1. **List the members of the following sets:** (i) Set defined by $x + 3 = 9$. - Step 1: Solve for $x$. $$x + 3 = 9$$ - Step 2: Subtract 3 from both sides. $$x = 9 - 3$$ $$x = 6$$ - So, the set is $\{6\}$. (ii) Set defined by $x^2 + 1 = 0$, where $x$ is a real number. - Step 1: Solve for $x^2$. $$x^2 = -1$$ - Step 2: Since $x^2$ cannot be negative for real numbers, there are no real solutions. - So, the set is empty: $\{\}$. (iii) Set defined by $x^2 = 16$ and $3x = 9$. - Step 1: Solve $3x = 9$. $$x = \frac{9}{3} = 3$$ - Step 2: Check if $x=3$ satisfies $x^2 = 16$. $$3^2 = 9 \neq 16$$ - Step 3: Since $x=3$ does not satisfy $x^2=16$, no element satisfies both conditions simultaneously. - So, the set is empty: $\{\}$. 2. **Find $n(P)$ given $n(P \cup Q) = 54$, $n(P \cap Q) = 8$, and $n(Q) = 27$.** - Use the formula for union of two sets: $$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$$ - Substitute known values: $$54 = n(P) + 27 - 8$$ - Simplify: $$54 = n(P) + 19$$ - Solve for $n(P)$: $$n(P) = 54 - 19 = 35$$ 3. **Find how many boys study both metal work and French given:** - Boys studying metal work: 8 - Boys studying French: 12 - Boys studying French or metal work (union): 15 - Use the formula: $$n(Metal \cup French) = n(Metal) + n(French) - n(Metal \cap French)$$ - Substitute values: $$15 = 8 + 12 - n(Metal \cap French)$$ - Simplify: $$15 = 20 - n(Metal \cap French)$$ - Solve for intersection: $$n(Metal \cap French) = 20 - 15 = 5$$ 4. **Find how many students like beans but not bread given:** - Total students: 20 - Students who like bread: 14 - Students who like bread but not beans: 9 - Step 1: Find students who like both bread and beans. $$n(Bread \cap Beans) = n(Bread) - n(Bread \text{ but not } Beans) = 14 - 9 = 5$$ - Step 2: Find students who like beans. Let $n(Beans)$ be total who like beans. - Step 3: Use total students to find those who like beans but not bread. $$n(Beans \text{ but not } Bread) = n(Beans) - n(Bread \cap Beans)$$ - Step 4: Since total students = 20, and those who like bread only or both are $9 + 5 = 14$, the rest either like beans only or neither. - Step 5: We need more info to find $n(Beans)$, but assuming all students like either bread or beans or both, then: $$n(Beans \text{ but not } Bread) = 20 - 14 = 6$$ 5. **Find how many students take neither history nor economics given:** - Total students: 35 - Students taking history: 19 - Students taking economics: 12 - Students taking both: 5 - Use formula for union: $$n(History \cup Economics) = n(History) + n(Economics) - n(Both) = 19 + 12 - 5 = 26$$ - Students taking neither: $$35 - 26 = 9$$ **Final answers:** 1. (i) $\{6\}$, (ii) $\{\}$, (iii) $\{\}$ 2. $n(P) = 35$ 3. $5$ boys study both subjects 4. $6$ students like beans but not bread 5. $9$ students take neither subject