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

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29 Aug 2013, 21:57

Attachment:

tarek99.png

First of all, hats off, I have tried myself building such approaches but could never get a full proof version that can cover all possible types, urs do. I have one doubt,

using (i) , x-y<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region the region marked grey will come in < condition using(ii) y-x<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region which looks ok to me.

Point out if I am interpreting it wrongly. Thanks.

First of all, hats off, I have tried myself building such approaches but could never get a full proof version that can cover all possible types, urs do. I have one doubt,

using (i) , x-y<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region the region marked grey will come in < condition using(ii) y-x<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region which looks ok to me.

Point out if I am interpreting it wrongly. Thanks.

Rewrite x-y<2 as y>x-2. True region is ABOVE (because of ">" sign) the line y=x-2. Rewrite y-x<2 as y<x+2. True region is BELOW (because of "<" sign) the line y=x+2.

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

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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 ?

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

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13 Sep 2013, 11:45

I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region. As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong. Thanks.

I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region. As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong. Thanks.

y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).
_________________

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

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15 Sep 2013, 10:27

Bunuel wrote:

ygdrasil24 wrote:

I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region. As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong. Thanks.

y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).

OK. If you refer to the link graphic-approach-to-problems-with-inequalities-68037.html by Walker, how is the graph x/y >2 drawn here? y < x/2 so graph should be all region less than y = x/2. I didnot get how this graph is drawn.

I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region. As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong. Thanks.

y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).

OK. If you refer to the link graphic-approach-to-problems-with-inequalities-68037.html by Walker, how is the graph x/y >2 drawn here? y < x/2 so graph should be all region less than y = x/2. I didnot get how this graph is drawn.

It's not a linear equation. So, the approach is different:

walker wrote:

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.

Attachment:

MSP12871igh5183c7965eh7000022622h2c27df8334.gif [ 2.71 KiB | Viewed 2411 times ]

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

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16 Sep 2013, 00:18

It's not a linear equation. So, the approach is different: Allow me for my misunderstanding since I am not in best of touch with QA, but this is a linear equation, If not loci cannot be a straight line.

"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.

Attachment:

MSP12871igh5183c7965eh7000022622h2c27df8334.gif

I am more keen to learn this graphic approach, I was going all well until I saw this graph and honestly I am not getting what you meant by "boundary". The only logic I can pick is once you have the region, you have to check for values against which the region should suffice. Here its Ist and 3rd quandrant. If we draw region for y < x. It would be all region below y =x , right ? or similarly for y < x/2 ? The only thing that changed is the slope from 1 to 1/2 , so line becomes more flat. In this region only Quadrant 1, 3,4 comes, since Quadrant 4 cannot be a right region as we need both x,y>0 or x,y<0, we are left with Quadrant 1 and 3. Now how come x =0 , limits this region and more so the shaded portion on the 3rd quadrant shows region y > x/2. Can you explain ?

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

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25 Sep 2013, 05:07

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.

That is a great tutorial, its the same approach i use to solve many inequality questions .. its not written anywhere how you choose which area to shade, I am assuming that everyone knows but that's a pretty big assumption .. so for people who don't know how to choose which region to shade .. here it is ..

I'll use the same inequalities that walker used:

3x+2y<18 .. you take origin as reference point and put its value in inequality i.e. x=0 and y=0 .. equation will be 0<18 which is true, hence we will shade the portion that contains origin .. pretty simple, eh ?

for the lines that pass through origin, you can take any point on X or Y axis as reference point and shade the portion in which that point lies if inequality is satisfied and shade the other region in which point doesn't lie if inequality is not satisfied ..

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

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15 Nov 2013, 04:28

Damn. I remember using this approach to solve inequalities in 8th grade. All those faint memories coming back. Great approach. Far better than the brute force number crunching approach that I have thus far followed.

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

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08 May 2014, 10:48

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

i still am not understanding which area to shade... please help... i am very confused.
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

\(\frac{x}{y}>2\) 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.

Hi Bunnel

I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient. when both x & y are negative then we cannot have x>2y right? so where are we considering this...
_________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

\(\frac{x}{y}>2\) 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.

Hi Bunnel

I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient. when both x & y are negative then we cannot have x>2y right? so where are we considering this...

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. _________________

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

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24 Jun 2014, 04:24

Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer. Please mention how I should approach in equalities.....

Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer. Please mention how I should approach in equalities.....

Graphic approach is not a silver bullet for inequality questions. Yes, it's handy for some of them but certainly not for all of them.

Hope the links below help to attack them more efficiently:

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