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

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11 Mar 2009, 11:12

walker wrote:

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

Hello walker, could you please help me get the final answer? I am clear with the diagram but do not understand how to deduce the answer from it. intersection of two straight lines is (4.5,2.25). From first option: x-y<2, I am trying to confirm that whether intersection (4.5,2.25) validates this eqiation. => x-y<2 => y > x-2 => y > 4.5 - 2 => y > 2.5 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.....

....thank you.
_________________

If You're Not Living On The Edge, You're Taking Up Too Much Space

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

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18 Mar 2009, 20:05

Ok this might be a dumb question, but I see both graphs 4 and 5 lines passing from the true region. How can you deduce which one is correct?

Also for graph 4 "walker" said, "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."

But I see some points that lie outside, how can "all" points satisfy this?

I m in waiting list, gotta take the test again to get in. Please please can somebody help me ?

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

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20 Mar 2009, 13:11

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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.

This approach works for linear equation or can we use it for equations like X^2 or X^3 also ??

This approach works for any equations that could be drawn under GMAT time conditions. Y = X^3 + 1 is easy to draw but Y = 0.34X^3 - 342X^2 + 6X - 3 is not.
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For me this approach has one great advantage over other ones: When picture is drawn, the answer is almost obvious. Moreover, drawing is pretty straightforward process: line by line, expression by expression. Nevertheless, the more approaches you know, the more confident you are.
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Hi Walker - for those of us who are artistically challenged, do you happen to know the way to solve this algebraically?

I will try to solve it using "construct example" approach:

x/y >2, 3x+2y <18?

a) Could we construct an example when 3x+2y<18 ? We need small x and y for witch x/y >2. Let's say we have x = 1 (I mean 1.000001) and y=0.5

1/0.5 > 2 - Ok 3*1+2*0.5 <18 Ok

first statement: 1-0.5<2 Ok second statement: -0.5+1<2 Ok

So, x=1.000001 and y=0.5 is an example that satisfies both statements and answer is YES (True) that 3x + 2y < 18

b) Could we construct an example when 3x+2y>=18 ? The condition x/y > 2 says that x and y can be both positive or negative. At negative x,y 3x+2y will be always negative. So, we need look for our example among x,y positive.

first statement: x-y<2. Under this statement we can say that positive x and y must be close to each other. So, let's consider the maximum difference: x-y = 2 ---> y+2/y >2 --> y=2 and x=4 (as earlier I mean x=3.9999999 y=1.99999999). At such x and y we have 3*4+ 2*2 = 16 < 18. So, we can't construct an example for witch 3x+2y >=18 and first statement is sufficient.

second statement: y-x < 2. Under this statement we can choose x=100000 and y =1 and 3x+2y >=18. So, we can construct an example for witch 3x+2y >=18 and second statement is insufficient.

My comments: This approach depends on concrete problem and your luck to see an examples. At the same time I love graph approach as it is always straightforward.
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Babyif19, you may believe may not, but this method worked always for me. Of course, it can't work for everyone that is why people here - to find methods and tricks that work for them. You know other fast solution? Please, post it in the forum and many people will thank you.
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Re: Graphic approach to problems with inequalities [#permalink]

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12 Nov 2009, 22:59

I actually don't know other quick ways to solve it yet...and I am an engineering/math major! I am very fascinated by your approach - most of the time i try to solve problems algebraically.

I got 49 on Quant on my last GMAT exam 2 years ago but sucked on the verbal section- retaking the test on Dec. 14th. Wish me luck. So far my averages on Manhattan CATs, Preps and Peterson are about 670 (highest 700, lowest 650). Quite disappointing but I am still trying to break the hurdle. Working full time is definitely not helping. I even deactivated facebook and turned into a complete anti-social being for the past few months
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I actually don't know other quick ways to solve it yet...and I am an engineering/math major! I am very fascinated by your approach - most of the time i try to solve problems algebraically.

I got 49 on Quant on my last GMAT exam 2 years ago but sucked on the verbal section- retaking the test on Dec. 14th. Wish me luck. So far my averages on Manhattan CATs, Preps and Peterson are about 670 (highest 700, lowest 650). Quite disappointing but I am still trying to break the hurdle. Working full time is definitely not helping. I even deactivated facebook and turned into a complete anti-social being for the past few months

Define your weaknesses and spend all your time on them. It is simple but many tend to do what they like but not what they need. If you have problem in SC, thy Manhattan SC book - it is a bible for sentence correction. But anyway, good luck with exam, you are very close to 700.
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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)
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