Also, number of students who play chess, carrom and not scrabble. Solution: Let A be the set of students who play chess B be the set of students who play scrabble C be the set of students who play carrom Therefore, We are given n(A ∪ B ∪ C) = 40, n(A) = 18, n(B) = 20 n(C) = 27, n(A ∩ B) = 7, n(C ∩ B) = 12 n(A ∩ B ∩ C) = 4 We have n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C) Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C ∩ A) + 4 40 = 69 – 19 - n(C ∩ A) 40 = 50 - n(C ∩ A) n(C ∩ A) = 50 - 40 n(C ∩ A) = 10 Therefore, Number of students who play chess and carrom are 10. Find the number of students who play (i)Ĭhess and carrom. 18 play chess, 20 play scrabble and 27 play carrom.ħ play chess and scrabble, 12 play scrabble and carrom and 4 playĬhess, carrom and scrabble. 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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