## Inequalities - is the way im solving correct?

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### Inequalities - is the way im solving correct?

by onesome » Sun Jul 06, 2008 10:58 am
Hi - Can someone please help in understanding the strategy to be used to solve the inequalities question mentioned below. I think i did reach the correct answer but i'm not sure if i am solving it correctly.

Question =>
If 3 < |x-5| < 7 , where x is an integer, how many possible values does x have?
A) 3
B) 4
C) 5
D) 6
E) 7

How did i solve:
Step 1: Assuming the modulus will result in +ve value :
3 < x-5 <7
8 < x < 12
Possible values = 9,10,11 , i.e. 3 different values..........(A)

Step 2: Assuming the modulus will result in -ve value :
3 < -(x-5) < 7
3 < -x+5 < 7
solving , 3-5 < -x+5-5 < 7-5
to be finally , -2 < x < 2
Possible values = -1,0,1 , i.e. 3 different values..........(B)

From (A) and (B)
Total 6 different possible values,

Kindly suggest if you think something is wrong above.

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by rs2010 » Sun Jul 06, 2008 1:02 pm
That is the correct way.

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by Ian Stewart » Sun Jul 06, 2008 1:39 pm
hemantsood wrote:That is the correct way.
I'd prefer to say "That is a correct way"!

The above solution is certainly correct, but as with most math problems, there is more than one solution. If you understand that |a - b| measures the distance between a and b on the number line, you can look at the problem as follows:

3 < |x-5| < 7

This is reallly two inequalities:

3 < |x-5|, so x is more than 3 away from 5, on the left or the right: x < 2 or x > 8.

|x-5| <7, so x is less than 7 away from 5, on the left or the right: x > -2, or x < 12.

Combining these, we have -2 < x < 2 or 8 < x < 12. If x is an integer, x can be -1, 0, 1, 9, 10 or 11. Drawing the number line makes this easier, of course.
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by rs2010 » Sun Jul 06, 2008 6:03 pm
Thanks Ian. I will remember it. Master | Next Rank: 500 Posts
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by ildude02 » Wed Jul 09, 2008 1:18 pm
Ian,

Why do we have to consider the negative value for |x -5| ? Since it's given that | x- 5| is greater than 3 and and less then 7, doesn't it mean it lies with a postive range? So, (x-5) can only be postive, implying (x-5) > 0. So, why to consider negative values for |x-5| as in x-5 < 0? As a rule of thumb, do we always need to consider -ve value as well for a variable within the MOD.. I know you use distance relation to sovle MOD's, but it would be too much for me to digest this late But I still hope you could answer this, I would really appreciate your response.

Ian Stewart wrote:
hemantsood wrote:That is the correct way.
I'd prefer to say "That is a correct way"!

The above solution is certainly correct, but as with most math problems, there is more than one solution. If you understand that |a - b| measures the distance between a and b on the number line, you can look at the problem as follows:

3 < |x-5| < 7

This is reallly two inequalities:

3 < |x-5|, so x is more than 3 away from 5, on the left or the right: x < 2 or x > 8.

|x-5| <7, so x is less than 7 away from 5, on the left or the right: x > -2, or x < 12.

Combining these, we have -2 < x < 2 or 8 < x < 12. If x is an integer, x can be -1, 0, 1, 9, 10 or 11. Drawing the number line makes this easier, of course.

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by [email protected] » Wed Jul 09, 2008 1:28 pm
My "too much math" alarm is sounding! I've always been a fan of the quick and dirty solution to GMAT questions.
If 3 < |x-5| < 7 , where x is an integer, how many possible values does x have?
All we need to do is count the number of values of x. Since we don't care what the values are, we can answer this question with minimal math.

Once we see that the "-5" is just going to shift our place on the number line (and since only integers are involved, we don't have to worry about anything wacky happening), we realize that it's irrelevant to answering the question.

Let's rephrase the question as:
If 3 < |x| < 7 , where x is an integer, how many possible values does x have?
How many integers are there between 3 and 7? 4, 5 and 6, so "three".

Because it's absolute value, we know there will be two solution sets. So, -4, -5 and -6 will also work. That's a total of 6: choose (d). Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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by Ian Stewart » Wed Jul 09, 2008 9:01 pm
Stuart Kovinsky wrote: Well, I guess we have a very different approach to the GMAT.
No, not at all- from what you say above, we have a very similar approach. In particular, this:
Stuart Kovinsky wrote: The key to GMAT success is to really understand the concepts underlying the math so that you can find creative solutions to problems.
really does capture what is essential on test day.

I'd take issue with only one point:
Stuart Kovinsky wrote: So, when I post a suggested solution to a problem on these boards, if someone else has already detailed how to do the algebra, I'm going to offer an alternative approach. You posted the textbook solution,
The textbook solution is the algebraic one. I posted a two-line solution based on the distance interpretation of absolute value- not the textbook solution, no algebra, and it takes ten seconds to do, if you understand the underlying concept. I like your solution as well- also very fast, and there's a lot of value in seeing different solutions to these problems. I posted what I did only because I saw the 'too much math' comment after my post, and I would not want people reading this thread to think there was 'too much math' involved when interpreting absolute value as a distance; indeed, if you're aiming for a 50 or 51 on the quant section, and you see an absolute value question on your test, you're almost certainly going to need to understand that |x-5| measures the distance between x and 5.
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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