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|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

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

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

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