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:

Official Solution states Statement 1 is insufficient because it renders y = -1/3 or y = -1. However, I believe that -1/3 is not a possible value for y because the question states x^2 = y^2.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

Official Answer: C My Answer: A

Please help me to understand why is my position not correct.

Re: If x^2 = y^2, is it true that x > 0? [#permalink]

Show Tags

20 May 2013, 10:50

stormbind wrote:

Bunuel wrote:

Please refer to the solutions provided.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

This part is not correct. x = 2y+1. If y = -1/3, x is not equal to -2/3.
_________________

Official Solution states Statement 1 is insufficient because it renders y = -1/3 or y = -1. However, I believe that -1/3 is not a possible value for y because the question states x^2 = y^2.

Explanation: If y = -1/3 then x = -2/3, which contradicts the question. Thus -1/3 is rejected, leaving only y = -1. This would make Statement 1 sufficient.

Official Answer: C My Answer: A

Please help me to understand why is my position not correct.

Why do we have to solve for 'y' when we are looking for the value of x? I know how to get x=-1 and x=1/3 but I'm still not sure why the values of y are relevant.

Bunuel wrote:

kotela wrote:

Bunuel wrote:

If x^2 = y^2, is true that x>0?

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunnel,

I need a clarification here, when i put statement 1 in X^2 = Y^2 i get the below

3Y^2= -4Y-1,

and since LHS is positive so the RHS needs to be positive so in order to RHS to be positive Y should be "-ve" and when Y is negative, Now

considering statement 1 we can get to know that X is negative.....

So my point is again statement 2 needed?

Please clarify me if i am wrong?

How did you get that if y is negative x must be positive?

For (1) we have: \(x^2 = y^2\) and \(x=2y+1\). Solving gives: \(x=-1\) and \(y=-1\) OR \(x=\frac{1}{3}\) and \(y=-\frac{1}{3}\), just substitute these values to check that they satisfy both equations.

Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)

2. y≤-1 Not sufficent

1+2 2. says that y≤-1. In #1, the only case where y≤-1 is y=2y+1

-y=1 y=-1

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

Here we get an answer of x=-1 which is obviously ≠ to 0.

Insufficient because we have two values for x. (however, using the same reasoning to arrive at c. cant we say that both answers we arrive at show that x ≠ 0 and thus, a is sufficient?)

2. y≤-1 Not sufficent

1+2 2. says that y≤-1. In #1, the only case where y≤-1 is y=2y+1

-y=1 y=-1

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

Here we get an answer of x=-1 which is obviously ≠ to 0.

Everything you did is correct except that you misunderstood the question. The question is:

Is x positive? Is x > 0? It does not ask you whether x is equal to 0.

Statement I tells you that x could be positive or negative. So not sufficient. Both statements together tell you that x is negative. Hence it is not positive. It answers the question with 'No'. Sufficient.
_________________

for #2.) we are given y<=-1. This is not sufficient because of the following:

|x| = |y|

x=y OR x=-y

x=-y OR x=-(-y) x=y

Correct?

I am not really sure what you have done here. The 4 cases will be x = y x = -y -x = y -x = -y which are equivalent to just two cases: x = y or x = -y. Statement 2 is not sufficient because all we know now is that y is negative. IF x = y, x is negative. If x = -y, x is positive. So we still don't know whether x is positive or not.

WholeLottaLove wrote:

Another thing that has bothered me is this. If x=-y, and for example, y=5, then would x=(-5)?

As always, thanks for the help!

Yes. If x = -y and y = 5, then x = -5
_________________

If x^2 = y^2, then we can get the square root of both sides: √x^2 = √y^2 |x| = |y| x=y OR x=-y

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

1) x=2y+1 So, we are looking for the value of x. x=y or -x=y

I.) x=2y+1 x=2x+1 -x=1 x=-1

II.) x=2y+1 x=-2x+1 3x=1 x=1/3

x could be greater than zero or less than zero. NOT SUFFICIENT

(2) y<= -1

x could = y or -y. For example. |x| = |y| |2| = |-2| OR |-2| = |-2|

NOT SUFFICIENT

1 + 2) x=2y+1 and y<= -1

In 1) we have two cases, where x is negative and where x is positive. However, when x is negative y is positive and when x is positive y is negative. 2) tells us that y is negative meaning x must be positive or greater than zero. SUFFICIENT

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

If x^2=y^2, is it true that X>0? |x|=|y| ---> x=y or x=-y

1. x=2y+1, x is also equal to y. --> 2y+1=y or y=-1. If y=-1, x=2(-1)+1=-2+1=-1 x=2y+1, x is also equal to -y. --> 2y+1=-y or y=-(1/3). If y=-(1/3), x=-(2/3)+1=(-2+3)/3=1/3 we see that x is positive in second case and not in the first. NOT SUFFICIENT

2. y<=-1. This one is easy to notice that x can be anything as long as its magnitude is same as y's.

1+2: y can only be -1 from combining these two statement. This is also our case 1 from statement 1. Hence, x=-1<0. SUfficient

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Why cant answer be B

From x^2=y^2 x and y can take the following values: 1 and -1 or -1 and 1 as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong?

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Why cant answer be B

From x^2=y^2 x and y can take the following values: 1 and -1 or -1 and 1 as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong?

x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number.

So, for (2) it's possible that x = y = -1, or x= 1 and y = -1, or x = y = -2, or x = y = -1.5, or x = 100 and y = -100...
_________________

\(x^2 = y^2\) --> \(|x|=|y|\) --> either \(y=x\) or \(y=-x\).

(1) x=2y+1 --> if \(y=x\) then we would have: \(x=2x+1\) --> \(x=-1<0\) (notice that in this case \(y=x=-1\)) but if \(y=-x\) then we would have: \(x=-2x+1\) --> \(x=\frac{1}{3}>0\) (notice that in this case \(y=-x=-\frac{1}{3}\)). Not sufficient.

(2) y<= -1. Clearly insufficient.

(1)+(2) Since from (2) \(y\leq{-1}\) then from (1) \(y=x=-1\), so the answer to the question is NO. Sufficient.

Answer: C.

Hope it's clear.

Hi Bunuel

Why cant answer be B

From x^2=y^2 x and y can take the following values: 1 and -1 or -1 and 1 as x and y different so not considering same integers like both positive or both negative

from 1st statement none of the values satisfy the given equation from 2nd statement, y is negative which makes x as positive. Where am I wrong?

x and y can be the same number. Generally, unless it is explicitly stated otherwise, different variables CAN represent the same number.

So, for (2) it's possible that x = y = -1, or x= 1 and y = -1, or x = y = -2, or x = y = -1.5, or x = 100 and y = -100...

Agreed Bunuel. Tried that too. But by that logic, Statement 1 holds true only when both x and y are negative and hence we could get A as the answer as we would know that y is negative along with x.

gmatclubot

Re: If x^2 = y^2, is true that x>0?
[#permalink]
12 Nov 2014, 10:39

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...