In a town of 10,000 families,it was found tha | vedclass

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(B) Given total families $= 10,000$.$n(A) = 40\% 	ext{ of } 10,000 = 4,000$$n(B) = 20\% 	ext{ of } 10,000 = 2,000$$n(C) = 10\% 	ext{ of } 10,000 =

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$3300$

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(B) Given total families $= 10,000$.$n(A) = 40\% 	ext{ of } 10,000 = 4,000$$n(B) = 20\% 	ext{ of } 10,000 = 2,000$$n(C) = 10\% 	ext{ of } 10,000 = 1,000$$n(A \cap B) = 5\% 	ext{ of } 10,000 = 500$$n(B \cap C) = 3\% 	ext{ of } 10,000 = 300$$n(A \cap C) = 4\% 	ext{ of } 10,000 = 400$$n(A \cap B \cap C) = 2\% 	ext{ of } 10,000 = 200$To find the number of families that buy newspaper $A$ only,we use the formula:$n(A 	ext{ only}) = n(A) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C)]$$n(A 	ext{ only}) = 4000 - [500 + 400 - 200]$$n(A 	ext{ only}) = 4000 - [700] = 3300.$

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