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 cold and hot drinks?
Answer: B
Let C be the set of people who like cold drinks and H be the set of people who like hot drinks.
n(C) = 27, n(H) = 42.
Since each person likes at least one drink, the union is the total number of people: n(C ∪ H) = 60.
Using the formula n(C ∪ H) = n(C) + n(H) - n(C ∩ H):
60 = 27 + 42 - n(C ∩ H)
60 = 69 - n(C ∩ H)
n(C ∩ H) = 69 - 60 = 9.
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If A and B are disjoint sets, then n(A ∩ B) is:
Answer: A
Disjoint sets are sets that have no elements in common. The intersection of disjoint sets is the empty set (∅).
Therefore, the cardinality of their intersection, n(A ∩ B), is 0.
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In a survey of 100 people, 70 people like coffee, 60 people like tea, and 40 people like both. How many people like only coffee?
Answer: B
Let C be the set of people who like coffee and T be the set of people who like tea.
n(C) = 70, n(T) = 60, n(C ∩ T) = 40.
The number of people who like only coffee is given by n(C) - n(C ∩ T).
Number of people who like only coffee = 70 - 40 = 30.
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De Morgan's Law states that (A ∪ B)' is equal to:
Answer: B
De Morgan's Laws are fundamental principles in set theory.
The first law states that the complement of the union of two sets is the intersection of their complements: (A ∪ B)' = A' ∩ B'.
The second law states that the complement of the intersection of two sets is the union of their complements: (A ∩ B)' = A' ∪ B'.
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In a town, 40% of the people have brown hair, 25% have brown eyes, and 10% have both brown hair and brown eyes. What percentage of people in the town have neither brown hair nor brown eyes?
Answer: B
Let H be the set for brown hair and E for brown eyes.
n(H) = 40%, n(E) = 25%, n(H ∩ E) = 10%.
Percentage of people with at least one trait = n(H ∪ E) = n(H) + n(E) - n(H ∩ E) = 40% + 25% - 10% = 55%.
Percentage of people with neither trait = 100% - n(H ∪ E) = 100% - 55% = 45%.
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In a school, 300 students play cricket and 250 play hockey. If 110 students play both games, how many students play either cricket or hockey?
Answer: B
Let C be the set for cricket and H for hockey.
n(C) = 300, n(H) = 250, n(C ∩ H) = 110.
The number of students who play either game is the union, n(C ∪ H).
n(C ∪ H) = n(C) + n(H) - n(C ∩ H)
n(C ∪ H) = 300 + 250 - 110 = 550 - 110 = 440.
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In a group of 80 people, 35 speak only Hindi and 20 speak only English. If everyone speaks at least one language, how many speak both?
Answer: C
The total number of people is the sum of those who speak only Hindi, only English, and both.
Total = (Only Hindi) + (Only English) + (Both)
80 = 35 + 20 + (Both)
80 = 55 + (Both)
Both = 80 - 55 = 25.
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In a group of 52 people, 16 drink tea but not coffee and 33 drink tea. How many drink coffee but not tea?
Answer: B
Let T be for tea and C for coffee. n(T only) = 16, n(T) = 33. Total = 52.
We can find the number who drink both: n(T ∩ C) = n(T) - n(T only) = 33 - 16 = 17.
The union of the sets (people who drink tea or coffee or both) is n(T ∪ C) = n(T only) + n(C only) + n(both) = 16 + n(C only) + 17.
Since everyone is assumed to drink something, total = n(T ∪ C). So, 52 = 16 + n(C only) + 17 = 33 + n(C only).
n(C only) = 52 - 33 = 19.
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If a set A has 5 elements, what is the number of elements in its power set, n(P(A))?
Answer: D
The power set of a set A, denoted P(A), is the set of all subsets of A. If a set A has 'n' elements, its power set has \(2^n\) elements.
Here, n = 5.
Number of elements in P(A) = \(2^5 = 32\).
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Let U = {1, 2, 3, 4, 5, 6, 7, 8} be the universal set and A = {1, 3, 5, 7}. Find A'.
Answer: C
The complement of a set A, denoted A', consists of all elements in the universal set U that are not in A.
U = {1, 2, 3, 4, 5, 6, 7, 8}
A = {1, 3, 5, 7}
A' = {2, 4, 6, 8}
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