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

Question

Does x = y ?

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

asethi · Answer

(1) x^2 - y^2 = 0
 x^2 = y^2, x can be + or - and y can also be + or -. Insufficient
(2) (x + y)^2 = 0
 0^2 = 0 , so x+y=0 or x=-y, sufficient

Ans: B

camlan1990 · Answer

(1) x^2 - y^2 = 0=> x=y or x=-y

Insufficient

(2) (x + y)^2 = 0
=> x+y=0
=> x=-y
But when x = 0, y=0 => x=y=0
When x not equal 0 => x = -y

Insufficient

(1) (2) => X+ Y = 0

INSUFFICENT

Ans: E

ENGRTOMBA2018 · Answer

Per statement 1,  x^2=y^2  --->  x= \pm y  . Thus this statement is not sufficient to answer "is x=y".

Per statement 2, (x+y)^2=0 ---> either x=y=0 , then a "yes" for is x=y but if x=1 and y=-1, then a"no" for "is x=y". This statement is not sufficient.

Combining, still get x=y=0 and x=1 and y=-1 satisfying both the statements ( x=\pm y ) giving you an insufficient answer.

E is the correct answer.

Yce · Answer

why is A not sufficient? i suppose x²-y² is equal to x²=y² which equals x=y 
my understanding is that -ve y² turns +ve when it goes to the side after = sign

hykhan · Answer

Plugging in numbers is an easier way to understand.

If x and y = 0 then x^2 - y^2 = 0 and the answer to the question is yes. But if X = 1 and Y = -1 then x^2 - y^2 = 0 but the answer to the question is no.

Since you get a Yes and a No answer, statement 1 is insufficient

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Ikemen · Answer

from stmt 1: (x+y)(x-y)=0 --> x+y = 0 or x-y = 0 --> x+y=0 has two cases --> case 1--> x=y, case 2 --> x=-y vise versa, or x=y=0 insuff

from stmt 2: same thing logic from stmt 1 applies here --> x+y=0, x=-y, or x=y=0. insuff

combining stmt 1&2 --> no new info. insuff

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

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