1. **Problem:** Find the complement of the intersection of sets A and B, i.e., $(A \cap B)'$. Given: - $A = \{1, 2, 3, 5\}$ - $B = \{2, 4, 6, 8\}$ - Universe $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ 2. **Step:** Find $A \cap B$: - Common elements in A and B are $\{2\}$ 3. **Step:** The complement $(A \cap B)'$ means all elements in $U$ not in $A \cap B$: - So, $(A \cap B)' = U \setminus \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9, 10\}$ 4. **Answer:** $(A \cap B)' = \{1, 3, 4, 5, 6, 7, 8, 9, 10\}$ --- 1. **Problem:** Find $(C' \cup A) \cap B$. 2. **Step:** Find $C'$, the complement of $C$ in $U$: - $C = \{1, 5, 7, 8\}$ - So, $C' = U \setminus C = \{2, 3, 4, 6, 9, 10\}$ 3. **Step:** Calculate $C' \cup A$: - $C' \cup A = \{2, 3, 4, 6, 9, 10\} \cup \{1, 2, 3, 5\} = \{1, 2, 3, 4, 5, 6, 9, 10\}$ 4. **Step:** Find $(C' \cup A) \cap B$: - $B = \{2, 4, 6, 8\}$ - Intersection is $\{2, 4, 6\}$ 5. **Answer:** $(C' \cup A) \cap B = \{2, 4, 6\}$ --- 1. **Problem:** Find the complement of $(B \cap C')$, i.e., $(B \cap C')'$. 2. **Step:** Find $C'$ as before: $\{2, 3, 4, 6, 9, 10\}$. 3. **Step:** Calculate $B \cap C'$: - $B = \{2, 4, 6, 8\}$ - Intersection $= \{2, 4, 6\}$ 4. **Step:** Find the complement $(B \cap C')' = U \setminus \{2, 4, 6\} = \{1, 3, 5, 7, 8, 9, 10\}$ 5. **Answer:** $(B \cap C')' = \{1, 3, 5, 7, 8, 9, 10\}$ --- 1. **Problem:** Find $B \cup (A' \cap C)'$. 2. **Step:** Find $A'$: - $A = \{1, 2, 3, 5\}$ - $A' = U \setminus A = \{4, 6, 7, 8, 9, 10\}$ 3. **Step:** Find $A' \cap C$: - $C = \{1, 5, 7, 8\}$ - Intersection is $\{7, 8\}$ 4. **Step:** Find the complement $(A' \cap C)' = U \setminus \{7, 8\} = \{1, 2, 3, 4, 5, 6, 9, 10\}$ 5. **Step:** Find $B \cup (A' \cap C)'$: - $B = \{2, 4, 6, 8\}$ - Union $= \{1, 2, 3, 4, 5, 6, 8, 9, 10\}$ 6. **Answer:** $B \cup (A' \cap C)' = \{1, 2, 3, 4, 5, 6, 8, 9, 10\}$ --- 1. **Problem:** Write all subsets of the set $\{a, b, c, d, e\}$. 2. **Step:** A set with $n$ elements has $2^n$ subsets. - Here, $n = 5$, so total subsets = $2^5 = 32$. 3. **Step:** These subsets include: - Empty set $\emptyset$ - All singletons: $\{a\}, \{b\}, \{c\}, \{d\}, \{e\}$ - All pairs, triples, quadruples, and the full set 4. **Answer:** The 32 subsets (generally listed by size) are: $\emptyset$, $\{a\}$, $\{b\}$, $\{c\}$, $\{d\}$, $\{e\}$, $\{a,b\}$, $\{a,c\}$, $\{a,d\}$, $\{a,e\}$, $\{b,c\}$, $\{b,d\}$, $\{b,e\}$, $\{c,d\}$, $\{c,e\}$, $\{d,e\}$, $\{a,b,c\}$, $\{a,b,d\}$, $\{a,b,e\}$, $\{a,c,d\}$, $\{a,c,e\}$, $\{a,d,e\}$, $\{b,c,d\}$, $\{b,c,e\}$, $\{b,d,e\}$, $\{c,d,e\}$, $\{a,b,c,d\}$, $\{a,b,c,e\}$, $\{a,b,d,e\}$, $\{a,c,d,e\}$, $\{b,c,d,e\}$, $\{a,b,c,d,e\}$. --- 1. **Problem:** Find the rational number that is one fourth of the way from $\frac{1}{5}$ to $\frac{7}{10}$. 2. **Step:** Calculate the difference: - $\frac{7}{10} - \frac{1}{5} = \frac{7}{10} - \frac{2}{10} = \frac{5}{10} = \frac{1}{2}$ 3. **Step:** One fourth of the way is adding one fourth of the difference to the start point: - $\frac{1}{5} + \frac{1}{4} \times \frac{1}{2} = \frac{1}{5} + \frac{1}{8} = \frac{8}{40} + \frac{5}{40} = \frac{13}{40}$ 4. **Answer:** The rational number one fourth of the way is $\frac{13}{40}$. --- 1. **Problem:** Given 26 boys with numbers of textbooks owned as follows: - English (E): 19 - French (F): 23 - Mathematics (M): 15 - English & French (E \cap F): 16 - French & Mathematics (F \cap M): 14 - Mathematics & English (M \cap E): 13 2. **Part a:** Illustrate info in a Venn diagram (conceptual only here). 3. **Part b:** Calculate: (i) All three books (E \cap F \cap M): Let this be $x$. By inclusion-exclusion, number with all three books: $$x = |E| + |F| + |M| - |E \cap F| - |F \cap M| - |M \cap E| + x$$ But total boys = 26, so: $$26 = |E| + |F| + |M| - |E \cap F| - |F \cap M| - |M \cap E| + x$$ $$26 = 19 + 23 + 15 - 16 - 14 - 13 + x$$ $$26 = 57 - 43 + x$$ $$26 = 14 + x \implies x = 12$$ (ii) Two books only: Number with exactly two books = $$(|E \cap F| - x) + (|F \cap M| - x) + (|M \cap E| - x) = (16-12) + (14-12) + (13-12) = 4 + 2 + 1 = 7$$ (iii) Only one book: $$|E| - (|E \cap F| + |M \cap E| - x) = 19 - (16 + 13 - 12) = 19 - 17 = 2$$ $$|F| - (|E \cap F| + |F \cap M| - x) = 23 - (16 + 14 - 12) = 23 - 18 = 5$$ $$|M| - (|M \cap E| + |F \cap M| - x) = 15 - (13 + 14 - 12) = 15 - 15 = 0$$ Total only one book = $2 + 5 + 0 = 7$ (iv) English and French but not Mathematics: - $|E \cap F| - x = 16 - 12 = 4$ (v) Only French books: - From above, only French = 5 **Answer:** - (i) 12 - (ii) 7 - (iii) 7 - (iv) 4 - (v) 5