Explanation: (D) Let \(A_{10}\) be the average of 10 quizzes, and \(T_{10}\) be the total marks of 10 quizzes. \(T_{10} = 10 \cdot A_{10}\). Let \(A_9\) be the average of the best 9 quizzes, and \(T_9\) be the total marks of the best 9 quizzes. \(T_9 = 9 \cdot A_9\). Given: \(A_9 = A_{10} \cdot (1 + 0.10) = 1.1 \cdot A_{10}\). We need to find \(\frac{T_9}{T_{10}} \times 100\%\). \(\frac{T_9}{T_{10}} = \frac{9 \cdot A_9}{10 \cdot A_{10}} = \frac{9 \cdot (1.1 \cdot A_{10})}{10 \cdot A_{10}} = \frac{9 \cdot 1.1}{10} = \frac{9.9}{10} = 0.99\). Percentage = \(0.99 \times 100\% = 99\%\).
