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 500,000 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:

So can't we say that St1 is suff. given that we have an equation 2 on LHS that is equal to RHS. i.e eq'n 2 is always true...

Amit, what's LHS and RHS? confused...

LHS = Left hand side of the equation RHS = Right hand side of the equation.

The equation here is 2.

If I understand you correctly, you can't consider the equations separately. Both equations must be true and they are only true for certain values of x, not all.

For example, for equation 2, you can pick any value of x and it will work. Great!, However, this is not the case for equation 1. I think of it like when you have two ranges of x values. Say, first range is 1<x<10, second range is 2<x<5. If you want to find valid range of x that is true for both, you must pick smaller one.

As bkk145 mentioned, The question asks whether |x-3| = 3 - x or not ?

Now, |x-3| = (x -3 ) i.e When (x-3) is +ve i.e (x-3) > 0 or
|x-3| = -(x-3) i.e When (x-3) is -ve i.e (x-3) < 0

So as asked in the Q stem, |x-3| = 3 - x only when |x-3| = - (x-3) or in other words when x-3 is -ve which is possible only if x < 3.

As per the St1, x ≠ 3 which implies x can be <3> 3. Hence St-1 is in-suff.

St2 : – x | x | > 0
=> This statement is only true |x| = -x since -x * -x = x ^2 > 0. If |x| = x then -x * x < 0 which negates st2.

Therefore x < 0. Hence X < 3. Hence we can conclude that (x - 3) <0> |x-3| = 3 - x. hence St2 is suff to answer this question.

The another method to solve this question is by using the distance logic.

Since we are asked whether |x-3| = 3 - x which in other words means whether distance between 3 and x = -ve of distance between x and 3. And the same logic can be applied here that distance only be a negative number x < 3.

Thanks bkk145. It was a simple concept but I getting lost in my confusion. I went though the doc again and spent some time. Modulus and inequality problems are the hot favs of GMAT test takers.

11. Is root (x-3)^2 = 3-x ? (1) x ≠ 3 (2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.

As bkk145 mentioned, The question asks whether |x-3| = 3 - x or not ?

Now, |x-3| = (x -3 ) i.e When (x-3) is +ve i.e (x-3) > 0 or |x-3| = -(x-3) i.e When (x-3) is -ve i.e (x-3) < 0

So as asked in the Q stem, |x-3| = 3 - x only when |x-3| = - (x-3) or in other words when x-3 is -ve which is possible only if x < 3.

As per the St1, x ≠ 3 which implies x can be <3> 3. Hence St-1 is in-suff.

St2 : – x | x | > 0 => This statement is only true |x| = -x since -x * -x = x ^2 > 0. If |x| = x then -x * x < 0 which negates st2.

Therefore x < 0. Hence X < 3. Hence we can conclude that (x - 3) <0> |x-3| = 3 - x. hence St2 is suff to answer this question.

The another method to solve this question is by using the distance logic.

Since we are asked whether |x-3| = 3 - x which in other words means whether distance between 3 and x = -ve of distance between x and 3. And the same logic can be applied here that distance only be a negative number x < 3.

Thanks bkk145. It was a simple concept but I getting lost in my confusion. I went though the doc again and spent some time. Modulus and inequality problems are the hot favs of GMAT test takers.

Not that I wanna be rude, but is this Jimmy Kimmel with a turban on the picture??

11. Is root (x-3)^2 = 3-x ? (1) x ≠ 3 (2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.

Andr359, can you pls. explain how we know that 3-x is greater than 0? If 3-x is less than 0 then the concept above will not be helpful right?

11. Is root (x-3)^2 = 3-x ? (1) x ≠ 3 (2) – x | x | > 0

Just one simple concept will give you the right answer quickly:

The function sqrt(x) is positive; that is, sqrt(x)>0 for any x >=0. With this in mind: sqrt((x-3)^2)=3-x if and only if 3>x. So, what the question is actually asking is, "is 3>x?" Just by simple inspection you'll notice that only (2) has enough data to solve the problem, therefore B.

Andr359, can you pls. explain how we know that 3-x is greater than 0? If 3-x is less than 0 then the concept above will not be helpful right?

It will be helpful because u switch to imaginay number after that... which is out of the GMAT ... So, yes, it cannot be that Sqrt( A ) < 0 so we conclude :

Sqrt((x-3)^2) >= 0
<=> 3-x >= 0
<=> x =< 3 : the question is boiled down to this