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?

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