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:

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is root{(x-3)^2}=3-x?

(1) x≠3 (2) −x|x|>0

When you modify the original condition and the question, it becomes n-th power root (A^n)=|A| when n=even, and |A|=A when A>=0, |A|=-A when A<0. So, |x-3|=3-x=-(x-3)? becomes x-3<0?, x<3?. There is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. For 1), x=/3-> x=2 yes, x=4 no, which is not sufficient. For 2), -x|x|>0 -> x<0<3, which is yes and sufficient. Therefore, the answer is B.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

I am having a hard time grasping why we cannot simplify the problem as:

((X-3)^2)^1/2 = (3 - X) to (X - 3)^2 = (3 - X)^2?

I know I have worked problems before where we have been able to solve it by squaring both sides, but when done in this scenario the answer is entirely different than the OA.

Please help.

That is because if the question says: Is 5 = -5? And you do not know but you square both sides and get 25 = 25 Can you say then that 5 = -5? No!

I understand that you would have successfully used the technique of squaring both sides before but that would be in conditions like these: Given equation: \(\sqrt{X} = 3\) Squaring both sides: X = 9 Here you already know that the equation holds so you can square it. It will still hold. This is like saying: It is given that 5 = 5. Squaring both sides, 25 = 25 which is true.

This question is similar to the first case. It is asked whether \(\sqrt{((X-3)^2)} = (3 - X)\)?

LHS is positive because \(\sqrt{((X-3)^2)} = |X-3|\) and by definition of mod, we know that |X| = X if X is positive or zero and -X if X is negative (or zero). Since |X-3| = - (X - 3), we can say that X - 3 <= 0 or that X <= 3 So the question is: Is X <= 3? Stmnt 1 not sufficient. But stmnt 2 says -X|X| > 0 This means -X|X| is positive. Since |X| will be positive, X must be negative to get rid of the extra negative sign in front. So this statement tells us that X < 0. Then X must be definitely less than 3. Sufficient.

I am having a hard time grasping why we cannot simplify the problem as:

((X-3)^2)^1/2 = (3 - X) to (X - 3)^2 = (3 - X)^2?

I know I have worked problems before where we have been able to solve it by squaring both sides, but when done in this scenario the answer is entirely different than the OA.

Please help.

That is because if the question says: Is 5 = -5? And you do not know but you square both sides and get 25 = 25 Can you say then that 5 = -5? No!

I understand that you would have successfully used the technique of squaring both sides before but that would be in conditions like these: Given equation: \(\sqrt{X} = 3\) Squaring both sides: X = 9 Here you already know that the equation holds so you can square it. It will still hold. This is like saying: It is given that 5 = 5. Squaring both sides, 25 = 25 which is true.

This question is similar to the first case. It is asked whether \(\sqrt{((X-3)^2)} = (3 - X)\)?

LHS is positive because \(\sqrt{((X-3)^2)} = |X-3|\) and by definition of mod, we know that |X| = X if X is positive or zero and -X if X is negative (or zero). Since |X-3| = - (X - 3), we can say that X - 3 <= 0 or that X <= 3 So the question is: Is X <= 3? Stmnt 1 not sufficient. But stmnt 2 says -X|X| > 0 This means -X|X| is positive. Since |X| will be positive, X must be negative to get rid of the extra negative sign in front. So this statement tells us that X < 0. Then X must be definitely less than 3. Sufficient.

How did you get this?

No, we are not establishing/using this. We are just simplifying the question. Since the question is:

Is \(|X-3| = - (X - 3)\) ? Is \((X - 3) <= 0\)? Is \(X <= 3\)?

And then we go on to evaluate each statement.
_________________

quite simple , as it is known that root could not have -ve values so we are sure about the LHS

now go through the statements- statement 1. x is not equal to 3 so x can be anything except 3 , try plugging in say x=5 then LHS(2) not equal to RHS(-2) so we get A No

say x=-5 then LHS(8) = RHS(8) so we get A yes . so not sufficient

statement 2. it says that x must be -ve so we get A yes so sufficient

\(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=-x\), if \(x<0\).

With this understanding I proceeded a bit differently and want to know if my approach is correct:

If I look closely at part inside sq root in LHS, then RHS = -(LHS)

With this thinking clearly i need term inside |x|to be negative since - (x-3) = 3 - x since |x| = \sqrt{\(x^2\)}

Proceeding further for statement 2, -x * |x| is a positive value, this is possible only when product is of two negative no so essentially |x| = -x or in other words - (x-3) = 3-x which is our RHS Does this makes sense?
_________________