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(1) \(xy=0\) --> either \(x=0\) or \(y=0\): if \(x=0\), then \(x^2+y^2=y^2=1\) and \(y=1\) or \(y=-1\), so \(x+y=0+1=1\) (answer YES) or \(x+y=0-1=-1\) (answer NO); if \(y=0\), then \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO); Two different answers. No sufficient.

(2) \(y=0\) --> \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

(1)+(2) \(xy=0\) and \(y=0\) --> \(y=0\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

From statement 1, xy=0 we get (x+y)^2 = 1 From statement 2, y=0 we get (x+y)^2 = 1

Thus (x+y)=1 -----> Square root of both sides

Both statements are sufficient.

The answer to this question is E, not D.

Consider two sets of numbers, which satisfy stem, as well as both statements and give different values of x+y: If \(y=0\) and \(x=1\) then \(x+y=1+0=1\); If \(y=0\) and \(x=-1\) then \(x+y=-1+0=-1\).

Two different answers. No sufficient.

Answer: E.

Now, the problem in your solution (the red part) is that (x+y)^2=1 means that x+y=1 OR x+y=-1 (you forgot to consider negative root). Basically the same way as x^2=4 means that x=2 or x=-2.

Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 Thanks [#permalink]

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19 Feb 2012, 08:42

Yes Bunuel, what you mention is correct and I also thought about it that way and this would hold true if ther question would have been phrased differently- perhaps something like : "What is the value of x?" However the question simply asks: is x+y=1? And based on my post above, the answer to that question is Yes using both statements independently.

Not sure if my thinking is correct, guess I have been doing alot of critical reasoning questions so my mind is working in a different way.

Yes Bunuel, what you mention is correct and I also thought about it that way and this would hold true if ther question would have been phrased differently- perhaps something like : "What is the value of x?" However the question simply asks: is x+y=1? And based on my post above, the answer to that question is Yes using both statements independently.

Not sure if my thinking is correct, guess I have been doing alot of critical reasoning questions so my mind is working in a different way.

Any thoughts?

No, your thinking is not correct. It's seems that you have some problem with this type of DS question. It's a YES/NO DS question. In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

Now, we have that even when statements are taken together x+y can equal to 1 as well as -1. So, both statements are not sufficient to give definite YES or definite NO answer to the question whether x+y=1.

(1) \(xy=0\) --> either \(x=0\) or \(y=0\): if \(x=0\), then \(x^2+y^2=y^2=1\) and \(y=1\) or \(y=-1\), so \(x+y=0+1=1\) (answer YES) or \(x+y=0-1=-1\) (answer NO); if \(y=0\), then \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO); Two different answers. No sufficient.

(2) \(y=0\) --> \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

(1)+(2) \(xy=0\) and \(y=0\) --> \(y=0\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

Answer: E.

Hi Bunuel,

In one of posts, I read that "square root function can not give negative result"

So in the solution above, is it ok to assume that Under root Y Square (or X Square) will have 2 values: one positive and one negative.

(1) \(xy=0\) --> either \(x=0\) or \(y=0\): if \(x=0\), then \(x^2+y^2=y^2=1\) and \(y=1\) or \(y=-1\), so \(x+y=0+1=1\) (answer YES) or \(x+y=0-1=-1\) (answer NO); if \(y=0\), then \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO); Two different answers. No sufficient.

(2) \(y=0\) --> \(x^2+y^2=x^2=1\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

(1)+(2) \(xy=0\) and \(y=0\) --> \(y=0\) and \(x=1\) or \(x=-1\), so \(x+y=1+0=1\) (answer YES) or \(x+y=-1+0=-1\) (answer NO). Two different answers. No sufficient.

Answer: E.

Hi Bunuel,

In one of posts, I read that "square root function can not give negative result"

So in the solution above, is it ok to assume that Under root Y Square (or X Square) will have 2 values: one positive and one negative.

Regards

Rohan

I guess you are confused by the part where we have \(x=1\) or \(x=-1\) from \(x^2=1\).

Square root function can not give negative result --> \(\sqrt{some \ expression}\geq{0}\), for example \(\sqrt{x^2}\geq{0}\) --> \(\sqrt{4}=2\) (not +2 and -2).

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5, because both 5^2 and (-5)^2 equal to 25.

So you mean that a square root operation results in 2 solution (positive and negative) only in case of an equation ? And otherwise (in case of non equation) there is only one solution i.e. positive ?

So you mean that a square root operation results in 2 solution (positive and negative) only in case of an equation ? And otherwise (in case of non equation) there is only one solution i.e. positive ?

Regards,

Rohan

Not sure I understand what you mean. Anyway:

\(x^2=4\) --> \(x=2\) or \(x=-2\).

\(\sqrt{x}=4\) --> \(x=16\). Or \(x=\sqrt{4}\) --> \(x=2\).
_________________

Statement 1 is clearly not sufficient, as y can be 1/2 or 0, so x can be +3/4 , -3/4 or +1/-1 Similar statement 2 alone is not sufficient

Even when you combine both

y = 1-x x+y =1 squaring both sides (x+y)^2 = 1 x^2 +y^2 + 2xy = 1

from (1), 1 + 2xy = 1, hence xy =0 so x could be 1, 2, 3... and y could be 0, not sufficient.

But is OA is C. I am not sure how

From xy=0, x=0, y=0 or both. But if x=0, then from y=1-x, we get that y=1 but we are told that y≠1, thus x≠0. Hence y=0 and from y=1-x, we get that x=1.

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