OG 13 #229

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OG 13 #229

by jeph86 » Sat Feb 27, 2016 12:57 pm

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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

i see the book has D as the answer and i had B. I don't understand the book explanation

thanks

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by [email protected] » Sat Feb 27, 2016 9:45 pm
Hi jeph86,,

This question can actually be solved with a bit of "brute force"; I'm going to give you some hints, then let you try it out to see how quickly you can solve it.

The question asks how many INTEGERS that are LESS THAN 5 fit the given inequality? From the answer choices, there is at least 1 and no more than 5 possible answers. Given the limitation that ALL of the answers are less than 5, how long would it take you to find them all?

Here's a suggestion:
-What happens when you plug in 4?
-What happens when you plug in 3?
Can you spot a pattern when you do this work?
Etc.

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by [email protected] » Sat Feb 27, 2016 9:58 pm
x = -2 and x = -3 both generate a result of 0 when plugged into the formula. So they both work, and the task becomes to determine whether there are any more values that work.

I would just start plugging in integers to see what happens.

Plugging in 4 generates (6)(7)/2 = 21. So that works.

Plugging in 3 generates (5)(6)/1 = 30. So that works, and we are up to 4 integers less than 5 that work.

2 won't work because x = 2 makes the denominator 0.

Any x such that -2 < x < 2 does not work, because you end up with a positive numerator divided by a negative denominator, which will always generate a negative number.

If x < -3, the values of x + 2, x + 3 and x - 2 are all negative. So the expression will always generate a negative number when x < -3.

So there are 4 integers that work, -2, -3, 3 and 4.

The correct answer is D.
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by chetan.sharma » Sun Apr 17, 2016 10:01 pm
jeph86 wrote: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

i see the book has D as the answer and i had B. I don't understand the book explanation

thanks
Hi,

If you were to do it algebrically..
(x+2)(x+3)/x-2 >=0..

lets first take the values satisfying 0..
(x+2)(x+3)/x-2 =0..
clearly (x+2)(x+3) should be 0.. x= -2 or -3..

lets see for > now
(x+2)(x+3)/x-2 >0..
two cases possible
1) Both (x+2)(x+3)>0 and x-2 >0..
so x-2>0 gives us x>2..
and (x+2)(x+3)>0 for all values of x>-2 and x<-3..
but since x>2 and x<5 is given, values which fit in are x=3 and x=4..

2) Both (x+2)(x+3)<0 and x-2 <0..
so x-2<0 gives us x<2..
and (x+2)(x+3)<0 for all values of x between 2 and 3, but x is an integer so NO values are possible..

so the values which fit in are -3, -2, 3, 4--------- 4 values
D

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by 800_or_bust » Mon Apr 18, 2016 5:49 am
jeph86 wrote: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

i see the book has D as the answer and i had B. I don't understand the book explanation

thanks
The key is to think of what we are solving for. In order for the value to be greater than or equal to zero, there are two possible cases - the sign of the numerator and denominator are the same (in which case, the function is positive), or the numerator evaluates to 0. Once you identify this, you can quickly rule out all values such that x is less than or equal to -4, because for those values the numerator will be positive and the denominator negative. At x = -2 and x = -3, the numerator will equal zero, so both of these are valid. There is a discontinuity at x = 2, since the function is undefined for this value (the denominator cannot equal zero), so we can toss this. That leaves -1, 0, 1, 3, and 4. Keeping in mind what I said in the first line, we can see for all x such that x is greater than or equal to 3, the numerator and denominator will both be positive. Therefore, 3 and 4 are valid solutions. -1, 0, and 1 can quickly be disproved mentally.
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by ceilidh.erickson » Mon Apr 18, 2016 12:36 pm
The other posters have provided good explanations. I just want to add something about questioning your assumptions...

If you thought that there were 2 explanations instead of 4, that suggests to be that you overlooked 1 of 2 things:

1) Perhaps you overlooked the "less than or equal to" and only counted "less than." This is a common mistake with inclusive sets. Pay attention to which inequality sign you have!

2) You may have interpreted "less than 5" as only counting integers from 1 to 5. Remember, less than 5 could also include any negatives! It's easy to forget about negative possibilities.

If you made either of these mistakes (both very common with GMAT students), make a note of it. You should be tracking your mistakes, to see if you're repeating certain errors over and over: https://www.manhattanprep.com/gmat/blog ... -studying/
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by [email protected] » Thu Sep 29, 2016 6:13 pm
Another quick approach is to try the integers in range.

x = 4 works
x = 3 works
x = 2 doesn't
x = 1 doesn't
x = 0 doesn't
x = -1 doesn't
x = -2 works
x = -3 works
x = -4 doesn't

And anything below x = -4 will have the same problem that x = -4 does: it'll give you (neg*neg)/neg, or pos/neg, which is negative.

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by crackverbal » Thu Oct 13, 2016 2:27 am
Hi Jeph86,

Plugging in numbers is not the only way of answering this question in 2 minutes. You can make use of the concept of a 'Quadratic Inequality'.

Let me first explain how to solve a Quadratic Inequality and then explain how to apply this concept in questions similar to question 229 of the OG.

Consider the Quadratic Inequality 3x^2 - 7x + 4 < 0
Factorizing the quadratic inequation we get 3x^2 - 7x + 4 <= 0 ---> (3x - 4)(x - 1) < 0

Now 1 and 4/3 here are referred to as critical points and we place them on number line

Image

The critical points divide the number line into three regions, now we can get 3 ranges of x

i) x < 1 (all values of x when substituted in (3x - 4)(x - 1) makes the product positive)

ii) 1 < x < 4/3 (all values of x when substituted in (3x - 4)(x - 1) makes the product negative)

iii) x > 4/3 (all values of x when substituted in (3x - 4)(x - 1) makes the product positive)

At this point we should understand that for the inequality (3x-4)(x-1) < 0 to hold true, exactly one of (3x-4) and (x-1) should be negative and other one be positive. The only region that gives us a negative product is the region in between 1 and 4/3

So the range of x that satisfies the inequality 3x^2 - 7x + 4 < 0 is 1 < x < 4/3

The steps to solve a quadratic inequation are as follows:

1. Isolate the variable and always keep the variable positive.
2. Maintain the Inequation in the form ax^2 + bx + c > 0 or < 0.
3. Obtain the critical points of the Inequation.
4. Place the critical points on the number line. The number line will get divided into the three regions.
5. Mark the rightmost region with + sign, the next region with a - sign and the third region with a + sign (alternating + and - starting from the rightmost region).
6. If the Inequation is of the form ax^2 + bx + c < 0, the region having the - sign will be the solution of the given quadratic inequality.
7. If the Inequation is of the form ax^2 + bx + c > 0, the region having the + sign will be the solution of the given quadratic inequality.

This procedure also works for a cubic inequality. the only difference is that we will have three critical points instead of two and 4 regions on the number line. Here again you take the rightmost region as positive and then alternate between the signs.

Now if we have an inequality of the form (x+2)(x+3)/x-2 >=0 then we can transform the question to (x+2)(x+3)(x-2) >= 0 since the conditions that we need to check for in the original inequation will be the same as the conditions that we need to check for in the transformed inequation. The only thing we need to keep in mind here is that in the solution we cannot have x = 2 as it would make the denominator 0.

Now the critical points here are -3, -2 and 2. Placing them on the number line and taking the positive regions we get x >= 2 and -3 <= x < = -2. In the solution of x >= 2 we ignore the =2 since the value makes the denominator 0.

Image

The values of x less than 5 will be 3, 4, -3 and -2.

Keep in mind that if you have similar questions to the one explained above, just transform the question into a product and then use the number line approach to solve. Just make sure you ignore the solution which makes the denominator 0.

To learn similar strategies to solve Inequality questions you can download the free ebook from the link given below.

https://gmat.crackverbal.com/free-resour ... k-library/

Hope this helps!

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Re: OG 13 #229

by GMATGuruNY » Tue Mar 08, 2022 11:24 am
jeph86 wrote:
Sat Feb 27, 2016 12:57 pm
GMAT Official Guide

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
One approach is to determine the CRITICAL POINTS: the values where the lefthand side is EQUAL TO 0 or is UNDEFINED.
The lefthand side is equal to 0 when x=-2 and x=-3.
The lefthand side is undefined when x=2.

We already know that x=-2 and x=-3 are valid solutions because they are where (x+2)(x+3) / x-2 = 0.

To determine the ranges where (x+2)(x+3) / x-2 > 0, test one value to the left and one value to the right of each critical point.

x>2:
Plugging x=3 into (x+2)(x+3) / x-2 > 0, we get:
(3+2)(3+3)/(3-2) > 0
30 > 0.
This works.
This means that ANY VALUE in this range will work.
There are two integer values in this range that are less than 5:
3 and 4

-2<x<2:
Plugging x=0 into (x+2)(x+3) / x-2 > 0, we get:
(0+2)(0+3)/(0-2) > 0
-3 > 0.
Doesn't work.
This means that no value in this range will work

-3<x<-2:
No integer values in this range.

x < -3:
If any value in this range is valid, then EVERY VALUE in this range will be valid, resulting in an INFINITE NUMBER OF SOLUTIONS.
The answer choices indicate that the number of valid integer solutions is at most five.
Implication:
x < -3 cannot be a valid range.

Thus, there are four integer values less than 5 that satisfy the inequality:
-3, -2, 3, 4

The correct answer is D.

Other problems that I've solved with the critical point approach:

https://www.beatthegmat.com/inequality-c ... 89518.html

https://www.beatthegmat.com/knewton-q-t89317.html

https://www.beatthegmat.com/which-is-true-t89111.html
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Re: OG 13 #229

by Ian Stewart » Sat Mar 19, 2022 3:28 am
The fraction will equal zero when the numerator (x+2)(x+3) is zero, so when x = -2 or -3, for two values of x.

The numerator (x+2)(x+3) is the product of two consecutive integers. If two consecutive integers are both nonzero, they have the same sign. So the numerator will always be positive when x is not -2 or -3. Since the numerator is positive, the fraction will be positive exactly when the denominator, x-2, is positive, or when x > 2. Since x < 5, the fraction is positive when x = 3 and x = 4, so there are four integer values of x that work.
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