1. **List the members of the following sets:** (i) Set: $\{x : x + 3 = 9\}$ - Step 1: Solve the equation $x + 3 = 9$. - Step 2: Subtract 3 from both sides: $x = 9 - 3$. - Step 3: Simplify: $x = 6$. - Step 4: Therefore, the set is $\{6\}$. (ii) Set: $\{x : x^2 + 1 = 0, x \text{ is real number}\}$ - Step 1: Solve $x^2 + 1 = 0$. - Step 2: Rearrange: $x^2 = -1$. - Step 3: Since $x^2$ cannot be negative for real $x$, there is no real solution. - Step 4: Therefore, the set is empty: $\emptyset$. (iii) Set: $\{x : x^2 = 16 \text{ and } 3x = 9\}$ - Step 1: Solve $3x = 9$ for $x$: $x = \frac{9}{3} = 3$. - Step 2: Check if $x=3$ satisfies $x^2 = 16$. - Step 3: Calculate $3^2 = 9 \neq 16$. - Step 4: No $x$ satisfies both conditions simultaneously. - Step 5: Therefore, the set is empty: $\emptyset$. 2. **Find $n(P)$ given $n(P \cup Q) = 54$, $n(P \cap Q) = 8$, and $n(Q) = 27$.** - Step 1: Use the formula for union of two sets: $$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$$ - Step 2: Substitute known values: $$54 = n(P) + 27 - 8$$ - Step 3: Simplify: $$54 = n(P) + 19$$ - Step 4: 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 (or both): 15 - Step 1: Use the formula: $$n(M \cup F) = n(M) + n(F) - n(M \cap F)$$ - Step 2: Substitute values: $$15 = 8 + 12 - n(M \cap F)$$ - Step 3: Simplify: $$15 = 20 - n(M \cap F)$$ - Step 4: Solve for $n(M \cap F)$: $$n(M \cap F) = 20 - 15 = 5$$ 4. **Find how many students like beans but not bread given:** - Total students: 20 - Students who like bread but not beans: 9 - Students who like bread: 14 - Step 1: Find students who like both bread and beans: $$n(\text{bread and beans}) = n(\text{bread}) - n(\text{bread but not beans}) = 14 - 9 = 5$$ - Step 2: Find students who like beans (total): $$n(\text{beans}) = n(\text{beans but not bread}) + n(\text{bread and beans})$$ - Step 3: Total students = 20, so: $$20 = n(\text{bread but not beans}) + n(\text{beans but not bread}) + n(\text{bread and beans})$$ - Step 4: Substitute known values: $$20 = 9 + n(\text{beans but not bread}) + 5$$ - Step 5: Simplify: $$20 = 14 + n(\text{beans but not bread})$$ - Step 6: Solve for $n(\text{beans but not bread})$: $$n(\text{beans but not bread}) = 20 - 14 = 6$$ 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 - Step 1: Use formula for union: $$n(H \cup E) = n(H) + n(E) - n(H \cap E) = 19 + 12 - 5 = 26$$ - Step 2: Students taking neither: $$n(\text{neither}) = \text{Total} - n(H \cup E) = 35 - 26 = 9$$ **Final answers:** 1. (i) $\{6\}$, (ii) $\emptyset$, (iii) $\emptyset$ 2. $n(P) = 35$ 3. Number of boys studying both subjects = 5 4. Number of students who like beans but not bread = 6 5. Number of students taking neither subject = 9