x^2+y^2=1

1.xy is zero hence either x=0 or y =0 but its not sufficent

2.y=0

so x^2+y^2=1

x^2+0=1
x^2=1
so X=1

so x+y=1

sufficeint So B

(1) is not sufficient because xy = 0 means either x is 0 or y is zero or both are zero.
Both will not be zero because it is given that x^2+y^2 = 1.
If x= 0, then y^2 = 1, this means y = =1 or -1.
So x+y =1 or -1
If y = 0 , then x^2 = 1, this means x = 1 or -1.
So x+y = -1 or 1.
From (2) y =0, this means x = 1 or -1. Or x+y = 1 or -1.
Not sufficient.
Combining both we have y = 0.
Again not sufficient.

Answer is (E).

By me Ans is:- D

FROM 1
x^2+y^2= (x+y)^2-2xy=1
Now by 1 it is given xy=0
So on substituting values we get (x+y)^2= 1 or we can say x+y=1

FROM 2
y=0
therefore So on substituting values in x^2+y^2=1 we get x^2=1 or x=1
therefore x+y= 1+0 =1

given x^2+y^2 = 1
to prove x+y =1 or not

1)xy = 0
2)y=0

I will start with 2) as it looks simpler

if y = 0 then x^2 = 1 => x = +1 or -1

which will make x+y = +1 or -1
now B is not the solution so we can get rid of D also

lets look at 1) xy = 0
x^2+ y^2 = 1
x^2+ y^2 +2xy = 1+2xy
(x+y)^2 = 1 (xy =0)
x+y = +1 or -1
hence A is not the solution
lets combine 1)+2) to check for C
xy = 0 and y=0 => x not equal to 0, does not help much

In my opinion answer should be E
Please do correct me if I have done something wrong.

jainrahul1985 wrote: If x^2 + y^2 =1, is x + y =1?
(1) xy =0
(2) y = 0


OA B
But I believe answer is wrong

Received a PM asking me to reply.

Statement 2:
y = 0
x^2 = 1
x = +1 or -1
x+y = +1 or -1
No fixed value

Insufficient

Statement 1:
xy = 0
Either x = 0 or y = 0

y=0; x+y = +1 or -1 =>>> No fixed value
x=0; x+y = +1 or -1 =>>> No fixed value

Insufficient

Statement 1 and 2
y = 0

Insufficient

Answer: E

What is the source of this problem Rahul?

I agree with E - becuase statement 1 says that either x or y is zero (can't both be zero becuase of the opening statement) but that means that either x^2 or y^2 is 1. But you can't rule out negative or positive so we don't know whether x+y=1 (could aldo be negative 1)

2. Statement 2 is a little better because now you know that x^2 =1 but you still don't know whether x is positive or negative so we don't know.

Putting them together yields no additional information so the answer is E.

A lot of the posters above have forgotten that anything squared can hide the sign.

For statement 1, we can reach (X+Y)^2=1. Thus, X+Y = + or - 1. Insufficient.

For statement 2, same thing, we can reach that X^2=1. However, X= + or - 1. So X+Y can be -1 or 1. Insufficient.

The answer should be E.

What's the source?

Either of the statement is sufficent
Since
if
xy =0 and x^2+y^2=1
Using these two statements if x=0 y must be 1 and vice versa.

Second statement is just y=0 we know what will be the value of x since x^2 +y^2=1.

So as for me either of two statement is sufficient

Unless x and y are non-negative, answer is E. Otherwise, answer would be D.

(x² + y²) = 1; Is x + y = 1 ?

Rephrasing:
x + y = 1 ?
(x + y)² = 1² ?
x² + 2xy + y² = 1 ?
(x² + y²) + 2xy = 1 ?
1 + 2xy = 1 ?
2xy = 0 ?
xy = 0 ?

1) xy = 0 is sufficient to answer (either x or y equals 1 and the other 0)
2) y = 0 is sufficient to answer (y = 0 thus x =1)

I got (D) but... I guess that what people say is right, if x = -1 or y = -1 things would be different. I think it would be (E) because we could have:

x = -1; y = 0
x = 1; y = 0

Square Root problems are always trap. Must consider + and - values to solve Square root problems.

x^2 + y^2 = 1
if xy = 0, means either 'x' or 'y' is 0, other value can be '-1' or '+1'.
but for x + y = 1, value should be '+1', means Insufficient.


if y = 0, means either 'x' is '-1' or '+1'
Lets put x = -1 and y = 0 in equation x^2 + y^2 = 1
(-1)^2 + (0) = 1
1 + 0 = 1 so x is -1

Lets put x = 1 and y = 0 in equation x^2 + y^2 = 1
(1)^2 + (0) = 1
1 + 0 = 1 so x is 1

So Insufficient.

Correct answer is E

jainrahul1985 wrote:If x^2 + y^2 =1, is x + y =1?
(1) xy =0
(2) y = 0


OA [spoiler]B
But I believe answer is wrong[/spoiler]

Another way to look at this: If (x,y) is on a circle with radius 1 centered at the origin, is x + y = 1?

(1)(x,y) is on one of the coordinate axes
(2) (x,y) is on the x-axis

None of these are sufficient.

+1 for E .
Good one .!