Q. Consider the following multiplication problem :
(PQ)×3=RQQ, where P, Q and R are different digits and R≠ 0.
What is the value of (P+R)÷Q?
(a) 1
(b) 2
(c) 5
(d) Cannot be determined due to insufficient data
Correct Answer: (b) 2
Question from UPSC Prelims 2021 CSAT Paper
Explanation :
PQ × 3 = RQQ
Or (10P + Q) × 3 = 100R + 10Q + Q
Or 30P + 3Q = 100R + 11Q
Or 30P = 100R + 8Q
The last digit of 30P will be 0, as well as that of 100R. So, the last digit of 8Q must also be 0.
So, the value of Q must be 5.
Hence, 30P = 100R + 8Q = 100R + 40
Or 3P = 10R + 4
If R = 1, then P = 14/3 (not an integer)
If R = 2, then P = 24/3 = 8
If R = 3, then P = 34/3 (not an integer, and in double digits)
So, P = 8, Q = 5, and R = 2
That is, 85 × 3 = 255
So, (P + R)/Q = (8 + 2)/5 = 10/5 = 2
