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:

Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.

from (2) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.

Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.

from (2) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.

Is \(\sqrt{(x-5)^2}=5-x\)?

First of all, recall that \(\sqrt{x^2}=|x|\).

Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so for \(-x|x|\) to be positive, \(-x\) must be positive \(-x>0\) --> \(x<0\). Sufficient.

For the question to be true. the right side of the equation has to be positive, (hence x has to be smaller than 5) because the left side of the equation is always positive.

s1 tells us that x is a negative number. So, it's sufficient s2 tells us that x is less than 5, so it sufficient.

The question basically wants to know if x<=5 else RHS will be x-5

Statement 1 -x|x|>0 or x|x|<0 (multiply by -1 both sides and reverse the sign) either x<0 or |x|<0 since |x| is always positive or 0 x<0 is true. if x<0 then x<5 hence sufficient

Statement 2 5-x>0 5>x This is what we are looking for hence sufficient

Since L.H.S is a square it will be always positive.and R.H.S 5-x will only be positive when x is less than or equal to 5. so in other words the question can be rephrase as is x<=5

stmt1: -x|x|>0

divide both side by |x| we get -x>0 it means x < 0 sufficient

Let me point out something here: You cannot square both sides to get Is \((\sqrt{(x-5)^2})^2 = (5-x)^2\) ?

People sometimes get confused here. Why can you not square it? It is a question similar to 'Is x = 5?' Can you square both sides here and change the question to 'Is \(x^2 = 25\)?' Please remember, they are not the same. x^2 can be 25 even if x is not 5 ( when x = -5, even then x^2 = 25). Only if it is given to you that x = 5, then you can say that x^2 = 25.

You can rephrase the question in the following manner (and many more ways)

Is \((\sqrt{(x-5)^2}) = (5-x)\) ? Is \(|x-5| = (5-x)\) ? or Is \(|5-x| = (5-x)\)? We know that |x| = x only when x >= 0 So \(|5-x| = (5-x)\) only when 5 - x >= 0 or when x <= 5

Stmnt 1: -x|x| > 0 Since |x| is always positive (or zero), -x must be positive too. So x must be negative. If x < 0, then x is obviously less than 5. Sufficient.

Just want to make sure that my line of thinking is right. I got D as well.

sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x

which means that if

x-5<0, -(x-5) = 5-x = no solution, therefore x<5

x-5>0, x-5 = 5-x, which means that x>5 and x = 5

from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.

from (2) 5-x>0, therefore x<5.

Can someone please point out if there is something wrong with my reasoning.

Is \(\sqrt{(x-5)^2}=5-x\)?

First of all, recall that \(\sqrt{x^2}=|x|\).

Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?

(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so for \(-x|x|\) to be positive, \(-x\) must be positive \(-x>0\) --> \(x<0\). Sufficient.

(2) \(5-x>0\) --> \(x<5\). Sufficient.

Answer: D.

Hi Bunuel,

I didn't understand from here : is |x−5|=5−x? --> is x−5≤0? --> is x≤5?.

Modulus always results in an positive value. If x-5>0 then |x−5| should be equal to x-5 However, If x-5<0 then modulus would result the positive value of it i.e. -(x-5)=5-x

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...