Set Notation & Operations - Finite Mathematics

1. For any three sets $A$, $B$, and $C$, which of the following expressions is equivalent to $A \setminus (B \cap C)$?
   Answer: $(A \setminus B) \cup (A \setminus C)$ When you encounter set difference problems involving intersections or unions, the key is understanding how De Morgan's laws and set operations interact. The expression $A \setminus (B \cap C)$ represents all elements in set $A$ that are NOT in both $B$ and $C$. To find what's equivalent to $A \setminus (B \cap C)$, think about it this way: an element $x$ belongs to this set if $x \in A$ AND $x \notin (B \cap C)$. For $x$ to not be in $(B \cap C)$, it must fail to be in at least one of $B$ or $C$. So $x$ is either not in $B$, or not in $C$, or not in both. This means $A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)$, which is answer choice B. An element is in this union if it's in $A$ but not in $B$, OR if it's in $A$ but not in $C$. Answer choice A, $(A \setminus B) \cap (A \setminus C)$, would only include elements in $A$ that are in neither $B$ nor $C$ - this is too restrictive. Answer choice C, $(A \cap B) \setminus (A \cap C)$, represents elements in both $A$ and $B$ but not in both $A$ and $C$ - this doesn't require the element to avoid $(B \cap C)$. Answer choice D uses complements incorrectly and doesn't maintain the set difference structure. Remember: when you see "not in an intersection," think "not in the first OR not in the second" - this typically leads to a union of differences.
2. Let $A$ and $B$ be two sets such that the number of elements in their intersection is 10 and the number of elements in their union is 40. If the number of elements in $A$ is equal to the number of elements in $B$, what is the number of elements in set $A$?
   Answer: 25 We use the Principle of Inclusion-Exclusion for two sets: $|A \cup B| = |A| + |B| - |A \cap B|$.\ We are given $|A \cup B| = 40$ and $|A \cap B| = 10$.\ We are also given that $|A| = |B|$. Let's denote this common value by $x$.\ Substituting the given values into the formula: $40 = x + x - 10$.\ Now, we solve for $x$: \ $40 = 2x - 10$\ $50 = 2x$\ $x = 25$.\ Therefore, the number of elements in set A is 25.\ Distractor A, 15, is the value of $|A \setminus B|$, since $|A \setminus B| = |A|-|A \cap B| = 25 - 10 = 15$.\ Distractor B, 20, comes from incorrectly assuming the sets are disjoint and solving $|A|+|B|=40$.\ Distractor D, 30, is the value of the symmetric difference $|A \Delta B| = |A \cup B| - |A \cap B| = 40-10=30$.
3. Consider the sets $P = \{x : x^2 - 5x + 6 = 0\}$ and $Q = \{x : x^2 - 7x + 12 = 0\}$. Which of the following statements about $P$ and $Q$ is correct?
   Answer: $P \cap Q = \{3\}$ and $P \cup Q = \{2, 3, 4\}$ Solving $x^2 - 5x + 6 = 0$: $(x-2)(x-3) = 0$, so $P = \{2, 3\}$. Solving $x^2 - 7x + 12 = 0$: $(x-3)(x-4) = 0$, so $Q = \{3, 4\}$. Therefore, $P \cap Q = \{3\}$ and $P \cup Q = \{2, 3, 4\}$.
4. In a survey of 200 students, 85 study mathematics, 70 study physics, 60 study chemistry, 30 study both mathematics and physics, 25 study both mathematics and chemistry, 20 study both physics and chemistry, and 10 study all three subjects. How many students study exactly two of these subjects?
   Answer: 45 students study exactly two subjects Students studying exactly two subjects = (Math and Physics only) + (Math and Chemistry only) + (Physics and Chemistry only). Math and Physics only = 30 - 10 = 20. Math and Chemistry only = 25 - 10 = 15. Physics and Chemistry only = 20 - 10 = 10. Total studying exactly two = 20 + 15 + 10 = 45.
5. Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$, $A = \{x \in U : x \text{ is odd}\}$, and $B = \{x \in U : x > 6\}$. If $C = (A \cup B)^c \cap (A \cap B)^c$, then $C$ equals:
   Answer: $\{2, 4, 6\}$ First, find $A = \{1, 3, 5, 7, 9\}$ and $B = \{7, 8, 9, 10\}$. Then $A \cup B = \{1, 3, 5, 7, 8, 9, 10\}$ and $A \cap B = \{7, 9\}$. So $(A \cup B)^c = \{2, 4, 6\}$ and $(A \cap B)^c = \{1, 2, 3, 4, 5, 6, 8, 10\}$. Therefore, $C = (A \cup B)^c \cap (A \cap B)^c = \{2, 4, 6\} \cap \{1, 2, 3, 4, 5, 6, 8, 10\} = \{2, 4, 6\}$.
6. Let $A$ and $B$ be subsets of a universal set $U$. Which of the following expressions is equivalent to $((A' \cup B)' \cup B)'$?
   Answer: $(A \cup B)'$ When you encounter complex set expressions with multiple complement and union operations, the key is to work systematically from the innermost parentheses outward, applying De Morgan's laws and basic set identities. Let's simplify $((A' \cup B)' \cup B)'$ step by step. Starting with the innermost expression $(A' \cup B)'$, we apply De Morgan's law: the complement of a union equals the intersection of complements. So $(A' \cup B)' = (A')' \cap B' = A \cap B'$. Now our expression becomes $(A \cap B' \cup B)'$. Next, we need to simplify $A \cap B' \cup B$. Using the distributive property in reverse, this equals $(A \cup B) \cap (B' \cup B)$. Since $B' \cup B = U$ (the universal set), we get $(A \cup B) \cap U = A \cup B$. Finally, taking the complement of this result: $(A \cup B)' = (A \cup B)'$, which is answer choice D. Let's check why the other options are wrong. Choice A suggests $A'$, but this ignores the influence of set $B$ entirely. Choice B claims the result is the empty set $\emptyset$, which would only be true if we had a contradiction like $X \cap X'$. Choice C proposes $A \cap B$, but this misses that we're taking the complement of $A \cup B$, not finding their intersection. Study tip: When simplifying complex set expressions, always work from inside out and remember that $(X \cup Y)' = X' \cap Y'$ and $(X \cap Y)' = X' \cup Y'$. Drawing Venn diagrams can also help verify your algebraic work.
7. A survey of 80 students regarding their subscriptions to three streaming services (labeled A, B, and C) revealed the following: 35 subscribe to A, 30 to B, and 40 to C. 15 subscribe to both A and B, 12 to both A and C, and 10 to both B and C. 5 students subscribe to all three services. How many of the surveyed students subscribe to none of these three services?
   Answer: 7 We need to find the number of students outside the union of the three sets, i.e., $|(A \cup B \cup C)'|$. This is equal to $|U| - |A \cup B \cup C|$.\ First, we use the Principle of Inclusion-Exclusion for three sets to find $|A \cup B \cup C|$: \ $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$\ Plugging in the given values:\ $|A \cup B \cup C| = 35 + 30 + 40 - (15 + 12 + 10) + 5$\ $|A \cup B \cup C| = 105 - (37) + 5$\ $|A \cup B \cup C| = 105 - 37 + 5 = 68 + 5 = 73$.\ This is the number of students who subscribe to at least one service. The number of students subscribing to none is:\ $|U| - |A \cup B \cup C| = 80 - 73 = 7$.
8. A club has 50 members. 30 members participate in activity A, and 25 members participate in activity B. If 12 members participate in both A and B, how many members participate in exactly one of the two activities?
   Answer: 31 We want to find the number of members who participate in activity A only, or activity B only. This is the cardinality of the symmetric difference, $|(A \setminus B) \cup (B \setminus A)|$.\ First, find the number of members in A only: $|A \setminus B| = |A| - |A \cap B| = 30 - 12 = 18$.\ Next, find the number of members in B only: $|B \setminus A| = |B| - |A \cap B| = 25 - 12 = 13$.\ The number of members participating in exactly one activity is the sum of these two values: $18 + 13 = 31$.\ Alternatively, one can use the formula $|A| + |B| - 2|A \cap B| = 30 + 25 - 2(12) = 55 - 24 = 31$.\ Distractor B is $|A \cup B| = |A|+|B|-|A \cap B| = 30+25-12 = 43$.\ Distractor C is $|A \setminus B|$.\ Distractor D is the number of members in neither activity: $50 - |A \cup B| = 50 - 43 = 7$.
9. Let $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$. Define $A = \{x \in U : x \text{ is divisible by 3}\}$ and $B = \{x \in U : x \text{ is prime}\}$. The set $(A \cup B)^c \cap (A \triangle B)$ equals:
   Answer: $\emptyset$ since the sets are disjoint by definition When you encounter set theory problems with multiple operations, work systematically through each step to avoid getting lost in the complexity. First, let's identify the sets. From $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$: - $A = \{3, 6, 9, 12\}$ (multiples of 3) - $B = \{2, 3, 5, 7, 11\}$ (prime numbers) Now work through $(A \cup B)^c \cap (A \triangle B)$: $A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12\}$ $(A \cup B)^c = \{1, 4, 8, 10\}$ (elements in $U$ but not in $A \cup B$) $A \triangle B = (A \cup B) - (A \cap B) = \{2, 5, 6, 7, 9, 11, 12\}$ (since $A \cap B = \{3\}$) Finally: $(A \cup B)^c \cap (A \triangle B) = \{1, 4, 8, 10\} \cap \{2, 5, 6, 7, 9, 11, 12\} = \emptyset$ These sets share no common elements, making the intersection empty. Choice A lists $\{1, 4, 8, 10\}$, which is $(A \cup B)^c$ alone, ignoring the intersection with $A \triangle B$. Choice B adds element 9, which isn't in $(A \cup B)^c$. Choice D claims $\{9\}$ satisfies both conditions, but 9 is in $A \triangle B$ yet not in $(A \cup B)^c$. Study tip: When dealing with compound set operations, calculate each component separately before combining them. Draw Venn diagrams if helpful, and always double-check that your final elements actually satisfy all the given conditions.
10. For any three sets $A$, $B$, and $C$ in a universal set $U$, which of the following statements is always true?
    Answer: $A \cap (B \setminus C) = (A \cap B) \setminus C$ We can analyze each statement using set properties:\ A) $A \setminus (B \cup C) = A \cap (B \cup C)' = A \cap (B' \cap C') = (A \cap B') \cap (A \cap C') = (A \setminus B) \cap (A \setminus C)$. The statement is false as it uses union instead of intersection.\ B) If $A \subseteq B$, it does not guarantee that $A \cup B' = U$. For example, let $U=\{1,2,3,4\}$, $B=\{1,2,3\}$, and $A=\{1,2\}$. Then $B'=\{4\}$. $A \cup B' = \{1,2,4\} \neq U$. The statement is false.\ C) This is an incorrect version of De Morgan's Laws. The correct law is $(A \cap B)' = A' \cup B'$. The statement is false.\ D) $A \cap (B \setminus C) = A \cap (B \cap C')$. Since set intersection is associative, this equals $(A \cap B) \cap C'$, which is the definition of $(A \cap B) \setminus C$. This statement is always true.
11. Given that $A \subseteq B \subseteq C$ and $|A| = 8$, $|B| = 12$, $|C| = 18$, which of the following statements about the sets $A^c \cap B$ and $B^c \cap C$ is necessarily true?
    Answer: $|A^c \cap B| = 4$ and $|B^c \cap C| = 6$ for any universe containing $C$ When you encounter set relationships with complements, focus on what the complement notation really means and whether it depends on a universal set context. Given $A \subseteq B \subseteq C$, let's analyze each expression. For $A^c \cap B$, this represents elements that are in $B$ but not in $A$. Since $A \subseteq B$, we can think of this as $B \setminus A$ (the relative complement of $A$ in $B$). This gives us $|A^c \cap B| = |B| - |A| = 12 - 8 = 4$, regardless of what universal set we choose, as long as it contains $B$. Similarly, $B^c \cap C$ represents elements in $C$ but not in $B$. Since $B \subseteq C$, this equals $C \setminus B$, giving us $|B^c \cap C| = |C| - |B| = 18 - 12 = 6$. Again, this is independent of the universal set choice. Looking at the answer choices: Choice A incorrectly suggests these values only hold when the universal set equals $C$. Choice B incorrectly claims $|B^c \cap C|$ depends on the universal set. Choice C incorrectly states both depend on the universal set. Choice D correctly recognizes both values are fixed for any universe containing $C$. The key insight is that when we have nested subsets like $A \subseteq B \subseteq C$, expressions like $A^c \cap B$ effectively become relative complements, making their cardinalities independent of the universal set choice. Study tip: When you see complement intersections with subset relationships, convert them to relative complements (set differences) to avoid universal set confusion.
12. In a survey of 100 consumers, it was found that 40 purchased product A, 30 purchased product B, and 55 purchased at least one of the two products. Let $U$ be the set of all consumers surveyed. Let $A$ be the set of consumers who purchased product A and $B$ be the set for product B. How many consumers purchased neither product A nor product B?
    Answer: 45 The set of consumers who purchased neither product A nor product B is represented by $A' \cap B'$. By De Morgan's Laws, this is equivalent to $(A \cup B)'$. The number of elements in this set is $|(A \cup B)'|$.\ We can find this by subtracting the number of consumers who purchased at least one product from the total number of consumers in the universal set.\ We are given $|U| = 100$ and $|A \cup B| = 55$ (the number who purchased at least one).\ Therefore, $|(A \cup B)'| = |U| - |A \cup B| = 100 - 55 = 45$.\ Distractor A is $|A \cap B| = |A| + |B| - |A \cup B| = 40 + 30 - 55 = 15$.\ Distractor B is $|U| - |A| - |B| = 100 - 40 - 30 = 30$, which incorrectly assumes the sets are disjoint.\ Distractor D is $|(A \cap B)'| = |U| - |A \cap B| = 100 - 15 = 85$.
13. Let $M$, $P$, and $C$ be the sets of students taking Mathematics, Physics, and Chemistry courses, respectively. Which set expression accurately represents the group of students who are taking Chemistry, and are also enrolled in either Mathematics or Physics, but not both?
    Answer: $C \cap ((M \cup P) \setminus (M \cap P))$ The problem describes students who satisfy two conditions simultaneously. The word 'and' implies an intersection.\ The first condition is 'taking Chemistry', which corresponds to the set $C$.\ The second condition is 'enrolled in either Mathematics or Physics, but not both'. This describes the symmetric difference of sets $M$ and $P$, which can be written as $(M \cup P) \setminus (M \cap P)$ or $(M \setminus P) \cup (P \setminus M)$.\ To find the set of students satisfying both conditions, we take the intersection of the sets representing each condition. This gives $C \cap ((M \cup P) \setminus (M \cap P))$.
14. Consider a universal set $U$ with three subsets $A$, $B$, and $C$. Which of the following set expressions represents the collection of elements that belong to exactly two of these three sets?
    Answer: $(A \cap B \setminus C) \cup (A \cap C \setminus B) \cup (B \cap C \setminus A)$ The elements belonging to 'exactly two' of the sets $A, B, C$ are those that are:\ - in $A$ and $B$, but not in $C$ ($A \cap B \setminus C$), OR\ - in $A$ and $C$, but not in $B$ ($A \cap C \setminus B$), OR\ - in $B$ and $C$, but not in $A$ ($B \cap C \setminus A$).\ The 'OR' corresponds to the union of these three disjoint sets. Therefore, the correct expression is $(A \cap B \setminus C) \cup (A \cap C \setminus B) \cup (B \cap C \setminus A)$.\ Distractor A represents elements in at least two sets. Distractor B represents elements in exactly one or exactly two sets. Distractor D does not represent the region correctly and is an invalid series of operations.
15. Let the universal set be $U = \{x \in \mathbb{Z} \mid 1 \le x \le 12\}$. Let $A$ be the set of prime numbers in $U$, $B$ be the set of even numbers in $U$, and $C$ be the set of multiples of 3 in $U$. What is the cardinality of the set $(A \cup B') \cap C'$?
    Answer: 5 First, we list the elements of each set:\ $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}$\ $A = \{2, 3, 5, 7, 11\}$\ $B = \{2, 4, 6, 8, 10, 12\}$\ $C = \{3, 6, 9, 12\}$\ Next, we perform the required operations in sequence:\ 1. Find the complement of $B$: $B' = U \setminus B = \{1, 3, 5, 7, 9, 11\}$.\ 2. Find the union of $A$ and $B'$: $A \cup B' = \{2, 3, 5, 7, 11\} \cup \{1, 3, 5, 7, 9, 11\} = \{1, 2, 3, 5, 7, 9, 11\}$.\ 3. Find the complement of $C$: $C' = U \setminus C = \{1, 2, 4, 5, 7, 8, 10, 11\}$.\ 4. Find the intersection of $(A \cup B')$ and $C'$: $(A \cup B') \cap C' = \{1, 2, 3, 5, 7, 9, 11\} \cap \{1, 2, 4, 5, 7, 8, 10, 11\} = \{1, 2, 5, 7, 11\}$.\ Finally, we find the cardinality (number of elements) of the resulting set, which is 5.

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