The mean of five two-digit numbers is 34. Among these five numbers, two are 'P' and 'Q'. If the digits of both 'P' and 'Q' are swapped, the new mean of the five numbers changes to 14.2. Alternatively, if the other three numbers (excluding 'P' and 'Q') are multiplied by '−1', the mean of the five numbers becomes 10. What is the sum of the tens digits of 'P' and 'Q'?

Let, P = 10a + b, Q = 10c + d & Sum of other 3 numbers = R Given P + Q + R = 34 × 5 = 170 --- (1) P + Q – R = 10 × 5 = 50 --- (2) Solving equation 1 & 2, we have, P + Q = (170 + 50)/2 = 110, So, R = 170 – 110 = 60 & (10a + b) + (10c + d) = 110 (b + d) = 110 – 10(a + c) --- (3) Again, given that, (10b + a) + (10d + c) + R = 5 × 14.2 (10b + a) + (10d + c) + 60 = 71 (10b + a) + (10d + c) = 11 10(b + d) + (a + c) = 11 10[110 – 10(a + c)] + (a + c) = 11 1100 – 100(a + c) + (a + c) = 11 99(a + c) = 1089 (a + c) = 11 = Sum of tens digit