Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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]

Show Tags

19 Feb 2012, 07:42

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.

Campus visits play a crucial role in the MBA application process. It’s one thing to be passionate about one school but another to actually visit the campus, talk...

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...