GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Apr 2019, 09:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Absolute fundamentals

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Director
Joined: 07 Jun 2004
Posts: 587
Location: PA
Absolute fundamentals  [#permalink]

### Show Tags

10 Mar 2011, 09:06
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 713
Re: Absolute fundamentals  [#permalink]

### Show Tags

11 Mar 2011, 11:01
Only two cases have to be considered.

case 1
------
3 + 2x > 4 - x
3x > 1
x > 1/3

case 2
------
3 + 2x > x - 4
x > -7

Taking the most restricted value x > 1/3

rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?
Math Expert
Joined: 02 Sep 2009
Posts: 54371
Re: Absolute fundamentals  [#permalink]

### Show Tags

11 Mar 2011, 12:53
2
rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?

Above solution is not correct.

$$|3+2x|> |4-x|$$"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are $$-\frac{3}{2}$$ and $$4$$. Hence we'll have three ranges to check:

A. $$x<-\frac{3}{2}$$ --> $$-(3+2x)>{4-x}$$ --> $$x<{-7}$$;

B. $$-\frac{3}{2}\leq{x}\leq{4}$$ --> $$3+2x>4-x$$ --> $$x>\frac{1}{3}$$, as we consider the range $$-\frac{3}{2}\leq{x}\leq{4}$$, then $$\frac{1}{3}<{x}\leq{4}$$;

C. $$x>4$$ --> $$3+2x>-(4-x)$$ --> $$x>{-7}$$, as we consider the range $$x>4$$, then $$x>4$$;

Ranges from A, B and C give us the solution as: $$x<{-7}$$ or $$x>\frac{1}{3}$$ (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html
_________________
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 713
Re: Absolute fundamentals  [#permalink]

### Show Tags

11 Mar 2011, 18:22
Bunuel

I got this thoroughly. Please enlighten me on these statements -

a) For a quadratic equation - the value is positive beyond the roots and negative between the roots. In other words the value of the quadratic alternates +/- in between the root intervals.
b) For an inequality - you must examine the root intervals. "sign" of the inequality MAY flip between the root intervals. To determine the exact sign - use the numberline.

I hope I have absorbed the information flawlessly.

thanks

Bunuel wrote:
rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?

Above solution is not correct.

$$|3+2x|> |4-x|$$"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are $$-\frac{3}{2}$$ and $$4$$. Hence we'll have three ranges to check:

A. $$x<-\frac{3}{2}$$ --> $$-(3+2x)>{4-x}$$ --> $$x<{-7}$$;

B. $$-\frac{3}{2}\leq{x}\leq{4}$$ --> $$3+2x>4-x$$ --> $$x>\frac{1}{3}$$, as we consider the range $$-\frac{3}{2}\leq{x}\leq{4}$$, then $$\frac{1}{3}<{x}\leq{4}$$;

C. $$x>4$$ --> $$3+2x>-(4-x)$$ --> $$x>{-7}$$, as we consider the range $$x>4$$, then $$x>4$$;

Ranges from A, B and C give us the solution as: $$x<{-7}$$ or $$x>\frac{1}{3}$$ (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: Absolute fundamentals  [#permalink]

### Show Tags

12 Mar 2011, 21:30
4
1
rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?

Bunuel has provided you the algebraic solution. Let me add the graphical approach. It may seem daunting at first, but if you take the effort of understanding it, it will seem very straight forward and easy.
Draw the graphs of both the mods
y = | 3 + 2x |
and y = |4 - x |
as shown below.

Attachment:

Ques2.jpg [ 14.42 KiB | Viewed 4927 times ]

Now where is the graph of | 3 + 2x | > the graph of |4 - x | ?
Where is the Red line above the Purple line?
Can I say it's so for the values of x as depicted by the green arrows?
Now all we need to do is find these points.
Point A: 2x+3 = 4-x which implies x = 1/3
Point B: -2x - 3 = 4-x which implies x = -7
So the given condition is satisfied when x > 1/3 or x < -7

Note 1: For more on how to draw graphs of mods, check:
http://www.veritasprep.com/blog/2011/01 ... h-to-mods/

Note 2: If you are wondering how do we decide whether we need to take (2x+3) or (-2x-3), (4-x) or (x-4) while finding points A and B, notice that we need to find the intersection of 2 lines to find these points.
When we make the graph of | 3 + 2x |, one line is (3+2x), and the other line is (-3-2x). When we make graph of |4 - x |, one line is (4-x) and the other is (x-4).
While finding point A, the Red line is going up from left to right so co-efficient of x must be positive hence we use (2x+3). The purple line is going down from left to right so the co-efficient of x must be negative so we use (4-x).
Similarly for point B, the red line is going down from left to right so co-efficient of x must be negative so we use (-2x-3). The purple line is going down from left to right so the co-efficient of x must be negative so we use (4-x).
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 713
Re: Absolute fundamentals  [#permalink]

### Show Tags

12 Mar 2011, 23:30
+1 This is epic. I wish you taught me inequalities back in school.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: Absolute fundamentals  [#permalink]

### Show Tags

13 Mar 2011, 08:15
1
gmat1220 wrote:
+1 This is epic. I wish you taught me inequalities back in school.

If you like alternative solutions, I suggest you check out:

http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

An even shorter and more intuitive approach to these questions.
Rephrase the question as:

| 2x + 3| - |4 - x | > 0

2| x + 3/2| - |x - 4 | > 0

The '2' outside the mod means 'twice the distance from -3/2'. Also, |4 - x | is the same as |x - 4 |.

Then see if you can figure out the answer from the number line. (I must tell you that it takes some effort to figure out the first time... A few people in my batches have done it though so its not terribly difficult...)
I could give an explanation later if you want to verify...
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager
Joined: 08 Nov 2010
Posts: 322
WE 1: Business Development
Re: Absolute fundamentals  [#permalink]

### Show Tags

13 Mar 2011, 12:50
Karishma, as usual - u are great. I would be more than happy to learn more about it.

thanks! +1
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: Absolute fundamentals  [#permalink]

### Show Tags

14 Mar 2011, 18:36
1
144144 wrote:
Karishma, as usual - u are great. I would be more than happy to learn more about it.

thanks! +1

Check out the inequality posts on my blog. I have discussed these methods there in posts titled 'Bagging the Graphs' and 'Holistic approach to mods'
They will help you get comfortable with the basics... Get back in case you have any doubts... Then try and solve this question.... I will give the solution using the method mentioned in the post if you like then....
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 713
Re: Absolute fundamentals  [#permalink]

### Show Tags

14 Mar 2011, 23:09
Karishma,
I have read your blog. Pls verify this - I am doing it intuitively.
If a - b > 0 this means at some point a = b. Lets determine that point. a = |2x+3| and b = |4-x|

Solving -
|2x+3| = |4-x|

Case 1
------
2x + 3 = 4 - x
3x = 1
x = 1/3
Therefore to make a > b, x > 1/3. i.e. move x further right of zero.

Case 2
-------
2x + 3 = x - 4
x + 7 = 0
x = -7

Therefore to make a > b, x < -7. i.e. move x further left of zero.

VeritasPrepKarishma wrote:
gmat1220 wrote:
+1 This is epic. I wish you taught me inequalities back in school.

If you like alternative solutions, I suggest you check out:

http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

An even shorter and more intuitive approach to these questions.
Rephrase the question as:

| 2x + 3| - |4 - x | > 0

2| x + 3/2| - |x - 4 | > 0

The '2' outside the mod means 'twice the distance from -3/2'. Also, |4 - x | is the same as |x - 4 |.

Then see if you can figure out the answer from the number line. (I must tell you that it takes some effort to figure out the first time... A few people in my batches have done it though so its not terribly difficult...)
I could give an explanation later if you want to verify...
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9125
Location: Pune, India
Re: Absolute fundamentals  [#permalink]

### Show Tags

15 Mar 2011, 19:10
2
gmat1220 wrote:
Karishma,
I have read your blog. Pls verify this - I am doing it intuitively.
If a - b > 0 this means at some point a = b. Lets determine that point. a = |2x+3| and b = |4-x|

Solving -
|2x+3| = |4-x|

Case 1
------
2x + 3 = 4 - x
3x = 1
x = 1/3
Therefore to make a > b, x > 1/3. i.e. move x further right of zero.

Case 2
-------
2x + 3 = x - 4
x + 7 = 0
x = -7

Therefore to make a > b, x < -7. i.e. move x further left of zero.

This is absolutely fine but I would not worry about equating... I would let common sense guide me... Let me tell you what I have in mind

Let's focus on just the x axis i.e. the number line

Since you have gone through the post, you know that mod is nothing but distance from 0 on the number line...
|x| = 2 means 'x is at a distance of 2 from point 0 on the number line'
|x-4| = 6 means 'the distance of x from 4 on the number line is 6' so x must be 10 or -2
These are the basics.

Let's take an easier example first. What does the following mean?
|x+2| > |x-4|
It means the value of x is such that distance from -2 is greater than distance from 4.
Attachment:

Ques3.jpg [ 3.01 KiB | Viewed 4684 times ]

Where is the distance from -2 equal to distance from 4? At point 1 on the number line, right? Red and green arrows will be equal.
So if x > 1, red arrow will be longer than green i.e. the distance of the point from -2 will be more than the distance from 4. So x > 1 satisfies this inequality.
What about points on the left of -2? Will there be any point such that its distance from -2 is equal to the distance from 4? Obviously not. All points will be closer to -2 than to 4.
Hence the only region is x > 1.

Now let's take the question at hand:
|2x+3| > |4-x|
2|x+3/2| > |x-4|
It means the value for x is such that twice the distance from -3/2 is more than the distance from 4.
Where will twice the distance from -1.5 be exactly equal to distance from 4?

Attachment:

Ques2.jpg [ 7.97 KiB | Viewed 4683 times ]

We should divide the distance of 5.5 between them in 3 equal parts to get 5.5/3
Now lets go 5.5/3 ahead of -1.5 to get -1.5+5.5/3 = 1/3. This is the point where twice the distance from -3/2 is equal to distance from 4. So you go to the right to make twice the distance from -3/2 greater than the distance from 4. So one solution is x > 1/3
Is there some other point where the same thing will happen?
Yes, at x = -7. How do I get it? because distance between -1.5 and 4 is 5.5
When I add this to the left of -1.5, I get point 7 which is where double the distance from -1.5 will be equal to distance from point 4. To the left of -7, twice the distance from -1.5 will be greater than the distance from 4.
So another solution is x < -7

It is much more intuitive and all you need to do is draw a number line and then reason it out. Let me warn you, it's not everyday that I come across people who are interested in and appreciate alternative strategies. So when I do get an audience, I tend to get a little out of hand... If it makes sense to you, go ahead and try it out.. let me know if you get stuck with anything.. if it doesn't make sense, ignore it...
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager
Joined: 18 Aug 2010
Posts: 79
Re: Absolute fundamentals  [#permalink]

### Show Tags

18 Mar 2011, 00:47
Bunuel wrote:
rxs0005 wrote:
Hi

if i have the below equation

| 3 + 2x | > |4 - x |

are there 2 cases o3 3 cases i need to consider to solve for x

3 + 2x > 4 - x

3 + 2x < -4 + x

is there another case i need to consider here ?

Above solution is not correct.

$$|3+2x|> |4-x|$$"

First you should determine the check points (key points are the values of x for which the expressions in absolute value equal to zero). So the key points are $$-\frac{3}{2}$$ and $$4$$. Hence we'll have three ranges to check:

A. $$x<-\frac{3}{2}$$ --> $$-(3+2x)>{4-x}$$ --> $$x<{-7}$$;

B. $$-\frac{3}{2}\leq{x}\leq{4}$$ --> $$3+2x>4-x$$ --> $$x>\frac{1}{3}$$, as we consider the range $$-\frac{3}{2}\leq{x}\leq{4}$$, then $$\frac{1}{3}<{x}\leq{4}$$;

C. $$x>4$$ --> $$3+2x>-(4-x)$$ --> $$x>{-7}$$, as we consider the range $$x>4$$, then $$x>4$$;

Ranges from A, B and C give us the solution as: $$x<{-7}$$ or $$x>\frac{1}{3}$$ (combined range from B and C).

Similar problem: inequalities-challenging-and-tricky-one-89266.html

Hello, here is my solution: testing all possibilities with signs

1) 3+2x>4-x
x>1/3
2) -3-2x>-4+x
x<1/3
3) -3-2x>4-x
-7>x
4)3+2x>4+x
x>-7

but i have read somewhere that we shall only test both positive and positive neg.
so we have solution x>1/3 x>-7 is this correct ?
thx
Manager
Status: One last try =,=
Joined: 11 Jun 2010
Posts: 124
Re: Absolute fundamentals  [#permalink]

### Show Tags

18 Mar 2011, 01:20
My solution:
Because both $$|3+2x|$$ and $$|4-x| >= 0$$, I can square the two sides:
$$(3+2x)^2 > (4-x)^2$$
$$9+12x+4x^2 > 16-8x+x^2$$
$$3x^2+20x-7 > 0$$
$$x1=\frac{1}{3}$$
$$x2={-7}$$
=> we have two ranges: $$x<{-7}$$ or $$x>\frac{1}{3}$$
_________________
There can be Miracles when you believe
Manager
Status: One last try =,=
Joined: 11 Jun 2010
Posts: 124
Re: Absolute fundamentals  [#permalink]

### Show Tags

18 Mar 2011, 02:16
I think this table will help illustrate the solution of Bunuel.
1. $$x<-\frac{3}{2}$$
$$|3+2x|= -(3+2x)$$ and $$|4-x|={4-x}$$

2. $$-\frac{3}{2}\leq{x}\leq{4}$$
$$|3+2x|={3+2x}$$ and $$|4-x|={4-x}$$

3. $$x>4$$
$$|3+2x|={3+2x}$$ and $$|4-x|=-(4-x)$$
Attachments

Table.jpg [ 6.54 KiB | Viewed 4262 times ]

_________________
There can be Miracles when you believe
Non-Human User
Joined: 09 Sep 2013
Posts: 10553
Re: Absolute fundamentals  [#permalink]

### Show Tags

01 Aug 2017, 11:28
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Absolute fundamentals   [#permalink] 01 Aug 2017, 11:28
Display posts from previous: Sort by

# Absolute fundamentals

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.