Suppose x & y are positive integers,x>y, and 3x+4y & 2x+3y when divided by 5,leave remainders 2&3 respectively. It follows that when (x-y) is divided by 5,the remainder necessarily equals
(a) 2 (b)1 (c) 4 (d) none of these

According to the problem,

(3x+2y)\equiv 2 \bmod{5}

(2x+3y)\equiv 3 \bmod{5}

subtracting the above 2 relations we get,

(x-y)\equiv (-1) \bmod{5}

i.e. (x-y)\equiv 4 \bmod{5}

Hence the remainder is 4