avada wrote:I am completely and utterly confused by this question.

If X and Y are positive integers is Y odd?

1. (Y+2)!/X! is odd integer

2. (Y+2)!/X! is greater than 2


Answer:C

Statement 1
1. (Y+2)!/X! is odd integer

Check when this can be possible--
Only when Y+2 = X+1 and Y+2 is odd

If Y+2 is odd, then Y is odd too.

Sufficient

Statement 2
2. (Y+2)!/X! is greater than 2

Case 1:
Y = 3
X = 2
(Y+2)!/X! = 5!/2! (greater than 2)
Y is odd

Case 2:
Y = 4
X = 2
(Y+2)!/X! = 6!/2! (greater than 2)
Y is even

Not sufficient


Option A

If X and Y are positive integers is Y odd?

1. (Y+2)!/X! is odd integer

Here, Statement 1 can also be proved insufficient

If (Y+2)!/X! is Odd then following two possibilities exist -

--> Y=odd, X=Y+1

Then, (Y+2)!/(Y+1) = (Y+2)
As Y is odd, Y+2 is also odd

--> (Y+2)!/X! = 1
(Y+2)! = X!
This can statement can be valid even if Y is even.
e.g. if Y=2, X=4 then (Y+2)!/X! = 1 = odd

Thus, statement 1 is insufficient.

Statement 2 is already proved insufficient by rijul007.

Now, when we combine both the statements,
then (Y+2)!/X! > 2

Thus, the possibility that (Y+2)!/X! = 1 can be ignored.

Only the other possibility assumed above that Y=odd exists.

Thus Y=odd

Ans: Option C

adthedaddy wrote:

If X and Y are positive integers is Y odd?

1. (Y+2)!/X! is odd integer

Here, Statement 1 can also be proved insufficient

If (Y+2)!/X! is Odd then following two possibilities exist -

--> Y=odd, X=Y+1

Then, (Y+2)!/(Y+1) = (Y+2)
As Y is odd, Y+2 is also odd

--> (Y+2)!/X! = 1
(Y+2)! = X!
This can statement can be valid even if Y is even.
e.g. if Y=2, X=4 then (Y+2)!/X! = 1 = odd

Thus, statement 1 is insufficient.

Statement 2 is already proved insufficient by rijul007.

Now, when we combine both the statements,
then (Y+2)!/X! > 2

Thus, the possibility that (Y+2)!/X! = 1 can be ignored.

Only the other possibility assumed above that Y=odd exists.

Thus Y=odd

Ans: Option C

Thanks for pointing that out!