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OG 13 #229 How many of the integers...

by wied81 » Thu May 03, 2012 12:47 pm
229. 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

OA: D

The OG does some funky number line diagrams in explaining this problem and I was wondering if there's another way. Just off the top of my head, to me, it seems like only solutions would be 3 and 4. The number 2 results in an undefined 0 in the denominator and the number 1 results in a negative denominator with a positive numerator, which would make the fraction negative and thus less than 0.

Any advice would be appreciated.

Thanks.
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by Stuart@KaplanGMAT » Thu May 03, 2012 1:51 pm
wied81 wrote:229. 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

OA: D

The OG does some funky number line diagrams in explaining this problem and I was wondering if there's another way. Just off the top of my head, to me, it seems like only solutions would be 3 and 4. The number 2 results in an undefined 0 in the denominator and the number 1 results in a negative denominator with a positive numerator, which would make the fraction negative and thus less than 0.

Any advice would be appreciated.

Thanks.
Here's the #1 thing to remember when using the OG: it's a great source of questions, but a horrible source of explanations. Especially for math, OG explanations are rarely the most efficient way to solve problems.

Since the biggest answer given is 5, plug 'n play is a great alternative way to solve this problem if you don't feel like doing all the algebra. We only care about integers less than 5 - and it's important to note that nowhere does it say we're limited to positive numbers - so let's start with 0, since it's easy to plug in.

If x=0, then we get:

2*3/-2 >= 0

Without calculation, we see that the left side is negative, so 0 doesn't work.
If x=1, we're also going to get +/-, so 1 is out.
If x=2, the denominator is 0, so that's right out.
If x=3, we have +/+.. we can quickly see that x=4 also gives +/+, so 3 and 4 both work.

A common mistake would be stopping without checking to see if any negative values also work.

If x=-1, then we have +/-.. no go.
If x=-2, then we have 0/- = 0... this works!
If x=-3, then we have 0/-... this works!
If x=-4, then we have +/-.. no go. We can also see that smaller values of x will lead to the same result.

So, the only values that fit are -3, -2, 3 and 4... choose D!

* * *

We could also use some math logic to solve. Let's break down into two cases:

1) (x+2)(x+3) / (x-2) = 0

and

2) (x+2)(x+3) / (x-2) > 0

For Case 1, the left side equals 0 if the numerator is 0, i.e. if x=-2 or -3.

For Case 2, the left side is greater than 0 if the numerator and denominator share the same sign (i.e. both - or both +). If x>2, then both will be positive... that gives us x=3 and x=4 as solutions. We can play around, but we'll soon see that there are no cases in which the top and bottom can both be negative, so there are no other solutions for case 2.

Each of cases 1 and 2 gives 2 possible values, so there are 4 solutions: choose D!
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by wied81 » Fri May 04, 2012 12:05 pm
Ahh..forgot to plug-in negative values. Silly me, I should never assume they are speaking about positive integers. Thanks for the breakdown Stuart.
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by GMATGuruNY » Fri May 04, 2012 1:22 pm
wied81 wrote:229. 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

OA: D

The OG does some funky number line diagrams in explaining this problem and I was wondering if there's another way. Just off the top of my head, to me, it seems like only solutions would be 3 and 4. The number 2 results in an undefined 0 in the denominator and the number 1 results in a negative denominator with a positive numerator, which would make the fraction negative and thus less than 0.

Any advice would be appreciated.

Thanks.
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 range where (x+2)(x+3) / x-2 > 0, try one integer value to the left and right of each critical point.

x < -3:
Plugging x=-4 into (x+2)(x+3) / x-2 > 0, we get:
(-4+2)(-4+3)/(-4-2) > 0
2/-6 > 0.
Doesn't work.
This means that no value less than -3 will work.

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

-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 between -2 and 2 will work.

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 greater than 2 will work.
There are only two integer values between 2 and 5:
3 and 4.

Thus, there are four integer values less than 5 that satisfy the inequality: -3, -2, 3 and 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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by gmattesttaker2 » Sun Mar 10, 2013 4:22 pm
Stuart Kovinsky wrote:
wied81 wrote:229. 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

OA: D

The OG does some funky number line diagrams in explaining this problem and I was wondering if there's another way. Just off the top of my head, to me, it seems like only solutions would be 3 and 4. The number 2 results in an undefined 0 in the denominator and the number 1 results in a negative denominator with a positive numerator, which would make the fraction negative and thus less than 0.

Any advice would be appreciated.

Thanks.
Here's the #1 thing to remember when using the OG: it's a great source of questions, but a horrible source of explanations. Especially for math, OG explanations are rarely the most efficient way to solve problems.

Since the biggest answer given is 5, plug 'n play is a great alternative way to solve this problem if you don't feel like doing all the algebra. We only care about integers less than 5 - and it's important to note that nowhere does it say we're limited to positive numbers - so let's start with 0, since it's easy to plug in.

If x=0, then we get:

2*3/-2 >= 0

Without calculation, we see that the left side is negative, so 0 doesn't work.
If x=1, we're also going to get +/-, so 1 is out.
If x=2, the denominator is 0, so that's right out.
If x=3, we have +/+.. we can quickly see that x=4 also gives +/+, so 3 and 4 both work.

A common mistake would be stopping without checking to see if any negative values also work.

If x=-1, then we have +/-.. no go.
If x=-2, then we have 0/- = 0... this works!
If x=-3, then we have 0/-... this works!
If x=-4, then we have +/-.. no go. We can also see that smaller values of x will lead to the same result.

So, the only values that fit are -3, -2, 3 and 4... choose D!

* * *

We could also use some math logic to solve. Let's break down into two cases:

1) (x+2)(x+3) / (x-2) = 0

and

2) (x+2)(x+3) / (x-2) > 0

For Case 1, the left side equals 0 if the numerator is 0, i.e. if x=-2 or -3.

For Case 2, the left side is greater than 0 if the numerator and denominator share the same sign (i.e. both - or both +). If x>2, then both will be positive... that gives us x=3 and x=4 as solutions. We can play around, but we'll soon see that there are no cases in which the top and bottom can both be negative, so there are no other solutions for case 2.

Each of cases 1 and 2 gives 2 possible values, so there are 4 solutions: choose D!
Hello Stuart,

Thanks a lot for the detailed explanation. I had one question though. Sorry if it's too trivial.

For x = 2:

We have (2 + 2)(2 + 3)/(2 - 2) = 4.5/0 = infinity >= 0

I was just wondering why we are not taking x = 2 as a valid integer then. Sorry again if this question is too basic. Thanks a lot for your help.

Best Regards,
Sri
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by paresh_patil » Mon Mar 11, 2013 2:18 am
Take a look at the denominator which is x-2
therefore, when we have x=2 the inequality will = 0
anything less than 2 will not satisfy the inequality.

So we have 4 integers less than 5 which satisfy the inequality.
the answer is D
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by thorin_oakenshield » Sat Jan 03, 2015 10:18 am
paresh_patil wrote:Take a look at the denominator which is x-2
therefore, when we have x=2 the inequality will = 0
anything less than 2 will not satisfy the inequality.

So we have 4 integers less than 5 which satisfy the inequality.
the answer is D
Hi Paresh,

How did you come up with 4 integers between less than 5?

TO
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by MartyMurray » Sat Jan 03, 2015 1:43 pm
thorin_oakenshield wrote:
paresh_patil wrote:Take a look at the denominator which is x-2
therefore, when we have x=2 the inequality will = 0
anything less than 2 will not satisfy the inequality.

So we have 4 integers less than 5 which satisfy the inequality.
the answer is D
Hi Paresh,

How did you come up with 4 integers between less than 5?

TO
Something about smoke and mirrors, or similar to a stopped clock being right twice a day, looks like to me anyway.

A couple of specific errors include the following. If x=2 the left side does not equal 0. Actually it is undefined or possibly equal to infinity. Also, as opposed to what Paresh said, the inequality can be satisfied with x values less than 2.

Hmm, alternatively maybe Paresh used acid trip math. Maybe if you smoke or drop the same stuff Paresh did you can get the answer the same way. LOL
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by Matt@VeritasPrep » Mon Jan 05, 2015 11:00 am
thorin_oakenshield wrote:
paresh_patil wrote:Take a look at the denominator which is x-2
therefore, when we have x=2 the inequality will = 0
anything less than 2 will not satisfy the inequality.

So we have 4 integers less than 5 which satisfy the inequality.
the answer is D
Hi Paresh,

How did you come up with 4 integers between less than 5?

TO
Incorrectly, unfortunately. If I follow his logic, x = 2 is a solution, since it "results" in 0. Since anything greater than 2 somehow results in this being negative, 2, 3, 4, and 5 are solutions. (Even though 5 is explicitly NOT a possibility.)

Needless to say, none of those statements are correct!

Another approach to this problem:

(x + 3)(x + 2)/(x - 2) ≥ 0

If either term in the numerator equals 0, then the left hand side equals 0. So (x + 3) = 0 and (x + 2) = 0 are both possibilities, which gives us TWO integer solutions, x = -3 and x = -2.

Now let's see what's happening with our inequality. If (x + 3)(x + 2) is positive AND (x - 2) is positive, then we'll have Pos*Pos / Pos, which is Positive. If x - 2 is positive, obviously x+2 and x+3 will also be positive, so anything that gives us x - 2 > 0 is a solution. This works for x = 3 and x = 4, so we have two more solutions.

Any other integers will mess us up somehow. If x = 2, then the denominator is 0 and we have an undefined number. If x = -1, 0, or 1, then we have (x - 2) is negative but (x + 2) and (x + 3) are positive, which gives us Pos*Pos/Neg, which is NOT positive.

If x is LESS THAN -3, then all of our terms are negative, which gives us Neg * Neg / Neg, or Pos/Neg, which is also not positive.

So we only have four possibilities: x = -3, x = -2, x = 3, and x = 4.
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