Is x an integer?

statement 1
x^2=2y
x=sqrt(2*integer) is not definitely an integer not suff.

statement 2
even number means that it is an integer, only integers have +-ve nature but anyway it's not suff
1 case: x=1 y=2
2 case x=1.25 y=4

statements 1&2
We can't get more information as above

P.S I'm very cautious about Chetan's questions, traps everywhere

To find: Is x an integer?

Statement 1:

\(X^2 = 2y\) and y is an integer.
y = 2 then x = +2 or -2. but yes an integer.
y = 4 then x = + or -\(2\sqrt{2}\). x is not an integer.
Statement 1 is insufficient.

Statement 2:

x*y is an even number.
x = 1/2 and y = 24. It satisfies even number condition but x is not an integer.
x = 2 and y = 4. It satisfies even number condition but x is an integer.
Statement 2 is insufficient.

From Statement 1 and Statement 2:
From \(X^2 = 2y\) we can say, \(X = \sqrt{2} * \sqrt{y}\)
And as, x*y is an even number. This implies x * y is even integer ( as even number cannot be a fraction ).
and we already know that y is an integer ( From statemnt 1 ).
We can rewrite x * y as \(\sqrt{2} * \sqrt{y} * y\)= even integer.
\(\sqrt{2y} * y\) = even integer.
\(\sqrt{2y}\) this should result in integer then only as a whole x * y will result as evn number.
This implies x should be an integer.
The correct choice is C.

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

(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

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

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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