Q. Consider the following addition problem : 3P+4P+PP+PP=PQ2; where P, Q and R are different digits.
What is the arithmetic mean of all such possible sums?
(a) 102
(b) 120
(c) 202
(d) 220
Correct Answer: (c) 202
Question from UPSC Prelims 2021 CSAT Paper
Explanation :
3P + 4P + PP + PP = RQ2
Or 30 + P + 40 + P + 10P + P + 10P + P = 100R + 10 Q + 2
Or 24P + 70 = 100R + 10 Q + 2
Or 20P + 70 + 4P = 100R + 10 Q + 2
The unit digit of the resultant is 2. It will be obtained when 4 is multiplied by P. So, P must be 3, or 8.
If P = 3, then:
24P + 70 = 24 × 3 + 70 = 72 + 70 = 142
If P = 8, then:
24P + 70 = 24 × 8 + 70 = 192 + 70 = 262
Arithmetic sum of 142 and 262 = (142 + 262)/2 = 202
