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Re: Algebra...Modulus--- Aproach? [#permalink]
08 Nov 2009, 13:24

papillon86 wrote:

|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

Re: Algebra...Modulus--- Aproach? [#permalink]
07 Dec 2010, 00:55

My answer is "D"

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
_________________

Re: Algebra...Modulus--- Aproach? [#permalink]
07 Dec 2010, 01:44

Quote:

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)

Re: Algebra...Modulus--- Aproach? [#permalink]
07 Dec 2010, 03:07

16

This post received KUDOS

Expert's post

papillon86 wrote:

|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}.

Re: Algebra...Modulus--- Aproach? [#permalink]
09 Dec 2010, 01:35

My Approach:

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