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(D) Let $n_{10}, n_{11}, n_{12}$ be the number of students selected from class $10, 11, 12$ respectively. We have $n_{10} + n_{11} + n_{12} = 10$,with

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$238$

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(D) Let $n_{10}, n_{11}, n_{12}$ be the number of students selected from class $10, 11, 12$ respectively. We have $n_{10} + n_{11} + n_{12} = 10$,with $n_{10} \ge 2, n_{11} \ge 2, n_{12} \ge 2$ and $n_{10} + n_{11} \le 5$.Possible cases $(n_{10}, n_{11}, n_{12})$:$1$. $(2, 2, 6): \binom{5}{2} 	imes \binom{6}{2} 	imes \binom{8}{6} = 10 	imes 15 	imes 28 = 4200$$2$. $(2, 3, 5): \binom{5}{2} 	imes \binom{6}{3} 	imes \binom{8}{5} = 10 	imes 20 	imes 56 = 11200$$3$. $(3, 2, 5): \binom{5}{3} 	imes \binom{6}{2} 	imes \binom{8}{5} = 10 	imes 15 	imes 56 = 8400$Total ways $= 4200 + 11200 + 8400 = 23800$.Given $100k = 23800$,therefore $k = 238$.

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