Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Aug 2014, 06:06

Close

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Evaluating Mods/Absolute Values Quicker

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 107
Followers: 1

Kudos [?]: 10 [0], given: 28

Evaluating Mods/Absolute Values Quicker [#permalink] New post 19 Jun 2013, 18:16
I have an equation that looks like this |x – 3| + |x + 1| + |x| = 10 and I want to find the x that satisfies the condition.

I know that -1,0, and 3 are my pivot points that I have to test.

x<-1
-1<=x<=0
0<x<=3

Knowing these are the three ranges where the function changes values is there a quick way to evaluate each subcomponent and know whether it will be positive or negative within that range? I always get stuck on this point because I have to evaluate every subcomponent and figure out what sign i takes with in that range.
For example, in the x<-1 range is each subcomponent evaluates too:
|x-3|=-(x-3) (because |x-3|=-(x-3) if x-3<0, ie, x<3. Because our range is x<-1, clearly within the range of x<3, we place a negative sign)
|x+1|=-(x+1)
and |x|=-x

Just trying to find a quicker way without checking the inequalities to know if I should place a (-) sign on the subcomponents or not.
Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 627
Followers: 41

Kudos [?]: 550 [0], given: 135

Premium Member
Re: Evaluating Mods/Absolute Values Quicker [#permalink] New post 22 Jun 2013, 07:13
Expert's post
alphabeta1234 wrote:
I have an equation that looks like this |x – 3| + |x + 1| + |x| = 10 and I want to find the x that satisfies the condition.

I know that -1,0, and 3 are my pivot points that I have to test.

x<-1
-1<=x<=0
0<x<=3

Knowing these are the three ranges where the function changes values is there a quick way to evaluate each subcomponent and know whether it will be positive or negative within that range? I always get stuck on this point because I have to evaluate every subcomponent and figure out what sign i takes with in that range.
For example, in the x<-1 range is each subcomponent evaluates too:
|x-3|=-(x-3) (because |x-3|=-(x-3) if x-3<0, ie, x<3. Because our range is x<-1, clearly within the range of x<3, we place a negative sign)
|x+1|=-(x+1)
and |x|=-x

Just trying to find a quicker way without checking the inequalities to know if I should place a (-) sign on the subcomponents or not.


We know that |x-0| is the distance of the point x from the origin. Representing all the pivotal points on the number line, the question finally boils down to this : For What value(s) of x, will the distance of x from the points -1,0 and 3 add upto to give 10 units?
Attachment:
Image.jpg
Image.jpg [ 8.44 KiB | Viewed 474 times ]


I. 0<=x<=3 Note that for any x,the expression |x|+|x-3| will always give you a CONSTANT sum of distance as 3 units only, i.e. |x|+|x-3| = 3. Also, assuming x to be farthest in this range i.e. at x= 3, the distance between x = 3 and x = -1 is 4 units. Thus, the sum total of all the distances,i.e.
|x – 3| + |x + 1| + |x| is at most 7 units.Thus, no x for [0,3] will ever add upto 10, for the above mentioned expression.

II. -1=<x<=0 Note the expression |x|+|x+1| will always give you a CONSTANT sum of distance as 1 unit only , i.e. |x|+|x+1| = 1. Again, just as above, assuming x to be farthest in the range i.e. at x = -1, the distance between x = -1 and x = 3 is 4 units. Thus, the sum total of all the distances,i.e.
|x – 3| + |x + 1| + |x| is at most 5 units.Thus, no x for [-1,0] will ever add upto 10, for the above mentioned expression.

III. For x>3 Notice that we have already found out that the maximum sum of distance for x = 3 is 7 units.Now, if i take any value of x>3, it would give a value which is more than 7 units;so now, we know that THERE IS one unique value of x for this range, which will give the sum of distance as 10 units.Thus, we need an extra 3 units. To find the exact value of x, note that the extra 3 units will be contributed by each of the 3 points (-1,0 and 3) EQUALLY, i.e. 1 unit from each of the 3 pivotal points. Thus, as x gets a distance of 1 unit from 3, hence x = 3+1 = 4. Thus, x = 4 is one solution.

II.For x<-1 Notice that we have already found the maximum sum of distance for x = -1 is 5 units.As per the problem, this total should be 10 units. Thus, we need an extra 5 units.Just as above, we know for sure that THERE IS another unique value of x, which will give a sum total of 10 units. To find the exact value of x, note that the extra 5 units will be contributed by each of the 3 points (-1,0 and 3) EQUALLY. Thus, as x is at a distance of \frac{5}{3} unit from -1 and also x<-1, hence x = \frac{-8}{3}.

Hope this helps.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4668
Location: Pune, India
Followers: 1072

Kudos [?]: 4777 [0], given: 163

Re: Evaluating Mods/Absolute Values Quicker [#permalink] New post 27 Jun 2013, 20:30
Expert's post
alphabeta1234 wrote:
I have an equation that looks like this |x – 3| + |x + 1| + |x| = 10 and I want to find the x that satisfies the condition.

I know that -1,0, and 3 are my pivot points that I have to test.

x<-1
-1<=x<=0
0<x<=3

Knowing these are the three ranges where the function changes values is there a quick way to evaluate each subcomponent and know whether it will be positive or negative within that range? I always get stuck on this point because I have to evaluate every subcomponent and figure out what sign i takes with in that range.
For example, in the x<-1 range is each subcomponent evaluates too:
|x-3|=-(x-3) (because |x-3|=-(x-3) if x-3<0, ie, x<3. Because our range is x<-1, clearly within the range of x<3, we place a negative sign)
|x+1|=-(x+1)
and |x|=-x

Just trying to find a quicker way without checking the inequalities to know if I should place a (-) sign on the subcomponents or not.


This question is from my post given here: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/
I have used a graphical approach here which is much faster. It will take some time to wrap your head around it initially but once you do, these questions become very simple.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Re: Evaluating Mods/Absolute Values Quicker   [#permalink] 27 Jun 2013, 20:30
    Similar topics Author Replies Last post
Similar
Topics:
2 Answer. Quicker. HarvardSean 3 14 Nov 2011, 10:48
Experts publish their posts in the topic Help with making math quicker vinviper 3 13 Mar 2008, 19:48
A quicker way ggarr 10 15 Sep 2007, 01:42
GMATPrep PS - Quicker way? MBAlad 6 02 Dec 2006, 18:47
ds: any quicker way? mirhaque 3 20 Apr 2005, 14:32
Display posts from previous: Sort by

Evaluating Mods/Absolute Values Quicker

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

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®.