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

This topic has expert replies
Junior | Next Rank: 30 Posts
Posts: 10
Joined: Wed Apr 18, 2012 8:03 am
Thanked: 1 times

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.

User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

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!
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course

Junior | Next Rank: 30 Posts
Posts: 10
Joined: Wed Apr 18, 2012 8:03 am
Thanked: 1 times

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.

User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

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
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Legendary Member
Posts: 641
Joined: Tue Feb 14, 2012 3:52 pm
Thanked: 11 times
Followed by:8 members

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

Senior | Next Rank: 100 Posts
Posts: 44
Joined: Thu Jan 03, 2013 10:30 pm
Thanked: 4 times

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

Newbie | Next Rank: 10 Posts
Posts: 1
Joined: Wed Dec 31, 2014 6:26 am

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

User avatar
Legendary Member
Posts: 2131
Joined: Mon Feb 03, 2014 9:26 am
Location: https://martymurraycoaching.com/
Thanked: 955 times
Followed by:140 members
GMAT Score:800

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

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

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.