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

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rraggio
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If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   Updated on: 17 May 2012, 01:19
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If x^2+y^2=1, is x+y=1 ?

(1) xy=0
(2) y=0
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E

Originally posted by rraggio on 15 Sep 2010, 07:37.
Last edited by Bunuel on 17 May 2012, 01:19, edited 1 time in total.
Edited the OA
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Re: If x^2+y^2=1, is x+y=1 ? [#permalink] New post   15 Sep 2010, 08:12
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rraggio wrote:
If x^2+y^2=1, is x+y=1 ?

(1) xy=0
(2) y=0

Thanks!


I think answer B is not correct.

(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.
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saxenashobhit
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Re: If x^2+y^2=1, is x+y=1 ? [#permalink] New post   15 Sep 2010, 21:29
D can be true only when problem says x and y are positive.
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If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   15 Sep 2010, 21:47
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saxenashobhit wrote:
D can be true only when problem says x and y are positive.


Both \(x\) and \(y\) cannot be positive as it would contradict the statements.

For D to be the answer the question should ask "is \(x+y>1\)?" instead of "is \(x+y=1\)?".
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 Thanks [#permalink] New post   18 Feb 2012, 23:44
IMO Answer is D.. Here's why:

x^2+y^2= 1

X^2+y^2 can be written as (x+y)^2-2xy

Therefore (x+y)^2-2xy = 1

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.
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 Thanks [#permalink] New post   19 Feb 2012, 00:52
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pratk wrote:
IMO Answer is D.. Here's why:

x^2+y^2= 1

X^2+y^2 can be written as (x+y)^2-2xy

Therefore (x+y)^2-2xy = 1

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.

Hope it's clear.
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 Thanks [#permalink] New post   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.

Any thoughts?
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 Thanks [#permalink] New post   19 Feb 2012, 09:30
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pratk wrote:
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.

Hope it's clear.
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   20 Feb 2012, 08:29
Thanks for the explanation. I get it now.
Guess I need to practice more of these Always Yes/No type.
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Re: If x^2+y^2=1, is x+y=1 ? [#permalink] New post   27 Jun 2013, 13:20
Bunuel wrote:
rraggio wrote:
If x^2+y^2=1, is x+y=1 ?

(1) xy=0
(2) y=0

Thanks!


I think answer B is not correct.

(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
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If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   27 Jun 2013, 13:24
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RohanKhera wrote:
Bunuel wrote:
rraggio wrote:
If x^2+y^2=1, is x+y=1 ?

(1) xy=0
(2) y=0

Thanks!


I think answer B is not correct.

(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 cannot 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.

Hope it's clear.
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Re: If x^2+y^2=1, is x+y=1 ? [#permalink] New post   27 Jun 2013, 13:32
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
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Re: If x^2+y^2=1, is x+y=1 ? [#permalink] New post   27 Jun 2013, 13:42
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RohanKhera wrote:
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 \(x=\sqrt{4}\) --> \(x=2\).
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   21 May 2014, 09:45
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Bunuel

Similar question but couldnt find any thread

If y is not equal to 1, is x=1?

(1) x^2 + y^2 = 1
(2) y= 1-x

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
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   22 May 2014, 03:07
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gaurav1418z wrote:
Bunuel

Similar question but couldnt find any thread

If y is not equal to 1, is x=1?

(1) x^2 + y^2 = 1
(2) y = 1 - x

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.

This question is discussed here: if-y-1-is-x-161421.html

Hope it helps.
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   22 May 2014, 03:12
Thanks as always Bunuel, yes it helps. Cheers and have a good day
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   20 Nov 2014, 15:31
Please tell me if my assumption is wrong:
Given: x^2+y^2=1
Assume: x^2+y^2+2xy = 1+2xy
(x+y)^2 = 1+2xy
x+y = sqrt(1+2xy)

Stmt1: xy = 0; So, plugging this in: x+y = sqrt(1) = 1 --> hence sufficient?!
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   21 Nov 2014, 04:28
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ParvathyRV wrote:
Please tell me if my assumption is wrong:
Given: x^2+y^2=1
Assume: x^2+y^2+2xy = 1+2xy
(x+y)^2 = 1+2xy
x+y = sqrt(1+2xy)

Stmt1: xy = 0; So, plugging this in: x+y = sqrt(1) = 1 --> hence sufficient?!


From \((x+y)^2 = 1+2xy\) we'd have \(x+y=-\sqrt{1+2xy}\) or \(x+y=\sqrt{1+2xy}\).
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Re: If x^2+y^2=1, is x+y=1 ? (1) xy=0 (2) y=0 [#permalink] New post   29 Jan 2023, 22:25
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