20) D

Number of pupils in first group = n
Mean of first group = m
Total sum of scores for the first group = nm

A new score of another pupil is added.
Total number of pupils now = n + 1
Mean score now = m – 2
Total score now = (n + 1) x (m – 2)

n (m – 2) + 1 (m – 2)
nm – 2n + m – 2         (eq 1)

However, it is mentioned that the student’s score = n
Therefore, total score now = nm + n         (eq 2)

eq 1 = eq 2

nm – 2n + m – 2 = nm + n

m – 2 = nm + n – nm + 2n

m – 2 = n + 2n

m – 2 = 3n

n m – 2
          3