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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: DS ... Absolute Value [#permalink]
11 Sep 2008, 13:27

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

1) | x + 1 | = 2| x -1 |

case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient 2) | x - 3 | not equal to 0 itself not suffcieint.

From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient.

C _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 00:08

x2suresh wrote:

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

1) | x + 1 | = 2| x -1 |

case I) x + 1 >0 and x -1>0 x+1= 2(x-1) --> x=3 case II) x + 1 >0 and x -1<0 x+1= 2(-x+1) --> x=1/3 case III) x + 1 <0 and x -1>0 --> not possible. case IV) x + 1 <0 and x -1<0 -x-1=2(-x+1) -> x=3 multiple solutions -->not suffcient

2) | x - 3 | not equal to 0 itself not suffcieint.

From statment 2: x<>3 from state 1 : x=1/3 or x=3 combined. x=1/3 sufficient.

C

Thanks Suresh. Thats awesome. _________________

----------------------------------------------------------- 'It's not the ride, it's the rider'

Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 21:40

4

This post received KUDOS

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

so i and ii combined say x=1/3. So answer is C _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: DS ... Absolute Value [#permalink]
13 Sep 2008, 22:05

1

This post received KUDOS

amitdgr wrote:

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

so i and ii combined say x=1/3. So answer is C

Good method !!! +1 for u this saves time and actually clears the fundamental _________________

Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 07:00

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

amitdgr wrote:

IgnitedMind wrote:

Is | x | < 1 ?

1) | x + 1 | = 2| x -1 |

2) | x - 3 | not equal to 0

There's another way to deal with absolute (in)equalities.

(i) | x + 1 | = 2| x -1 |

square on both sides x²+1+2x = 4 ( x² + 1 -2x) 3x²-10x+3=0 (3x-1)(x-3) = 0 so x can be 1/3 or 3 so (i) is insufficient

(ii) |x-3| is not 0, in other words x is not 3 . This is insufficient because x can be one of billions and billions of numbers except 3. so (ii) is not sufficient.

combining (i) and (ii) (i) says x can be 1/3 or 3 (ii) says x cannot be 3

Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 08:10

Just to add my late two cents perhaps sombody may find this method easier.

Statement 1: |x+1| = 2*|x-1|

The easiest way to see the various scenarios is to track the critical points of the two absolute value expressions on a number line. The critical point of the expression /x + 1/ is -1 and the critical point of the expression /x - 1/ is 1 How many distinct regions are there on the number line with the points -1 and/or 1 as boundaries? The answer is three: (-1<x), ( x<1) and (-1 < x < 1).

Now let’s test this scenario:

SCENARIO 1: when x < -1 (x+1) is –ve in this reason. So is (x-1) so the equation becomes: -x-1 = 2*(-x+1) => -x-1 = -2x+2 => x=3

SCENARIO 2: when -1 < x < 1, (x + 1) is positive in this region and (x - 1) is negative, so the equation becomes: (x+1) = 2*(-x+1) => x+1 = -2x+2 => 3x = 1 => X = 1/3

SCENARIO 3: when x > 1, (x + 1) is positive in this region and (x - 1) is also positive, so the equation becomes: x+1 = 2*(x-1) => x+1 = 2x-2 => x = 3

So we have 2 possible values of x. i.e. 1/3 and 3 Statement1 insufficient

Statement 2 is clearly insufficient

Combine: st.1 tells us that x is either 1/3 or 3 st. 2 tells us that x is not equal to 3 Then x must be 1/3 Sufficient IMO C

Re: DS ... Absolute Value [#permalink]
15 Sep 2008, 10:43

bigfernhead wrote:

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check.

I am no math Guru, I just happened to learn this method somewhere and this works for me _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: DS ... Absolute Value [#permalink]
16 Sep 2008, 06:15

Tracking point works, I agree, but it takes me way too long.

For the squaring, do you need absolute values on both side? Or this is this the time that you choose to apply it?

amitdgr wrote:

bigfernhead wrote:

Does this always work? Squaring both sides? absolute values, equalities, and inequalities?

When you have mods on both sides of an equality or inequality then this method is best, as squaring removes so many options that you would otherwise have to try. Plugging numbers or Tracking critical points works. But when you have so many scenarios, as worked out in above post, squaring is better in my opinion. Time yourself and check.

I am no math Guru, I just happened to learn this method somewhere and this works for me