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Best answer

Given that, the number of students in school $n(U) = 1000,$ the number of students who play chess $n(C) = 300, $ the number of students who play football $n(F) = 600,$ and the number of students who play both chess and football $n(C \cap F) = 50.$

Now, the number of students who play either chess or football $n(C \cup F) = n(C) + n(F) – n(C \cap F)$

$\implies n(C \cup F) = 300 + 600 – 50 = 850.$

$\therefore$ The number of students who play neither chess nor football $\overline{n(C \cup F)} = n(U) – n(C \cup F) = 1000 – 850 = 150.$

So, the correct answer is $(B).$

Now, the number of students who play either chess or football $n(C \cup F) = n(C) + n(F) – n(C \cap F)$

$\implies n(C \cup F) = 300 + 600 – 50 = 850.$

$\therefore$ The number of students who play neither chess nor football $\overline{n(C \cup F)} = n(U) – n(C \cup F) = 1000 – 850 = 150.$

So, the correct answer is $(B).$