![]() Given, n(A) = 36 n(B) = 12 n(C) = 18 n(A ∪ B ∪ C) = 45 n(A ∩ B ∩ C) = 4 We know that number of elements belonging to exactly two of the three sets A, B, C = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C) = n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4 …….(i) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C) Therefore, n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) From (i) required number = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12 = 36 + 12 + 18 + 4 - 45 - 12 = 70 - 57 = 13Īpply set operations to solve the word problems on sets:Įach student in a class of 40 plays at least one indoor game chess,Ĭarrom and scrabble. C = set of persons who got medals in music. B = set of persons who got medals in dramatics. Solution: Let A = set of persons who got medals in dance. Medals went to a total of 45 persons and only 4 persons got medals inĪll the three categories, how many received medals in exactly two of Medals in dance, 12 medals in dramatics and 18 medals in music. In a competition, a school awarded medals in different categories. Word problems on sets using the different properties (Union & Intersection): Given, n(A) = 72 n(B) = 43 n(A ∪ B) = 100 Now, n(A ∩ B) = n(A) + n(B) - n(A ∪ B) = 72 + 43 - 100 = 115 - 100 = 15 Therefore, Number of persons who speak both French and English = 15 n(A) = n(A - B) + n(A ∩ B) ⇒ n(A - B) = n(A) - n(A ∩ B) = 72 - 15 = 57 and n(B - A) = n(B) - n(A ∩ B) = 43 - 15 = 28 Therefore, Number of people speaking English only = 57 Number of people speaking French only = 28 A ∩ B be the set of people who speak both French and English. B - A be the set of people who speak French and not English. A - B be the set of people who speak English and not French. Solution: Let A be the set of people who speak English. How many can speak English only? How many can speak French onlyĪnd how many can speak both English and French? ![]() In a group of 100 persons, 72 people can speak English and 43 can speakįrench. B be the set of students in dance class.) (i) When 2 classes meet at different hours n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 35 + 57 - 12 = 92 - 12 = 80 (ii) When two classes meet at the same hour, A∩B = ∅ n (A ∪ B) = n(A) + n(B) - n(A ∩ B) = n(A) + n(B) = 35 + 57 = 92įurther concept to solve word problems on sets: Solution: n(A) = 35, n(B) = 57, n(A ∩ B) = 12 (Let A be the set of students in art class.
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