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:

|x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| + |y| is

a) 0.6 b) 0.2 c) 0.36 d) 0.4

Hi guys Need help to improve upon such topics which include Mod..... Can some one discuss aproach to be be followed for such question as well as some tips and variety on similar lines?

x-y=0.2(1) or y-x=0.2(2) (1) x^2 + (x-0.2)^2=0.1 x^2+x^2-0.4x+0.04=0.1 2x^2-0.4x=0.06 x^2-0.2x-0.03=0 (x-0.3).(x+0.1)=0 x=0.3 or x=-0.1 then y=0.1 or y=-0.3 then |x|+|y|=0.4 (2) x^2+(x+0.2)^2=0.1 2x^2+0.4x+0.04=0.1 x^2+0.2x-0.03=0 (x+0.3).(x-0.1)=0 x=-0.3 or x=0.1 then y=-0.1 or y=0.3 then |x|+|y|=0.4

Since |x^2 + y^2| = 0.1 and |x - y| = 0.2 Assume x = 0.1 and y = 0.3 ... only these values will satisfy given equations (search for two squares whose sum is 0.1 since it is a square of the numbers, both numbers should be positive and we need to find addition)

Therefore |x| + |y| = |0.1| + |0.3| = 0.4

Hence “D”

Not really a complicated question..... u can solve it by reverse approach as well i.e. go from choices to question .... take any choice break it and see whether it satisfies the given equations .... with trial and error you will come to answer

Assume x = 0.1 and y = 0.3 ... only these values will satisfy given equations (search for two squares whose sum is 0.1 since it is a square of the numbers, both numbers should be positive and we need to find addition)

|x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| + |y| is

a) 0.6 b) 0.2 c) 0.36 d) 0.4

First of all as x^2 and y^2 are non-negative, then |x^2 + y^2| = 0.1 is the same as x^2+y^2=0.1.

So given: \(x^2+y^2=\frac{1}{10}\) and \(|x-y|=\frac{1}{5}\). Question: \(|x|+|y|=?\)

Square \(|x-y|=\frac{1}{5}\) to get rid of modulus --> \(x^2-2xy+y^2=\frac{1}{25}\), as \(x^2+y^2=\frac{1}{10}\) then \(2xy=\frac{3}{50}\);

Square \(|x|+|y|\) --> \((|x|+|y|)^2=x^2+2|xy|+y^2\), as \(x^2+y^2=\frac{1}{10}\) and \(2xy=\frac{3}{50}\) (note that 2xy is posiitve, so 2|xy|=2xy) then \((|x|+|y|)^2=x^2+2|xy|+y^2=\frac{1}{10}+\frac{3}{50}=\frac{8}{50}=\frac{16}{100}\) --> so, \(|x|+|y|=\sqrt{\frac{16}{100}}=\frac{4}{10}\).

Given: mod(x^2 + y^2) = 0.1 and mod(x - y) = 0.2 ==> (x-y) = +/-(0.2) and ==> [ (x-y)^2 + 2xy] = +/-(0.1) substituting the value of (x-y) = +/-(0.2) in the above equation

0.04 + 2xy = +/-(0.1) ==> xy = 0.03 or -0.07. From xy = 0.03 (x = 0.1 and y = 0.3) as this satisfies mod(x-y) = 0.2

Re: |x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| + [#permalink]

Show Tags

09 Oct 2014, 00:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: |x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| + [#permalink]

Show Tags

25 Nov 2015, 09:25

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: |x^2 + y^2| = 0.1 and |x - y| = 0.2, then the value of |x| + [#permalink]

Show Tags

01 Jan 2017, 11:24

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...