1. **Problem statement:** Given sets $A = \{3, 7, a^2\}$ and $B = \{2, 4, a+1, a+b\}$ with all elements integers, and $A \cap B = \{4, 7^2\}$, find: a) values of $a$ and $b$ b) $A \cup B$ 2. **Problem statement:** Among 100 students, 72 like English and 43 like Mathematics. Find: a) Number who like English only b) Number who like Mathematics only --- ### Solution for problem 1: 1. From $A \cap B = \{4, 7^2\}$, note that $7^2 = 49$. 2. Since $7 \in A$ but $7 \notin B$, the element $7$ is not in the intersection, so $7^2=49$ must be in both sets. 3. Therefore, $49 \in A$ and $49 \in B$. 4. Since $49 \in A = \{3,7,a^2\}$, and $3,7$ are fixed, $a^2 = 49$. 5. So, $a = \pm 7$. Since elements are integers, both are possible, but check $a+1$ and $a+b$ in $B$. 6. $4 \in B$ and $4$ is not in $A$ except possibly $a+1$ or $a+b$. 7. Since $4 \in B$, and $2$ is fixed, $a+1$ or $a+b$ must be $4$ or $49$. 8. Try $a=7$: - $a+1 = 7+1=8$ (not 4 or 49) - $a+b$ must be $4$ or $49$. - If $a+b=4$, then $b=4-7=-3$. - If $a+b=49$, then $b=42$. 9. Check if $b=-3$ or $42$ makes sense. 10. Since $B=\{2,4,a+1,a+b\} = \{2,4,8,4\}$ if $b=-3$, but $4$ repeats, sets do not have duplicates, so $a+b=4$ is already in $B$. 11. So $B=\{2,4,8,4\}$ reduces to $\{2,4,8\}$ which contradicts the problem. 12. Try $a+b=49$, then $b=42$. 13. Then $B=\{2,4,8,49\}$ which matches $A \cap B = \{4,49\}$. 14. So $a=7$, $b=42$. 15. $A=\{3,7,49\}$, $B=\{2,4,8,49\}$. 16. $A \cup B = \{2,3,4,7,8,49\}$. --- ### Solution for problem 2: 1. Total students = 100 2. Students liking English = 72 3. Students liking Mathematics = 43 4. Let $x$ be students liking both subjects. 5. Using inclusion-exclusion: $$72 + 43 - x = \text{number liking English or Mathematics} \leq 100$$ 6. So, $$115 - x \leq 100 \implies x \geq 15$$ 7. Assuming all students like at least one subject, $x=15$. 8. English only = $72 - x = 72 - 15 = 57$ 9. Mathematics only = $43 - x = 43 - 15 = 28$ --- **Final answers:** a) $a=7$, $b=42$ b) $A \cup B = \{2,3,4,7,8,49\}$ c) English only = 57 d) Mathematics only = 28