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?

To solve the equation 3P + 4P + PP + PP = RQ2, we first express the terms in their expanded decimal forms.

3P is 30 + P 4P is 40 + P PP is 10P + P, which is 11P

The equation becomes: (30 + P) + (40 + P) + 11P + 11P = RQ2 70 + 24P = RQ2

Since the sum ends in 2, the unit digit of 24P must be 2. This happens when P is 3 or 8.

Case 1: P = 3 70 + 24 times 3 = 70 + 72 = 142 Here, R = 1 and Q = 4. All digits (P=3, Q=4, R=1) are different, so 142 is a valid sum.

Case 2: P = 8 70 + 24 times 8 = 70 + 192 = 262 Here, R = 2 and Q = 6. However, the unit digit of the sum is also 2. The problem states P, Q, and R are different digits. Since R = 2 and the unit digit is 2, this matches the requirement for the form RQ2, but we must check if P, Q, and R are distinct. Here P=8, Q=6, R=2. These are all different from each other. Thus, 262 is also a valid sum.

The possible sums are 142 and 262. The arithmetic mean is (142 + 262) divided by 2. 404 divided by 2 = 202.

Therefore, the correct answer is C.