generating new problem on sets any five questions
Answers
Answer:
1. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).
Solution:
Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
then n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 20 + 28 - 36
= 48 - 36
2. If n(A - B) = 18, n(A ∪ B) = 70 and n(A ∩ B) = 25, then find n(B).
Solution:
Using the formula n(A∪B) = n(A - B) + n(A ∩ B) + n(B - A)
70 = 18 + 25 + n(B - A)
70 = 43 + n(B - A)
n(B - A) = 70 - 43
n(B - A) = 27
Now n(B) = n(A ∩ B) + n(B - A)
= 25 + 27
= 52
3. In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. How many like both coffee and tea?
Solution:
Let A = Set of people who like cold drinks.
B = Set of people who like hot drinks.
Given
(A ∪ B) = 60 n(A) = 27 n(B) = 42 then;
n(A ∩ B) = n(A) + n(B) - n(A ∪ B)
= 27 + 42 - 60
= 69 - 60 = 9
= 9
Therefore, 9 people like both tea and coffee.
. There are 35 students in art class and 57 students in dance class. Find the number of students who are either in art class or in dance class.
• When two classes meet at different hours and 12 students are enrolled in both activities.
• When two classes meet at the same hour.
Solution:
n(A) = 35, n(B) = 57, n(A ∩ B) = 12
(Let A be the set of students in art class.
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
= 92
6. In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?
Solution:
Let A = set of persons who got medals in dance.
B = set of persons who got medals in dramatics.
C = set of persons who got medals in music.
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
= 13
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