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.

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:

Hehe you'll find this funny. Decided I had enough of gmat shortcuts and dusted off my high school algebra book. Anyway there is a theorem regarding this type of problems

it gave the example |2x-3|=|7-3x| According to the book you can simplify it by simply:
2x-3=7-3x or 2x-3=-(7-3x)

And the OA is C but OE wasnt given downloaded one of the paper base gmat tests from this site only gave the answer key

in this problem 4 cases exist
a.RHS +ve ,LHS +ve
b.RHS -ve,LHS +ve
c.LHS +ve,RHS -ve
d.RHS -ve,LHS -ve

we can observe a and d are alike and similarly,b and c are alike
therefore,
we get (x+1)=2(x+1) or (x+1)= -2(x+1),
=>x+1=2x+2 or x+1= -2x-2
=>x= -1 or x= -1
hence,we can say A is sufficient

Re: Inequality with absolute value DS [#permalink]
22 Aug 2006, 03:40

Guys sorry there is a typographical error. Question should read

Is |x|<1?

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

2) |x-3| not equal to zero
My fault the author of the test disabled the ability to cut and paste so I have to retype.
So from here C will stand.

And I really researched that particular theorem I was talking about. When face with two absolute values on either side of the equation. You dont need to break it down into its four scenarios. I did that as well and it took me too much time, which prompted me to research the topic. Thanks for the replies

for problem 2,
stmt 1 gives us 2 values of x -> 3, 1/3. thus we can't say if |x|<1
stmt 2 tells us that |x-3| <> 0, for which x can take any value other than 3.
combining these 2 stmts we can say that since x cannot be 3, it has to be 1/3 and |x| < 1 is true.
hence (C). _________________