## GMAT Prep :easier solution pls ?

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### GMAT Prep :easier solution pls ?

by himu » Mon Mar 11, 2013 8:52 pm
Thanks for your precious time !

I already found this one @https://www.beatthegmat.com/og-13-229-ho ... 11325.html
I too got this one right but both solutions here are lengthy .
Can anyone pls explain of they have a better / easier solution ?

How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?

A. 1
B. 2
C. 3
D. 4
E. 5

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by [email protected] » Mon Mar 11, 2013 10:14 pm
himu wrote:How many of the integers that satisfy the inequality (x+2)(x+3) / (x-2) >= 0 are less than 5?
I'll take the opportunity to explain the easiest and least time consuming method to solve these kind of problems which is popularly known as the Wavy Curve Method. The method may seem a bit of overkill or complex at first but trust me with little practice you will be able to solve these kind of problems (even more complex ones also) by spending less than 30 seconds.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Wavy curve method is used to find the solution set for a given inequality without actually solving the inequalities. The steps involved in this method are...
• 1. Factorize the numerator and denominator of the expression into linear factors.
2. Make the coefficients of x positive in all the linear factors.
3. Equate each linear factor to zero and find the value of x in each case. These values of x are called critical points.
4. Mark the critical points of the numerator with inked circles and the critical points of the denominator with cross on the number line.
5. Check the value of the expression for any real number greater than the right most marked number on the number line.
6. From right to left, beginning above the number line (if the value of the expression is positive in step 5, otherwise from below the number line if the value of the expression is negative in step 5), a wavy curve is drawn to pass through all the marked points so that when it passes through a critical point, the curve intersects the number line. But if the critical point is from a factor which is repeated even number of times, the curve will not intersect the number line but remain on the same side after touching it.
7. Continue drawing the curve as mentioned in step 6 to cover all the critical points and reach the extreme left on the number line.
8. Now the expression is positive whenever the curve is situated above the number line and negative whenever the curve is situated below the number line.
• 9A. If the question said expression > 0, positive parts of the wavy curve are our solution. For expression â‰¥ 0, the roots of the equation (critical point marked with inked circle) are also part of the final solution.
9B. If the question said expression < 0, positive parts of the wavy curve are our solution. For expression â‰¤ 0, the roots of the equation (critical point marked with inked circle) are also part of the final solution.
9C. The critical points marked with cross are never part of the solution as they make the expression undefined. So remember to exclude them from the final solution.
10. The union of different regions (either positive or negative) represent our final solution set.
Note #1: The wavy curve explained is NOT same as the graph of the given expression in strict sense. So, there is no need to draw a smooth curve or something. Any rough curve is fine as long as you follow the guidelines on when to cross the number line and when not to.
Note #2: In drawing the wavy curve, leave the end points floating, i.e. do not connect them to the number line as we do not know the exact positions of +âˆž or -âˆž.

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Now, let us solve the given problem following the above steps,
(x+2)(x+3)/(x-2)â‰¥ 0
• 1. Nothing to do as all the factors are linear factors.
2. Nothing to do as all the coefficients of x are positive in all the linear factors.
3. Critical points are -2, -3, and 2.
4. Mark the -2 and -3 with inked circles and 2 with cross on the number line.

5. Right most marked number on the number line is 2. Let us check the value of the expression for x = 3 ---> (3 + 2)(3 + 3)/(3 - 2) = 6*9 > 0
6. As the value of the expression in step 5 is positive, we will start drawing a wavy curve beginning above the number line as follows...

Now all the linear factors of the given expression occurs only once. Hence, while drawing the wavy curve whenever we reach a critical point, we will cross the number line at the critical point as follows...

7. Continue drawing the curve to cover all the critical points and reach the extreme left on the number line

8. Now the expression is positive whenever the curve is situated above the number line and negative whenever the curve is situated below the number line.
• 9A. As the problem said expression â‰¥ 0, the positive parts of the wavy curve, i.e. values between -3 and -2 and the values between 2 and positive infinity are our solution. Also the roots of the equation (critical point marked with inked circle), i.e. -3 and -2 are also part of the final solution.
9C. The critical points marked with cross, i.e. 2 are not part of the solution as they make the expression undefined.
10. Hence, our final solution is -3 â‰¤ x â‰¤ -2 and x > 2.
Now, the question asked how many integers less than satisfy the inequality.
Hence, we need to find out the number of integral values of of x less than 5.
Clearly the possible values are -3, -2, 3 and 4.
Hence, 4 integers.

Note : This method may seem complex and confusing to someone but this is a very good method to solve complex inequality problems in a matter of seconds. All you need is a little practice. I will post a few more examples with detailed solution using this method if anyone needs more clarification.
Anju Agarwal
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by sachin_yadav » Wed Apr 10, 2013 11:12 pm
Awesome explanation.

Thanks Anju.
Not at all complex and easy to understand. The only problem i face when i am not able to make factors such as this one xÂ² - x + 3, but i got your explanation in another post.
https://www.beatthegmat.com/inequality-d ... 16764.html

Can you please submit few more examples with the detailed explanation as you mentioned in your previous post. I am sure those will be helpful and good for practice.

Regards
Sachin
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### is looking for more examples

by [email protected] » Thu Apr 11, 2013 10:40 am
sachin_yadav wrote:Can you please submit few more examples with the detailed explanation as you mentioned in your previous post. I am sure those will be helpful and good for practice.
Here is another one which is a bit more complex but no inequality is that complex if you use this method.

(xâ�´ - xÂ³ - 2xÂ²)/(xÂ² + 3x - 4) > 0

1. Factorize the numerator and denominator of the expression into linear factors.
Numerator : (xâ�´ - xÂ³ - 2xÂ²) = xÂ²(xÂ² - x - 2) = xÂ²(x + 1)(x - 2)
Denominator : (xÂ² + 3x - 4) = (x - 1)(x + 4)
3. Nothing to do as all the coefficients of x are positive in all the linear factors.
3. Critical points are :
• For numerator : 0, -1, and 2
For denominator : 1 and -4
4. Mark 0, -1 and 2 with inked circles and 1 and -4 with cross on the number line.

5. Right most marked number on the number line is 2. Let us check the value of the expression for x = 3 ---> 3Â²(3 + 1)(3 - 2)/[(3 - 1)(3 + 4)] = 36/14 > 0
6. As the value of the expression in step 5 is positive, we will start drawing a wavy curve beginning above the number line as follows...

Now all the linear factors of the given expression except xÂ² occurs only once. Hence, while drawing the wavy curve, we won't cross the number line at 0, but we'll touch the line and remain on the same side. However for other critical points, we will cross the number line as follows...

7. Continue drawing the curve to cover all the critical points and reach the extreme left on the number line

8. Now the expression is positive whenever the curve is situated above the number line and negative whenever the curve is situated below the number line.

9A. As the problem said expression > 0, the positive parts of the wavy curve, i.e. values less than -4, values between -1 and 0, values between 0 and 1, and values greater than 2 are our solution. Note that -1, 0, and 2 are not part of our solution as they make the expression zero.
9C. The critical points marked with cross, i.e. -4 and 1 are not part of the solution as they make the expression undefined.
10. Hence, our final solution is x < -4, -1 < x < 0, 0 < x < 1, and x > 2.

Do post if you have any doubt in any step or any problem you are facing while solving other inequalities.

If someone is looking for more examples please have look at this post >> https://www.beatthegmat.com/range-of-ine ... tml#628087
Last edited by [email protected] on Wed Apr 17, 2013 11:14 am, edited 1 time in total.
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by [email protected] » Thu Apr 11, 2013 12:18 pm
sachin_yadav wrote:The only problem i face when i am not able to make factors such as this one xÂ² - x + 3, but i got your explanation in another post.
https://www.beatthegmat.com/inequality-d ... 16764.html
Some step-by-step tips while factoring a quadratic expression.
#1. Look at the constant term. The constant term in the factors are going to be factors of this constant term (either positive or negative). Because, (x - a)(x - b) = xÂ² - (a + b)x + ab. a and b are factors of ab.
#2. Try splitting the coefficient of x in the equation with factors of the constant term.
#3. If you don't succeed in doing so, check the following :
• Let's say the expression is of the form axÂ² + bx + c, then
#3a. If bÂ² - 4ac < 0 --> It cannot be factored. And the expression will be either positive or negative for any value of x. Try logic like I have did in that post to determine the sign.
#3b. If bÂ² - 4ac > 0 --> It can be factored. If couldn't factored it in step 2, then evaluate [-b Â± âˆš(bÂ² - 4ac)]/(2a). The factors are x - [-b + âˆš(bÂ² - 4ac)]/(2a) and x - [-b - âˆš(bÂ² - 4ac)]/(2a) [Note: This is really not required for GMAT but can be used as a last resort if you are not able to factor the equation.]
Hope that helps.
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by sachin_yadav » Sun Apr 14, 2013 1:51 am
[email protected] wrote:5. Right most marked number on the number line is 2. Let us check the value of the expression for x = 3 ---> 3Â²(3 + 1)(3 - 2)/[(3 - 1)(3 + 4)] = 36/14 > 0
I am not quite clear with this above step. I have been drawing all the curves beginning above the number line and selecting answers based on the inequality signs. Can you please elaborate more on this step.
[email protected] wrote:#3a. If bÂ² - 4ac < 0 --> It cannot be factored. And the expression will be either positive or negative for any value of x. Try logic like I have did in that post to determine the sign.
xÂ² - x + 3
With bÂ² - 4ac < 0, i get -11 < 0
and then i follow this step
xÂ² - x + 3 is always positive as

for x > 1, xÂ² > x --> (xÂ² - x) > 0 ---> (xÂ² - x + 3) > 0
for x < 1, -1 < (xÂ² - x) < 0 ---> (xÂ² - x + 3) > 0
For x = 0, (xÂ² - x + 3) = 3 > 0
So our effective inequality is 1/[x(x - 3)] â‰¤ 0 (complete question [xÂ² - x + 3]/[x(x - 3)] â‰¤ 0 )

Sachin
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by [email protected] » Tue Apr 23, 2013 2:46 am
[email protected] wrote:5. Right most marked number on the number line is 2. Let us check the value of the expression for x = 3 ---> 3Â²(3 + 1)(3 - 2)/[(3 - 1)(3 + 4)] = 36/14 > 0
I am not quite clear with this above step. I have been drawing all the curves beginning above the number line and selecting answers based on the inequality signs. Can you please elaborate more on this step.
Sorry for my late response.

Have a look at point #6 in my original post.
"From right to left, beginning above the number line (if the value of the expression is positive in step 5, otherwise from below the number line if the value of the expression is negative in step 5), a wavy curve is drawn..."

So, from where to start drawing the curve, above or below the number line, depends upon the sign of the expression for any real number greater than the right most marked number on the number line.

However, this is not relevant if you have followed step 2, i.e. all the coefficients of x are made positive. In that case you will always get a positive value of the expression in step 5 and the curve will start from above the number line.

sachin_yadav wrote:xÂ² - x + 3
With bÂ² - 4ac < 0, i get -11 < 0
and then i follow this step
xÂ² - x + 3 is always positive as

for |x| > 1, xÂ² > x --> (xÂ² - x) > 0 ---> (xÂ² - x + 3) > 0
for |x| < 1, -1 < (xÂ² - x) < 0 ---> (xÂ² - x + 3) > 0
For x = 0, (xÂ² - x + 3) = 3 > 0

So our effective inequality is 1/[x(x - 3)] â‰¤ 0 (complete question [xÂ² - x + 3]/[x(x - 3)] â‰¤ 0 )
Except for the small mistake corrected in blue, your analysis is fine.
But the bold part is redundant as you don't have to show anyone why (xÂ² - x + 3) is always positive. In the other post I did so to to explain the concept.

But once you get bÂ² - 4ac < 0, you know that the expression is either positive or negative for all values of x. To determine the sign just plug x = 0 or x = 1. Whatever sign you'll get, the expression is always of that sign.

In this case, for xÂ² - x + 3, bÂ² - 4ac = -11 < 0
So, xÂ² - x + 3 is either positive or negative for any value of x.
For x = 0, xÂ² - x + 3 = 3 > 0
So, xÂ² - x + 3 is always positive.

Hope that helps.
Anju Agarwal
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Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by jainpiyushjain » Wed May 22, 2013 9:58 am
[email protected] wrote: 9B. If the question said expression < 0, positive parts of the wavy curve are our solution. For expression â‰¤ 0, the roots of the equation (critical point marked with inked circle) are also part of the final solution.
Dear Anju
I think there is an error in your post quoted above. I believe it should be that when the expression is < 0, our solution is less than zero region.

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by Ishanvs » Tue Jan 27, 2015 8:48 am
@anju:Can you use this method for the below question ?If yes please can you solve it
If -2x > 3y, is x negative?
(1) y > 0
(2) 2x + 5y - 20 = 0

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by GMATGuruNY » Tue Jan 27, 2015 9:06 am
Generally, the wavy curve method (also known as the CRITICAL POINT approach) is appropriate for a quadratic inequality with one variable.
The problem below involves two variables and no quadratic, so I would use a different approach.
If -2x > 3y, is x negative ?

1) y>0

2) 2x+5y-20=0
Since -2x > 3y, x is negative only if yâ‰¥0.

Question stem, rephrased:
Is yâ‰¥0?

Statement 1: y>0
SUFFICIENT.

Statement 2: 2x+5y-20=0
2x+5y-20 = 0.
5y-20 = -2x.
Substituting 5y-20 for -2x in -2x>3y, we get:
5y-20>3y
2y>20
y>10.
SUFFICIENT.

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by [email protected] » Tue Jan 27, 2015 11:34 am
Ishanvs wrote: If -2x > 3y, is x negative?

(1) y > 0
(2) 2x + 5y - 20 = 0
Here's another approach:

Target question: Is x negative?

Given: -2x > 3y

Statement 1: y > 0
In other words, y is POSITIVE
This means that 3y is POSITIVE
It is given that -2x > 3y
Since 3y is POSITIVE, we can write: -2x > SOME POSITIVE #
If -2x is greater than SOME POSITIVE #, we know that -2x is POSITIVE
If -2x is POSITIVE, then x must be negative
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: 2x + 5y - 20 = 0
IMPORTANT: It is given that -2x > 3y
So, let's take 2x + 5y - 20 = 0 and rewrite it as 5y - 20 = -2x [I have isolated -2x, just like we have in the GIVEN information]
Now, we'll take -2x > 3y, and replace -2x with 5y - 20 to get: 5y - 20 > 3y
Subtract 3y from both sides: 2y - 20 > 0
Add 20 to both sides: 2y > 20
Solve: y > 10
This means that y is POSITIVE
We already saw in statement 1, that when y is positive, x must be negative
Since we can answer the target question with certainty, statement 2 is SUFFICIENT