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

Question

|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

Bunuel · Accepted Answer

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} .

Answer: D.

maliyeci · Answer

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

nonameee · Answer

Is there a faster way to solve this?

Also, I don't get how he got this:

x^2-0.2x-0.03=0
(x-0.3).(x+0.1)=0 ????

Thank you.

vyassaptarashi · Answer

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

nonameee · Answer

How did you guess those numbers?

nonameee · Answer

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  |x-y| , how sure were you that you wouldn't obtain "additional" solutions?

Bunuel · Answer

There are no "additional" solutions. |x|=3 --> x^2=9 --> x=+/-3, which is the solution of |x|=3.

nonameee · Answer

I got your point. You are right.

But I think that when you square:

|x-y|

you should get this:

(|x-y|)^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?

Bunuel · Answer

No that's not correct.  (|x|)^2=x^2  so  (|x-y|)^2=(x-y)^2 .

nonameee · Answer

Yes, you are right. I made a mistake. Thank you.

soundofmusic · Answer

My Approach:

Given: mod(x^2 + y^2) = 0.1 and mod(x - y) = 0.2
==> (x-y) = +/-(0.2) and 
==>   = +/-(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

==> mod(x) + mod(y) = 0.1 + 0.3 = 0.4

Answer D.

shahideh · Answer

My approach:

You just need to know that:  |x^2+y^2|= x^2+y^2

Based on the question:  |x-y|=0.2  (1)
Square both sides: x^2+y^2-2|x||y|=0.04 
So:  2|x||y|=0.06

|x+y|^2=x^2+y^2+2|x||y|= 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

ficklehead · Answer

Buunuel, can u please explain :

|x|+|y| --> (|x|+|y|)^2=x^2+2|xy|+y^2

and not

|x|+|y| --> (|x|+|y|)^2=x^2+2|x||y|+y^2

ie how ?
|xy| = |x||y|

Bunuel · Answer

Generally:  |xy|=|x|*|y| .

For example:  x=-2  and  y=3  -->  |xy|=|-6|=6  and  |x|*|y|=|-2|*|3|=6 .

Hope it's clear.

krrish · Answer

here My take: lx-yl = 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

WholeLottaLove · Answer

|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)

|x-y|=1/5 (x-y)^2=(1/5)^2 (x-y)*(x-y)=1/25 x^2-2xy+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 + 2|xy| + |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!

believerinthy · Answer

Can you quote the source?

I think while I was solving the problem even with the fastest approach, it took me around 1.5 to 2 mins just to get to the answer.
I think it might be a bit too much of calculations and equations for gmat.

Fdambro294 · Answer

How I got to the Answer, Please Correct me if I made any errors anywhere in the logic:

(1st) Rule: if the Input inside an Absolute Value Modulus MUST ALWAYS be NON-Negative, then the Modulus Bars are extraneous --- they can be removed

---> Adding 2 Squares of Any Value will NEVER result in a (-)Negative "Input"

thus:

= (X)^2 + (Y)^2

Given:

(X)^2 + (Y)^2 = 1/10****

(2nd) Squaring Both Sides of the 2nd Expression Given:

(   ) ^2 = (2/10) ^ 2

(X - Y)^2 = 4 / 100

(X - Y)^2 = (X)^2 + (Y)^2 - 2XY = 4 / 100

----Substituting in: (X)^2 + (Y)^2 = 1/10 -------

1/10 - 2XY = 4/100

(10/100) - (4/100) = 2XY

XY = 3 / 100***

(3rd) Using the Square of a Sum Quadratic Template

(X + Y)^2 = (X)^2 + (Y)^2 + 2XY

----Substituting in the R.H.S. for the Values found above----

(X + Y)^2 = 1/10 + 6/100 = 16/100

----taking the Square Root of Both Sides of the Equation----

sqrt{ (X + Y)^2 } = sqrt{ 16/100 }

= 4/10

X + Y = +4/10 or -(4/10)

and

= 2/10

X - Y = +2/10 or -(2/10)

Lastly: Combining (X + Y) + (X - Y) = 2X for the 4 Possibilities

2X can =

+4/10 + 2/10 = 6/10

+4/10 - 2/10 = 2/10

-(4/10) + 2/10 = -2/10

-(4/10) - (2/10) = -6/10

X can = +3/10 or -(3/10)---

or

X can = +1/10 or -(1/10) ----

(LASTLY)

Using the Given to Find Value of Y: (X)^2 + (Y)^2 = 1/10 = 10/100

CASE 1: (X)^2 = 9/100

(Y)^2 = 1/100

Y = 1/10

+   =   +   = 4/10 = .4

CASE 2: (X)^2 = 1/100

(Y)^2 = 9/100

Y = 3/10

+   =   +   = 4/10 = .4

Answer is .4 in either case

-D-

my head hurts...

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

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