Two distinct numbers x & y are chosen at random from the set {1,2,3,....,30}. The probability that x²-y² is divisible by 3 is:
(A) ½ (B) ¼
(C) 2/3 (D) None of these

Total no. of cases is 30*29= 870

For favourable cases, either both x & y should divisible by 3 or none of them should divisible by 3. In the second scenario, the remainder can be 1 or 2 if x and y have same remainder x-y would be divisible by 3 if x & y have different remainder then x+y would be divisible by 8 .

such cases are (10*9) +(20*19) = 470

probability will be 47/87