TeHCM wrote:

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.

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K S Baskar

Faculty - 4gmat.com