Venn Diagrams & Set Relations - Finite Mathematics

1. Consider sets $X$ and $Y$ where $X \subseteq Y \subseteq U$ and $|U| = 200$. If $|(X \cup Y)^c| = 60$ and $|X^c \cap Y| = 45$, what is $|X|$?
   Answer: $|X| = 95$ elements based on set complement relationships Since $X \subseteq Y$, we have $X \cup Y = Y$. Therefore, $|(X \cup Y)^c| = |Y^c| = 60$, which means $|Y| = 200 - 60 = 140$. The set $X^c \cap Y$ represents elements in $Y$ but not in $X$. Since $X \subseteq Y$, we have $Y = X \cup (X^c \cap Y)$ and these are disjoint. Therefore, $|Y| = |X| + |X^c \cap Y|$, which gives us $140 = |X| + 45$. Solving: $|X| = 95$. Choice A subtracts incorrectly. Choice C adds instead of subtracting. Choice D uses wrong intermediate values.
2. Given sets $A = \{x : x^2 - 5x + 6 \leq 0\}$ and $B = \{x : |x - 4| < 2\}$, what is $A \cap B$?
   Answer: $\{x : 2 < x \leq 3\}$ First, solve $x^2 - 5x + 6 \leq 0$. Factoring: $(x-2)(x-3) \leq 0$, so $A = [2,3]$. Next, solve $|x-4| < 2$, which gives $-2 < x-4 < 2$, so $2 < x < 6$, meaning $B = (2,6)$. The intersection $A \cap B = [2,3] \cap (2,6) = (2,3]$. Choice A incorrectly includes the endpoint 2. Choice C uses the wrong upper bound. Choice D excludes both correct endpoints.
3. For two sets, $A$ and $B$, the number of elements that belong to exactly one of the two sets is 30. If the total number of elements in $A \cup B$ is 42, how many elements are in $A \cap B$?
   Answer: 12 The set of elements in $A \cup B$ can be partitioned into three disjoint subsets: elements in $A$ only ($A \setminus B$), elements in $B$ only ($B \setminus A$), and elements in both ($A \cap B$). The number of elements that belong to 'exactly one' of the two sets is the sum of the cardinalities of the first two subsets: $|A \setminus B| + |B \setminus A|$. We are given this value is 30. The total number of elements in the union is the sum of the cardinalities of these three disjoint parts: $|A \cup B| = (|A \setminus B| + |B \setminus A|) + |A \cap B|$. We are given $|A \cup B| = 42$. Substituting the known values, we get $42 = 30 + |A \cap B|$. Solving for $|A \cap B|$ gives $42 - 30 = 12$.
4. Let $A$ and $B$ be subsets of a universal set $U$. Given $|U|=100$, $|A|=50$, and $|B|=65$, what is the minimum possible value of $|A \cap B'|$?
   Answer: 0 The set $A \cap B'$ represents the elements that are in $A$ but not in $B$, which is also written as $A \setminus B$. The cardinality is given by $|A \setminus B| = |A| - |A \cap B|$. To minimize $|A \setminus B|$, we must maximize the term being subtracted, $|A \cap B|$. The maximum possible size of the intersection of two sets is the size of the smaller set. In this case, max($|A \cap B|$) = min($|A|$, $|B|$) = min(50, 65) = 50$. This configuration is possible if $A$ is a subset of $B$. We must check if this fits within the universal set: if $A \subseteq B$, then $|A \cup B| = |B| = 65$, which is less than $|U|=100$. So, this is a valid configuration. Therefore, the maximum possible value of $|A \cap B|$ is 50. The minimum value of $|A \setminus B|$ is then $50 - 50 = 0$.
5. Based on the survey data, how many business owners use exactly two of these three types of online platforms?
   Answer: 65 Let $W$, $S$, and $E$ be the sets of owners with a Website, Social Media, and E-commerce platform, respectively. First, find the number of owners with at least one platform: $|W \cup S \cup E| = 200 - 15 = 185$. Next, use the Principle of Inclusion-Exclusion to find the number with all three, $|W \cap S \cap E|$. $|W \cup S \cup E| = |W| + |S| + |E| - (|W \cap S| + |W \cap E| + |S \cap E|) + |W \cap S \cap E|$. $185 = 120 + 90 + 70 - (50 + 35 + 25) + |W \cap S \cap E|$. $185 = 280 - 110 + |W \cap S \cap E|$, which gives $|W \cap S \cap E| = 15$. The number of owners with exactly two platforms is the sum of those with (W and S only), (W and E only), and (S and E only). Exactly W and S: $|W \cap S| - |W \cap S \cap E| = 50 - 15 = 35$. Exactly W and E: $|W \cap E| - |W \cap S \cap E| = 35 - 15 = 20$. Exactly S and E: $|S \cap E| - |W \cap S \cap E| = 25 - 15 = 10$. The total is $35 + 20 + 10 = 65$.
6. A Venn diagram for sets $A$, $B$, and $C$ is partially labeled with the cardinalities of its regions in terms of a variable $x$. The cardinalities for the regions representing elements in only one set are $|A \text{ only}| = 10$, $|B \text{ only}| = 15$, and $|C \text{ only}| = 20$. The cardinalities for the intersection regions are $|A \cap B \cap C'| = x$, $|A \cap C \cap B'| = 2x$, $|B \cap C \cap A'| = 3x$, and $|A \cap B \cap C| = 5$. If the total number of elements in set $A$ is 21, what is the total number of elements in the set $B \cup C$?
   Answer: 52 First, use the given total cardinality of set $A$ to find the value of $x$. The set $A$ consists of four disjoint regions: $|A| = |A \text{ only}| + |A \cap B \cap C'| + |A \cap C \cap B'| + |A \cap B \cap C|$. Plugging in the given values: $21 = 10 + x + 2x + 5$. This simplifies to $21 = 15 + 3x$, which gives $3x = 6$, so $x=2$. Now, we can find the cardinality of $B \cup C$. This is the sum of all disjoint regions that are part of $B$ or $C$: $|B \cup C| = |B \text{ only}| + |C \text{ only}| + |A \cap B \cap C'| + |A \cap C \cap B'| + |B \cap C \cap A'| + |A \cap B \cap C|$. Substituting the values: $|B \cup C| = 15 + 20 + x + 2x + 3x + 5 = 40 + 6x$. Now substitute $x=2$ into this expression: $|B \cup C| = 40 + 6(2) = 40 + 12 = 52$.
7. How many products in the inventory are Organic and Vegan, but not Handmade?
   Answer: 13 We need to find the number of products that are in the set 'Organic and Vegan' but not in the set 'Handmade'. This corresponds to the set cardinality $|(O \cap V) \setminus H|$ or $|O \cap V \cap H'|$. This can be calculated by taking the total number of products that are Organic and Vegan, $|O \cap V|$, and subtracting the number of products that are Organic, Vegan, AND Handmade, $|O \cap V \cap H|$. From the data provided, we have $|O \cap V| = 25$ and $|H \cap O \cap V| = 12$. Therefore, the number of products that are Organic and Vegan but not Handmade is $25 - 12 = 13$.
8. In a survey of smartphone users, 70% have app A installed, and 55% have app B installed. If 10% of users have neither app A nor app B installed, what percentage of users have both app A and app B installed?
   Answer: 35% Let $A$ be the set of users with app A and $B$ be the set with app B. We are given $|A|=70\%$ and $|B|=55\%$. The percentage of users with neither app is $|(A \cup B)'| = 10\%$. This means the percentage with at least one of the apps is $|A \cup B| = 100\% - 10\% = 90\%$. We can use the Principle of Inclusion-Exclusion, $|A \cup B| = |A| + |B| - |A \cap B|$, to find the percentage with both apps, $|A \cap B|$. Plugging in the values: $90\% = 70\% + 55\% - |A \cap B|$. This simplifies to $90\% = 125\% - |A \cap B|$. Solving for $|A \cap B|$ gives $|A \cap B| = 125\% - 90\% = 35\%$.
9. At a technology conference of 500 attendees, 350 use cloud service X, and 280 use cloud service Y. If every attendee uses at least one of these two services, how many attendees use service X but not service Y?
   Answer: 220 Let $X$ be the set of attendees using service X and $Y$ be the set for service Y. We are given $|X|=350$, $|Y|=280$. The condition 'every attendee uses at least one' means $|X \cup Y| = 500$. We want to find the number of attendees who use X but not Y, which is $|X \setminus Y|$. We can calculate this as $|X \setminus Y| = |X| - |X \cap Y|$. First, we must find $|X \cap Y|$ using the Principle of Inclusion-Exclusion: $|X \cup Y| = |X| + |Y| - |X \cap Y|$. Substituting the known values: $500 = 350 + 280 - |X \cap Y|$. This gives $500 = 630 - |X \cap Y|$, so $|X \cap Y| = 130$. Now we can find $|X \setminus Y|$: $|X \setminus Y| = |X| - |X \cap Y| = 350 - 130 = 220$.
10. If $A \subseteq B$ and $B \subseteq C$, and we know that $|A| = 12$, $|C| = 30$, and $|A \cup C| = 30$, what can we conclude about $|B|$?
    Answer: $|B|$ can be any value with $12 \leq |B| \leq 30$ Since $A \subseteq B \subseteq C$ and $|A \cup C| = |C| = 30$, this confirms that $A \subseteq C$ (which we already knew from transitivity). The subset relationships give us $|A| \leq |B| \leq |C|$, so $12 \leq |B| \leq 30$. The condition $|A \cup C| = 30$ doesn't provide additional constraints on $|B|$ beyond what the subset relationships already give us. Choice A assumes a specific value without justification. Choice C assumes $B = C$. Choice D incorrectly restricts the upper bound.
11. Let $A$, $B$, and $C$ be subsets of a universal set $U$. If $A \setminus B \subseteq C$, which of the following statements must be true?
    Answer: $A \subseteq B \cup C$ The condition $A \setminus B \subseteq C$ means that any element in $A$ but not in $B$ must also be in $C$. To check if $A \subseteq B \cup C$ must be true, consider an arbitrary element $x \in A$. There are two possibilities for $x$ with respect to set $B$: either $x \in B$ or $x \notin B$. Case 1: If $x \in B$, then $x$ is certainly in $B \cup C$. Case 2: If $x \notin B$, then since $x \in A$, we know $x \in A \setminus B$. By the given condition, this implies $x \in C$. If $x \in C$, then $x$ is also in $B \cup C$. In both cases, any element of $A$ is also an element of $B \cup C$. Therefore, $A \subseteq B \cup C$ must be true.

Start this quiz