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**