We can calculate the number of students enrolled in both schools during the given period of time,
- The total number of students enrolled in school $P\text{(in 5 years)} = (3 + 5 + 5 + 6 + 4) \times 1000 = 23000$
- Average number of students enrolled in school $P = \frac{23000}{5}$
- And, the total number of students enrolled in school $Q\text{(in 5 years)} = (4 + 7 + 8 + 7 + 5) \times 1000 = 31000$
- Average number of students enrolled in school $Q = \frac{31000}{5}$
- Average of the difference of the number of students enrolled in school $P$ and $Q = \frac{31000}{5} – \frac{23000}{5} = \frac{8000}{5}$
Now the required ratio $ = \dfrac{\frac{23000}{8}}{\frac{8000}{5}} = \frac{23}{8}.$
$\therefore$ The ratio of the average number of the students enrolled in school $P$ to the average of the difference of the number of students enrolled in schools $P$ and $Q= 23:8.$
$\textbf{Short Method:}$ While calculating the ratio, common entities are cancelled.
- The total number of students enrolled in school $P=3 + 5+ 5 + 6 + 4 = 23$
- The total number of students enrolled in school $Q=4 + 7 + 8 + 7 + 5 = 31$
The difference between the number of students enrolled in school $P$ and school $Q=31 - 23 = 8$
Now the required ratio $ = 23:8.$
So, the correct answer is $(B).$