Consider the following addition problem :
3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.
What is the arithmetic mean of all such possible sums ?

Correct Answer: (C) 202

The given equation is $3P + 4P + PP + PP = RQ2$, which expands to $(30 + P) + (40 + P) + (11P) + (11P) = 70 + 24P = RQ2$. The units digit of $24P$ must end in 2, meaning $4P$$\equiv$ 2 \pmod{10}, so $P = 3$ or $P = 8$. For $P = 3$: Sum $= 70 + 24(3) = 142$ (giving distinct digits $P=3, Q=4, R=1$). For $P = 8$: Sum $= 70 + 24(8) = 262$ (giving distinct digits $P=8, Q=6, R=2$). The arithmetic mean of all possible sums is $\frac{142+262}{2}=202$$\frac{142 + 262}{2}$ = 202.