Concur with C.
These kind of questions, is there a easier way than plugging in numbers?
Yes it is possible, although substituting numbers might get the answer quicker.
It is evident that if x - y is divisible by 3, then "x" may or may not be divisible by 3.
Similarly it is evident for the info given in statement 2 i.e, x - 2y is divisible by 3. Then "x" may or may not be divisible by "3".
Now, when we try for "C", i.e., the information given in both the statements is true.
i.e, x - y is divisible by 3 and so is x - 2y.
For any two integers A and B, if both A and B are divisible by 3, then any combination of the two numbers say, A + B or A - B or 2A + B or 2A - B etc will also be divisible by 3.
IN this case then if one were to use B - 2A as the combination of the two numbers
i.e., x - 2y - 2(x - y) will be divisble by 3
or x - 2x is divisible by 3 or -x is divisible by 3.
So, "C" is the answer.
K S Baskar
Faculty - 4gmat.com