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Statement 1

\(x^2 = 2y\), y is an integer

y = 2 , \(x^2=4\), integer
y=3, \(x^2=6\), not integer ............ (Not Sufficient)

Statement 2

x*y is an even integer

Possible pairs (2,2), (3,2), (1/3,18) ........ (Not sufficient)

Combine both:

\(x = \sqrt{2y}\)

y is an integer and x*y is an even integer
\(\sqrt{2y}*y\) is an even integer

Hence, \(\sqrt{2y}\) must be an integer, or x is an integer

Hence Answer C
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(1)
(x,y) can be (\(\sqrt{2}\),1) or (2,2)
Not suff

(2)
(x,y) can be (1,2) or (\(\frac{1}{2}\),4)
Not suff

(1)+(2)
x*y is even and y is an integer. So x*integer=even. x cant be a fraction because Statement 1 would not hold, i.e. 2y would be an integer and \(x^2\) would be a fraction. Also, x cant be a root because then y would also have to be a root, but y is an integer. So x has to be an integer.

Answer: C
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Statement 1:
x^2 = 2*y

Hence y can be 2 => x is an integer
y can be 3 => x is not an integer

Statement 1 not suff

Statement 2:
x*y is even NUMBER

x can be 1/2 or 2

(1)+(2)

from 2nd y=Even /x (integer)

using it in (1)

x^2 = 2*Even/x
x^3 = 2*Even=> Even Integer

Correct ans C
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chetan2u
Is x an integer?
(1) \(x^2=2y\), where y is an integer.
(2) x*y is an even number.

New question


Be careful in analysing the information given to you in each statement

(1) \(x^2=2y\), where y is an integer.
If y is 2 X=2 YES X is an integer..
If y is 3, x^2=2*3=6.....X=√6....NO X is not an integer
Insufficient

(2) x*y is an even number
if y is 4, X can be 1/2.....4*1/2=2
y as 4, X can be any integer..4*3=12
Insufficient

Combined
X*y is even and x^2=2y.....X=√(2y)
So y*√(2y) is even but since y is integer √(2y) has to be an integer..
So X is an integer
Sufficient

C
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