If x+2/y = 3, is xy an integer?

The answer is C for me as well......

as you can find out the value of 1 variable x or y if you have the value of one variable on hand. ..& then you can find out whether xy is an integer or not.

stmt 1... is clearly insufficient as it says that x<3.. & from this the value of xy can be non-integer or integer. Therefore, Insufficient
stmt 2... is clearly insufficient as it says that x>-13..& from this as well the value of xy can be non-integer or integer. Therefore, Insufficient.

From 1+2 ...... we have a a definite range I,e.; \(-1<x<3\). & for all the value of x in this range the value of y must be integer as well. & as we know Integer * Integer = Integer. Therefore, C is sufficient.

Hi manishuol, your answer is correct. But not your solution

Statement 1 says that x is an integer, read carefully

From 1+2 it not true that we have int*int. If x=0 then y is not an integer. xy is an integer because x is 0.

Hope this clarifies

Hi manishuol, your answer is correct. But not your solution

Statement 1 says that x is an integer, read carefully

From 1+2 it not true that we have int*int. If x=0 then y is not an integer. xy is an integer because x is 0.

Hope this clarifies

If \(x+\frac{2}{y}= 3\),is xy an integer?

(1) x is an integer less than 3
(2) x >-1

I really understand this, why my method doesn't solve this problem

x + 2/y = 3
so, xy + 2 =3y
so, xy= 3y - 2

We need to know if xy is integer.

(1) x is an integer less than 3
xy = 3y -2
So, this depends upon value of y,
y=0.1, xy= -1.7 ( Not an integer)
y=1, xy=1 (Integer)
Not sufficient
(2) x >-1
xy = 3y -2
So, this depends upon value of y,
y=0.1, xy= -1.7 ( Not an integer)
y=1, xy=1 (Integer)
Not sufficient

(1)+(2) give the same solution. E is my answer.

Please let me know what is wrong with my method.