x\) and \(y\) are positive.

In I quadrant, as \(x\) and \(y\) are both positive, we can rewrite \(\frac{x}{y}>2\) as \(x>2y>0\) (remember \(x>0\) and \(y>0\)).

So basically question becomes: If \(x>0\) and \(y>0\) and \(x>2y>0\), is \(3x+2y<18\)?

(1) \(x-y<2\).

Subtract inequalities \(x>2y\) and \(x-y<2\) (we can do this as signs are in opposite direction) --> \(x-(x-y)>2y-2\) --> \(y<2\).

Now add inequalities \(x-y<2\) and \(y<2\) (we can do this as signs are in the same direction) --> \(x-y+y<2+2\) --> \(x<4\).

We got \(y<2\) and \(x<4\). If we take maximum values \(x=4\) and \(y=2\) and substitute in \(3x+2y<18\), we'll get \(12+4=16<18\).

Sufficient.

(2) \(y-x<2\) and \(x>2y\): \(x=3\) and \(y=1\) --> \(3x+2y=11<18\) true. \(x=11\) and \(y=5\) --> \(3x+2y=43<18\) false.

Not sufficient.

Answer: A.

Sorry for being naive. Can you please to me the second part of the solution when you conclude it as "Not sufficient". How did you get the numbers 3, 1, 11,5 ?

Those were helpful indeed . Thanks. Your Quarter Wit, Quarter Wisdom blog is an eye opener. However i feel that question could have been solved faster by an algebric approach, but only because we were provided that both x and y are positive. In absence of such constraints i think the graphical approach will be faster. Another query though. I found this mentioned at the end of the post. I hope you have come to appreciate the wide range of applicability of graphs. Next time, I will introduce a graphical way of working with Modulus and Inequalities. Could I get a link for that. Thnx in advance And i faintly remember learning some tricks about plotting graphs in my high school. Something to the tune of-"Graph of y=kx can be drawn by expanding the graph of y=x by k times and y=K+x can be drawn by shifting y=x by k units(to the left or right?? ) Do i need to revisit those for the GMAT??

Sure, you can use either method - it depends on what you are more comfortable with. I find working with equations/inequalities way too cumbersome and have developed an ease with graphs (with practice of course). I prefer to take a holistic view and figure out the answer since GMAT questions are basically logic based, (and hence the ample use of graphs). You might find that graphs slow you down initially but with practice, they can save you a lot of time. Anyway, both the methods work perfectly fine so choose whichever you like more. I have many posts on Mods and inequalities peppered in-between other posts on my blog. I would suggest you to start from the bottom of the last page and go upwards checking out the posts that catch your fancy: http://www.veritasprep.com/blog/categor ... om/page/3/

They are not essential to know if you plan to use algebra for most questions. They can be quite helpful if you plan on working out the questions using the holistic approaches.
_________________

Re: Graphic approach to problems with inequalities [#permalink]

Show Tags

30 Aug 2013, 06:26

1

This post received KUDOS

For most of the cases - yes (for example for lines or parabolas).[/quote] One more thing in questions involving greater or less than, a solutions can be infinite.

In the attached image when y > x, and x < 1, we can have infinite solutions, right ? if y < x , x < 1 again infinte solutions, right ?

So can we safely say if there are 'two' inequalities having less than or greater than condition, we can never have finite solutions , or number of solutions is indeterminate ?

Can you please tell me how do you know which region to shade? I mean , to the left? or right? above or below the x axis?

How do you find out

So you draw the line showing the equation represented by the inequality. How do you decide which side of the line does the inequality represent. Usually, you can do that by plugging in (0, 0) in the equation. The point (0, 0) will lie on one side of the line. Put x = 0 and y = 0 in your inequality. If it holds, it means the inequality holds for point (0, 0) and hence will hold for that entire side of the line. So you shade the side where (0, 0) lies. If the inequality does not hold when you put (0, 0), it means (0, 0) is not a solution of the inequality and hence the inequality holds for the opposite side so you shade the opposite side.

If the line passes through (0, 0) try any other point which obviously lies on one side of the line.
_________________

I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that y-x=2 never pass the red area. How am I supposed to interpret that graph?

All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient" - but the last figure doesn't go through red zone either, yet still insufficient?

Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either?

When the thread gets to the 6th page with, presumably, multiple solutions, you need to quote the post to which you are referring or give its link.

Figure 5. Original post. The line y-x=2 never goes through the red zone - yet "insufficient"

It has nothing to do with whether it passes from the red region or not.

You are given that y - x < 2. When you draw y - x = 2 in figure 5, you get a line. But what is y - x < 2? It is the area to the right of the line y - x = 2. You can find this by putting in (0, 0) in the inequality y - x < 2. You get 0 - 0 < 2 that is 0 < 2 which is true. Since (0, 0) lies to the right of the line and satisfies the inequality, it means the right of the line is y - x < 2 and the left of the line is y - x > 2.

So you are given that points to the right of y - x = 2 are feasible. Look at all the points to the right of y - x= 2 and that satisfy x/y > 2 (i.e. the green and red regions). Note that of all the points to the right of y - x = 2, some lie in the green region and some lie in the red region (ignoring those that lie in neither region). The ones that lie in green region are those for which 3x+2y < 18. For those that lie in the red region, 3x + 2y > 18. So can we say whether 3x + 2y is less than 18? No. Because some points that satisfy y - x > 2 lie in the reg region and some lie in the green region.

When will an inequality be sufficient. It will be sufficient when the region it points to has points lying only in one region - either red or green.

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...