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x^2 + y^2 = 0.1 and x  y = 0.2, then the value of x +
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08 Nov 2009, 13:08
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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
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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:07
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 nonnegative, 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 \(xy=\frac{1}{5}\). Question: \(x+y=?\) Square \(xy=\frac{1}{5}\) to get rid of modulus > \(x^22xy+y^2=\frac{1}{25}\), as \(x^2+y^2=\frac{1}{10}\) then \(2xy=\frac{3}{50}\); Square \(x+y\) > \((x+y)^2=x^2+2xy+y^2\), as \(x^2+y^2=\frac{1}{10}\) and \(2xy=\frac{3}{50}\) (note that 2xy is posiitve, so 2xy=2xy) then \((x+y)^2=x^2+2xy+y^2=\frac{1}{10}+\frac{3}{50}=\frac{8}{50}=\frac{16}{100}\) > so, \(x+y=\sqrt{\frac{16}{100}}=\frac{4}{10}\). Answer: D.
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Re: Algebra...Modulus Aproach?
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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? xy=0.2(1) or yx=0.2(2) (1) x^2 + (x0.2)^2=0.1 x^2+x^20.4x+0.04=0.1 2x^20.4x=0.06 x^20.2x0.03=0 (x0.3).(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+0.2x0.03=0 (x+0.3).(x0.1)=0 x=0.3 or x=0.1 then y=0.1 or y=0.3 then x+y=0.4



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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 00:22
Is there a faster way to solve this?
Also, I don't get how he got this:
x^20.2x0.03=0 (x0.3).(x+0.1)=0 ????
Thank you.



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Re: Algebra...Modulus Aproach?
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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



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Re: Algebra...Modulus Aproach?
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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) How did you guess those numbers?



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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:14
Thank you, Bunuel, for your explanation.
Quick question:
When squaring absolute value, can you get "additional" solutions that are not solutions of the original equation?
E.g., when you squared \(xy\), how sure were you that you wouldn't obtain "additional" solutions?



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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:20



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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:27
I got your point. You are right.
But I think that when you square:
\(xy\)
you should get this:
\((xy)^2 = x^2  2*x*y + y^2 = x^2  2*x*y + y^2\)
I don't think that you can just write: 2xy.
What do you think?



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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:43



Director
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Re: Algebra...Modulus Aproach?
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07 Dec 2010, 03:47
Yes, you are right. I made a mistake. Thank you.



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Re: Algebra...Modulus Aproach?
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09 Dec 2010, 01:35
My Approach:
Given: mod(x^2 + y^2) = 0.1 and mod(x  y) = 0.2 ==> (xy) = +/(0.2) and ==> [ (xy)^2 + 2xy] = +/(0.1) substituting the value of (xy) = +/(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(xy) = 0.2
==> mod(x) + mod(y) = 0.1 + 0.3 = 0.4
Answer D.



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Re: Algebra...Modulus Aproach?
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18 Nov 2011, 01:20
My approach:
You just need to know that: \(x^2+y^2= x^2+y^2\)
Based on the question: \(xy=0.2\) (1) Square both sides:\(x^2+y^22xy=0.04\) So: \(2xy=0.06\)
\(x+y^2=x^2+y^2+2xy= 0.1+0.06=0.16\)
So \(x+y=0.4\) (2)
From (1) and (2): \(x=0.3\) or \(x=0.1\) then \(y=0.1\) or \(y=0.3\)
So D



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Re: x^2 + y^2 = 0.1 and x  y = 0.2, then the value of x +
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15 Mar 2012, 08:13
Buunuel, can u please explain :
x+y > (x+y)^2=x^2+2xy+y^2
and not
x+y > (x+y)^2=x^2+2xy+y^2
ie how ? xy = xy



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Re: x^2 + y^2 = 0.1 and x  y = 0.2, then the value of x +
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15 Mar 2012, 08:19



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Re: x^2 + y^2 = 0.1 and x  y = 0.2, then the value of x +
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03 Jul 2013, 08:34
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 here My take: lxyl = 0.2 square on both sides now substitute the given euation in the aboge obtained solution we get equation of the form a^2 + b^2 +2ab , so now take square root on both side of the eqn x^2 + y^2 + 2xy = 0.16, taking square root on both sides we get (x + y)^2 =0.4^2 this re=sults down to only one option hope u find it easy, if any assistance is needed please PM basics + logic = > Magic in Gmat



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Re: x^2 + y^2 = 0.1 and x  y = 0.2, then the value of x +
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03 Jul 2013, 12:15
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
x^2 + y^2 = 1/10 (because a square added to another square cannot be less than zero)
xy=1/5 (xy)^2=(1/5)^2 (xy)*(xy)=1/25 x^22xy+y^2=1/25
2xy+x^2+y^2=1/25 2xy+(1/10)=(1/25) 2xy=(4/100)(10/100) 2xy=6/100 2xy=3/50
So, now that we have that: Let' find the value of x+y:
Rather than square x and y square the sum of (x+y) to get a figure of xy (as we got in the above sum of 2xy=3/50)
(x+y)^2 (x+y)*(x+y) x^2 + 2xy + y^2 x^2+2xy+y^2
Before, we got a sum for x^2+y^2 which = 1/10 so plug it again back into the previous equation:
x^2+y^2=1/10 and 2xy=3/50
x^2+2xy+y^2 ===> x^2+y^2+2xy
1/10 + 3/50 = (10/100) + (6/100) = 16/100...just one problem...why do we take the square root of that?
Thanks!



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