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(B) Let the total number of people in the office be $100$.Given that at least $50 \%$ of the people read an e-newspaper,so the number of readers $N \g

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At least $37.5 \%$ read exactly one e-paper.

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(B) Let the total number of people in the office be $100$.Given that at least $50 \%$ of the people read an e-newspaper,so the number of readers $N \ge 50$.Among those who read an e-newspaper,at most $25 \%$ read more than one e-paper.Let $M$ be the number of people who read more than one e-paper. Then $M \le 0.25 	imes N$.Let $S$ be the number of people who read exactly one e-paper. Then $S = N - M$.Since $M \le 0.25 	imes N$,we have $S = N - M \ge N - 0.25 	imes N = 0.75 	imes N$.Since $N \ge 50$,we have $S \ge 0.75 	imes 50 = 37.5$.Thus,at least $37.5 \%$ of the people read exactly one e-paper.

In my office,at least $50 \%$ of the people read an e-newspaper. Among those who read an e-newspaper,at most $25 \%$ read more than one e-paper. Only one of the following statements follows from the statements given below. Which one is it?

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