GMAT PREP inequalities

This topic has expert replies
Source: — Problem Solving |

Senior | Next Rank: 100 Posts
Posts: 84
Joined: Tue Feb 05, 2008 8:52 pm
Thanked: 8 times

by luvaduva » Wed Apr 23, 2008 8:13 pm
Remember that you are plotting values for X such that the expression holds true.

Answer D will have a number line with two segments: One between -2 and -5 and a second between 2 and 5.

It IS finite, but isn't one segment.

Haha, I tried making a graph, but the spacing wouldn't work.

Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

by camitava » Thu Apr 24, 2008 12:15 am
Can anyone pls let me know how the option-1 to 4 represent a line segment of fixed length? I am not getting the approach ...
Correct me If I am wrong


Regards,

Amitava

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 Apr 24, 2008 11:41 am
camitava wrote:Can anyone pls let me know how the option-1 to 4 represent a line segment of fixed length? I am not getting the approach ...
The problem is that most of them do NOT represent a line segment of finite length.

(1) x^4 >= 1 is true for all x < -1 and all x > 1: two infinite lines.

(2) x^3 <= 27 is true for all x <= 3: an infinite line.

(3) x^2 >= 16 is true for all x >= 4 and all x <= -4: two infinite lines.

(4) 2 <= |x| <= 5 is true for -5 <=x <= -2 AND 2 <= x <= 5: TWO finite line segments.

However:

(5) 2 <= 3x + 4 <= 6 is true for -(2/3) <= x <= 2/3: a finite line segment.
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

Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Mar 29, 2007 3:15 am
Thanked: 7 times
Followed by:1 members

by anju » Fri Aug 22, 2008 2:00 pm
Stuart Kovinsky wrote:
camitava wrote:Can anyone pls let me know how the option-1 to 4 represent a line segment of fixed length? I am not getting the approach ...
The problem is that most of them do NOT represent a line segment of finite length.

(1) x^4 >= 1 is true for all x < -1 and all x > 1: two infinite lines.

(2) x^3 <= 27 is true for all x <= 3: an infinite line.

(3) x^2 >= 16 is true for all x >= 4 and all x <= -4: two infinite lines.

(4) 2 <= |x| <= 5 is true for -5 <=x <= -2 AND 2 <= x <= 5: TWO finite line segments.

However:

(5) 2 <= 3x + 4 <= 6 is true for -(2/3) <= x <= 2/3: a finite line segment.
Hi Stuart,

I did not understand this. I marked D as an option reasoning because x is between 2 and 5. i understand |x| can have + or - values, but x has to be between or equat to 2 and 5 and hence this has a finite line. Pls. help me understand.

Thanks,

Master | Next Rank: 500 Posts
Posts: 105
Joined: Tue Jul 15, 2008 7:03 pm
Thanked: 5 times

by pre-gmat » Fri Aug 22, 2008 3:07 pm
|X| is where X could be negative value or a positive value.

Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Mar 29, 2007 3:15 am
Thanked: 7 times
Followed by:1 members

by anju » Fri Aug 22, 2008 9:39 pm
pre-gmat wrote:|X| is where X could be negative value or a positive value.
If x is negative then x cannot be between 2 and 5 and hence i mentioned my question still remains same as how come option D represent 2 finitie lines.

Master | Next Rank: 500 Posts
Posts: 105
Joined: Tue Jul 15, 2008 7:03 pm
Thanked: 5 times

by pre-gmat » Fri Aug 22, 2008 10:58 pm
You need to solve the inequality 2<=|x|

If X is positive, then 2<=x

If x is negative then X<=-2

Similarly

|x|<=5

if x is positive then X<=5
If x is negative then X>=-5

so as Stuart points out correctly there are two line segments:
-5 <=x <= -2 AND 2 <= x <= 5

Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Mar 29, 2007 3:15 am
Thanked: 7 times
Followed by:1 members

by anju » Mon Aug 25, 2008 10:20 am
pre-gmat wrote:You need to solve the inequality 2<=|x|

If X is positive, then 2<=x

If x is negative then X<=-2

Similarly

|x|<=5

if x is positive then X<=5
If x is negative then X>=-5

so as Stuart points out correctly there are two line segments:
-5 <=x <= -2 AND 2 <= x <= 5
ah... now i got what you are saying...
the above equations can also be written as
If x is negative then X<=-2 OR -x>=2
If x is negative then X>=-5 OR -x<=5
I did visualize the first part but didn't notice the OR part where -x can also fall between 2 and 5 since x is negative and for a negative value the inqualities behave differently.

Thanks all!