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I am confused with the answer; and the explanation as well.

My answer is D.

Because, in statement 1, if we add y^2 on both sides, we get x^2 = y^2. Then, if we square root both the sides of the equation, we get x = y (no matter whether they are positive or negative). So, this is sufficient and is eliminating option B, C, and E.

Then, in the statement 2, if we square root both the sides of the equation, we get x - y = 0, or x = y (no matter whether they are positive or negative).

Therefore, my answer is D. However, Kaplan Premier 2011 shows the answer as B and explains it thereby.

square root of x^2 = could be x or -x ( not just x)

in other words square root of x^2 = |x| = +-x

so when you take square root on both sides , you will have +- on both sides. not enough to find the sign of the variable.

=> x = y or x = -y

Eg: \sqrt{(x^2)} = 25 => |x| = 5 =>x = +5 or -5

Hope it helps.

Schawjibb wrote:

@ Shubhabsh Da'

I am confused with the answer; and the explanation as well.

My answer is D.

Because, in statement 1, if we add y^2 on both sides, we get x^2 = y^2. Then, if we square root both the sides of the equation, we get x = y (no matter whether they are positive or negative). So, this is sufficient and is eliminating option B, C, and E.

Then, in the statement 2, if we square root both the sides of the equation, we get x - y = 0, or x = y (no matter whether they are positive or negative).

Therefore, my answer is D. However, Kaplan Premier 2011 shows the answer as B and explains it thereby.

I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?

Q. Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0

Thanks in advance!

Solution: Statement 1: \(x^2-y^2=0\) \(x^2=y^2\) Squaring both sides, would result to two solutions: x= y OR x=-y. So, INSUFFICIENT!

Statement 2: \((x-y)^2=0\) To get a 0, x-y must be equal to 0. \(x=y\) SUFFICIENT!

Could you please help me in one regard? I need a resource (book/forum discussion) from where I can really learn how to solve a DS problem systematically and in minimum time.

Background: I have started going through the DS section from Kaplan Premier 2011 yesterday. And the result of yesterday's (first-time) practice was devastating. 13 out of 20 were incorrect. And no question was answered in less than 2 minutes! I am just drowning into frustration.......

My 2 cents on this.

See the best resources are the OG's, GMAT Club Math Forum and the MGMAT tests.

Seeing the problem that you are facing right now,probably you should brush up your basics first.Then start with OG and subsequently move on to higher difficulty level numerical s. _________________

Thanks a lot! I think my basic problem pertaining to DS is with Number Properties. Because, in other fields I feel myself comfortable as long as the math part is concerned. Do you think MGMAT Number Properties book will suffice? Suggestions are highly appreciated?

Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]

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15 Oct 2013, 09:31

Schawjibb wrote:

I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?

Q. Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0

Thanks in advance!

Hi there, let me take a bite a this one.

So does x = y?

From Statement 1:

x^2 - y^2 = 0

This means that Abs (X) = Abs (Y) But could be that x = -y, or x = y Hence Insuff

From Statement 2

(x-y)^2 = 0 (x-y) = 0 So x = y Suff

Answer is B

Cheers J

Last edited by jlgdr on 15 Oct 2013, 10:09, edited 1 time in total.

Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]

Show Tags

15 Oct 2013, 09:38

Expert's post

jlgdr wrote:

Schawjibb wrote:

I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?

Q. Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0

Thanks in advance!

Hi there, let me take a bite a this one.

So does x = y?

From Statement 1:

x^2 - y^2 = 0

This means that Abs (X) = Abs (Y) But could be that x = -y, or x = y Hence Insuff

From Statement 2

(x-y)^2 = 0 (x+y)(x-y) = 0 x = -y or x = y So Insuff

From Statements 1 and 2 together: We have exactly the same information, hence IMHO E is the correct answer

Cheers J

That's not correct.

Does x = y?

(1) x^2 - y^2 = 0 --> x^2=y^2 --> x=y or x=-y. Not sufficient.

Re: Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0 [#permalink]

Show Tags

15 Oct 2013, 10:08

Bunuel wrote:

jlgdr wrote:

Schawjibb wrote:

I found this question in Kaplan Premier 2011. I am confused with their answer and its explanation (or, may be I am wrong). Can anyone come up with an answer supported by precise explanation?

Q. Does x = y? (1) x^2 - y^2 = 0 (2) (x - y)^2 = 0

Thanks in advance!

Hi there, let me take a bite a this one.

So does x = y?

From Statement 1:

x^2 - y^2 = 0

This means that Abs (X) = Abs (Y) But could be that x = -y, or x = y Hence Insuff

From Statement 2

(x-y)^2 = 0 (x+y)(x-y) = 0 x = -y or x = y So Insuff

From Statements 1 and 2 together: We have exactly the same information, hence IMHO E is the correct answer

Cheers J

That's not correct.

Does x = y?

(1) x^2 - y^2 = 0 --> x^2=y^2 --> x=y or x=-y. Not sufficient.

(2) (x - y)^2 = 0 --> x-y=0 --> x=y. Sufficient.

Answer: B.

Apologies for that. Yes that's in fact correct. Thanks for pointing out. Cheers J

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