Correct Answer: (C) To determine the new average score of the class, we need to track the total score and the number of students at each stage of the problem. \textbf{Step 1: Calculate the initial total score of the class.} Initially, we have: Number of students ($N_{\text{initial}}$) $= 40$ Average score ($A_{\text{initial}}$) $= 75$ The total score of the class ($T_{\text{initial}}$) is given by the product of the number of students and their average score: $T_{\text{initial}} = N_{\text{initial}} \times A_{\text{initial}} = 40 \times 75 = 3000$ \textbf{Step 2: Account for students leaving the class.} $10$ students leave the class, and their average score was $60$. Number of students leaving ($N_{\text{leave}}$) $= 10$ Average score of leaving students ($A_{\text{leave}}$) $= 60$ The total score contributed by the students who left ($T_{\text{leave}}$) is: $T_{\text{leave}} = N_{\text{leave}} \times A_{\text{leave}} = 10 \times 60 = 600$ After these students leave: New number of students ($N_{\text{after leave}}$) $= N_{\text{initial}} - N_{\text{leave}} = 40 - 10 = 30$ New total score of the class ($T_{\text{after leave}}$) $= T_{\text{initial}} - T_{\text{leave}} = 3000 - 600 = 2400$ \textbf{Step 3: Account for new students joining the class.} $5$ new students join the class, and their average score was $85$. Number of students joining ($N_{\text{join}}$) $= 5$ Average score of joining students ($A_{\text{join}}$) $= 85$ The total score contributed by the new students ($T_{\text{join}}$) is: $T_{\text{join}} = N_{\text{join}} \times A_{\text{join}} = 5 \times 85 = 425$ After these new students join: Final number of students ($N_{\text{final}}$) $= N_{\text{after leave}} + N_{\text{join}} = 30 + 5 = 35$ Final total score of the class ($T_{\text{final}}$) $= T_{\text{after leave}} + T_{\text{join}} = 2400 + 425 = 2825$ \textbf{Step 4: Calculate the new average score of the class.} The new average score ($A_{\text{final}}$) is the final total score divided by the final number of students: $A_{\text{final}} = \frac{T_{\text{final}}}{N_{\text{final}}} = \frac{2825}{35}$ To simplify the fraction, we can divide both the numerator and the denominator by their greatest common divisor, which is $5$: $A_{\text{final}} = \frac{2825 \div 5}{35 \div 5} = \frac{565}{7}$ Performing the division: $565 \div 7 \approx 80.71428... Rounding the result to two decimal places, the new average score is $80.71$. Therefore, the correct option is (C).