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

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Does x = y ? (1) x^2 - y^2 = 0 (2) (x + y)^2 = 0 [#permalink] New post 15 May 2011, 02:17
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Question Stats:

22% (01:20) correct 77% (00:19) wrong based on 22 sessions
Guys
The answer sounds unconvincing. Can you verify this?

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

[Reveal] Spoiler:
The OA is E. But I think C is the way to go. Pls verify

S1 tells us that x = -y or x = y
Insufficient

S2 tells us that x = -y or x = y = 0
Insufficient

1) + 2)
x = -y or x = y
x = -y or x = y = 0

The root x = -y satisfy both the equations. Now the trap x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement about whether x = y

Last edited by gmat1220 on 15 May 2011, 02:27, edited 1 time in total.
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Re: Simple yet tripping ! [#permalink] New post 15 May 2011, 02:27
I think this could be the reason..

Basically question is x-y=0 ?

From S1 and S2
x = -y or x = y
x = -y or x = y = 0

So either x+y=0 or x-y=0. hence cannot be determined.
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Re: Simple yet tripping ! [#permalink] New post 15 May 2011, 02:31
I think this is the trap. x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement whether x = y

jamifahad wrote:
I think this could be the reason..

Basically question is x-y=0 ?

From S1 and S2
x = -y or x = y
x = -y or x = y = 0

So either x+y=0 or x-y=0. hence cannot be determined.
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Re: Simple yet tripping ! [#permalink] New post 15 May 2011, 02:39
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gmat1220 wrote:
Guys
The answer sounds unconvincing. Can you verify this?

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


Try to solve it using substitution and it will get clearer:

Q: Does x=y?

Two cases;
Case I: x != y; x=-1, y=1
Case II: x = y; x=0, y=0

(1) x^2=y^2
Case I satisfies.
Case II satisfies.
So, we have two sets both satisfying this statement. However, in one case x=y and in other x!=y. We can't conclude that x=y.

(1) x=-y
Case I satisfies.
Case II satisfies.
So, we have two sets both satisfying this statement. However, in one case x=y and in other x!=y. We can't conclude that x=y.

Combining;
The same two cases satisfy. Nothing conclusive.
Ans: "E"

If the questions said:
"Does x=y if xy!=0". Then "B" would be the answer.
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Re: Simple yet tripping ! [#permalink] New post 16 May 2011, 09:35
gmat1220 wrote:
Guys
The answer sounds unconvincing. Can you verify this?

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

[Reveal] Spoiler:
The OA is E. But I think C is the way to go. Pls verify

S1 tells us that x = -y or x = y
Insufficient

S2 tells us that x = -y or x = y = 0
Insufficient

1) + 2)
x = -y or x = y
x = -y or x = y = 0

The root x = -y satisfy both the equations. Now the trap x = -y does not preclude y from being zero. Hence x = -y can mean x = y = 0 in which case the answer is YES or it can mean x = -y in which case the answer is NO. So we cant make a conclusive statement about whether x = y


Statement 1: tells |x|=|y|. this gives us 3 values:
a. x=y (xy not equal to 0)
b. x=-y
c. x=y=0

x and y can take infinite values.

Statement 2: tells us x=-y or x=y=0 So answer can be both "No" and "Yes"

Both statements combined, we get x=-y or x=y=0. No new information. So insufficient.

Answer: E

Admittedly, I almost picked B. But suddenly I remembered I had pledged that I will ALWAYS check for zero in algebra questions. Glad it paid off :)
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Re: Simple yet tripping ! [#permalink] New post 16 May 2011, 10:40
yup we need to check for 0 too here.

x= -y, x=y=0 both creeps in, in a and b.

hence E.
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Re: Simple yet tripping ! [#permalink] New post 17 May 2011, 11:07
very tricky ...

I also pledged that I will ALWAYS check for zero in algebra questions.
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Re: Simple yet tripping ! [#permalink] New post 03 Nov 2011, 13:05
Should E. tricky zeros
Re: Simple yet tripping !   [#permalink] 03 Nov 2011, 13:05
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