Solution: Let the number of candidates in groups P, Q, and R be 3k, 5k, and 4k, respectively, where ‘k’ is a constant of proportionality.

The average (arithmetic mean) scores for the three groups were 64, 84, and 72, respectively. The combined average for three groups is

=> \(\frac{\left[\left(64\cdot3k\right)\ +\ \left(84\cdot5k\right)\ +\ \left(72\cdot4k\right)\right]}{\left[\left(3k\ +\ 5k\ +\ 4k\right)\right]}\)

=> \(\frac{\left[192k\ +\ 420k+\ 288k\right]}{\left[12k\right]}\)

=> \(\frac{\left[900k\right]}{\left[12k\right]}\)= 75

Therefore, B is the correct answer.

Answer B