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\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.

Re: Graphic approach to problems with inequalities [#permalink]
27 Oct 2010, 22:24

Apologies for asking such a basic question - I'm afraid i haven't yet figured out how to draw these graphs accurately in the actual exam- I've heard you aren't permitted to take a ruler into the center.

How do you draw the figure for eg. x-y < 2 accurately enough to see that the equation passes through point P in your diag. ?

Re: Graphic approach to problems with inequalities [#permalink]
28 Oct 2010, 03:24

Expert's post

akshey2021 wrote:

Apologies for asking such a basic question - I'm afraid i haven't yet figured out how to draw these graphs accurately in the actual exam- I've heard you aren't permitted to take a ruler into the center.

How do you draw the figure for eg. x-y < 2 accurately enough to see that the equation passes through point P in your diag. ?

Yes, you can't use a ruler. In many cases even if you don't draw accurate, you will see key points and areas. Then (like in 4. here) you can calculate for a couple of questionable points their precise positions. _________________

Re: Graphic approach to problems with inequalities [#permalink]
17 Jan 2011, 10:16

I m not able to understand here one thing at while testing condition's one or two how came at conclusion that it is not sufficient or sufficient ??????? pls help me its really a bouncer to me

Re: Graphic approach to problems with inequalities [#permalink]
25 May 2011, 11:56

jsahni123 wrote:

Hi,

Maybe some help for these two questions (graphical method)

i) Is x.|y|>y^2 ?

a)x>y b)y>0

ii) If x and y are non-zero integers and |x| + |y| = 32 what is xy?

a)-4x -12y = 0 b)|x| - |y| = 16

Thanks

Personally I think that if you an inequality with an absolute value, it is probably easier to just do it without the graphical method. In my opinion, if you really did want to graph an absolute value, I think you would have to keep in mind that |x| (where x is any value) is equal to sq root ((x)^2). With this being stated, I would rewrite the inequalities to reflect this and graph from there.

As an example, if I were to calculate the first question, this would be my approach:

1. Is x.|y|>y^2 ?

i. x>y ii. y>0

In order for the inequality x|y| > y^2 to be true, either (a) y is -ve, and x is +ve or (b) both x and y are +ve. If (a) is the case, |x|>|y|. If (b) is the case, x>y. For an answer A, B, C, D, the statements (i) and (ii) must satisfy or not satisfy both statements (a) and (b). If statements (i) and (ii) satisfy only (a) OR (b), and not both, then the answer is E.

Statement (i): x>y. This is always true for case (b), but not (a). If y is -5 and x is -4, x>y, this violates the requirement that |x|>|y|, and the inequality is not true. Therefore not sufficient.

Statement (ii): Not sufficient because this does not tell us anything about x.

The answer would be C - x>y>0, as this satisfies the requirements we set out in (a) and (b).

Re: Graphic approach to problems with inequalities [#permalink]
02 Jul 2011, 03:55

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's all Regards, Serg a.k.a. Walker

Attachment:

The attachment tarek99.png is no longer available

i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y and followed nach0's rules... i got this graph...what am i missing here? whats wrong...?

Attachments

graph.png [ 12.14 KiB | Viewed 1914 times ]

_________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

Re: Graphic approach to problems with inequalities [#permalink]
02 Jul 2011, 05:28

1

This post received KUDOS

Expert's post

fivedaysleft wrote:

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. _________________

Re: Graphic approach to problems with inequalities [#permalink]
02 Jul 2011, 10:48

so thats what i was missing! thanks a lot! _________________

It matters not how strait the gate, How charged with punishments the scroll, I am the master of my fate : I am the captain of my soul. ~ William Ernest Henley

\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 ?

Re: Graphic approach to problems with inequalities [#permalink]
28 Jul 2011, 23:16

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.

[/url]

Hello Walker and everyone I might sound really dumb here but just wanted to clarify a doubt as to WHY NOT any point in quadrant 1 and quadrant 3 relating to (X,Y) will give x/y>2. How did you calculate that the points "x/y>2 lies between line x/y=2 and x-axis." I understood that both need to be negative or both the variables need to be Positive. But these values seem to be satisfied by points in whole of quadrant 1 and quadrant 3

Re: Graphic approach to problems with inequalities [#permalink]
23 Sep 2011, 16:33

Is x > y?

(1) > y

(2) x3 > y

can some one solve this non linear which has x3 and sqrt (x) with graph method. i need to decide on choosing on the graph method finally for my test. please help me..

if this is not feasible please suggest alternative options