1. Given the universal set $\mu = \{1, 2, 3, 4, 5, 6\}$, sets $A = \{2, 4, 6\}$ and $B = \{1, 2, 3, 4\}$. Find $A \cup B$ (the union). Step 1: The union of two sets combines all unique elements from both sets. Step 2: List elements in $A$: $2, 4, 6$. Step 3: List elements in $B$: $1, 2, 3, 4$. Step 4: Combine without duplicates: $\{1, 2, 3, 4, 6\}$. This matches none of the options exactly, but closest is option D $\{1, 2, 3, 4, 5\}$ missing $5$ instead of $6$. Likely correct answer: $\{1, 2, 3, 4, 6\}$. --- 2. Total students = $30$, $18$ take Mathematics (M), $5$ take both Mathematics and Biology (B), $8$ take neither. Find number taking Biology. Step 1: Students taking Math or Bio or both = $30 - 8 = 22$. Step 2: Let $x$ be number taking Biology. Use inclusion-exclusion: $$|M \cup B| = |M| + |B| - |M \cap B|$$ Step 3: Substitute known values: $$22 = 18 + x - 5$$ Step 4: Solve for $x$: $$22 = 13 + x$$ $$x = 9$$ Answer is option C: $9$. --- 3. Simplify $$ \frac{2}{a^2 - 2b} + \frac{1}{a + 2b} $$ Step 1: Notice $a^2 - 2b$ is not easily factorable with $a+2b$; verify if problem meant $a^2 - 4b^2$. Assuming denominator is $a^2 - 4b^2 = (a-2b)(a+2b)$ (likely typo in question). Step 2: Rewrite: $$ \frac{2}{a^2 - 4b^2} + \frac{1}{a + 2b} = \frac{2}{(a-2b)(a+2b)} + \frac{1}{a+2b} $$ Step 3: Get common denominator $(a - 2b)(a + 2b)$: $$ \frac{2}{(a-2b)(a+2b)} + \frac{1 \cdot (a-2b)}{(a+2b)(a-2b)} = \frac{2 + a - 2b}{(a-2b)(a+2b)} = \frac{a + 2 - 2b}{a^2 - 4b^2} $$ Step 4: Since $a + 2 - 2b$ does not match given choices, check for error. Possibly $2$ should be $0$. Alternatively, check option A: $\frac{3a + 2b}{a^2 - 4b^2}$ or B: $\frac{a + 6b}{a^2 - 4b^2}$. Recalculate numerator: $$2 + a - 2b = a + 2 - 2b$$ is not matching options. Considering possible original expression typo, best matching answer is A. --- 4. Factorise $2xy - 6pq - 3py + 4qx$. Step 1: Group terms: $$(2xy - 3py) + (-6pq + 4qx)$$ Step 2: Factor each group: $$y(2x - 3p) + 2q(-3p + 2x)$$ Step 3: Note $-3p + 2x = -(3p - 2x)$, rewrite second group: $$y(2x - 3p) - 2q(3p - 2x)$$ Or rewrite as: $$y(2x - 3p) + 2q(x - 1.5p)$$ not matching form. Step 4: Swap terms or reorder factors: Try factoring as $(y + 2q)(2x - 3p)$ Expand $(y + 2q)(2x -3p) = 2xy - 3py + 4qx - 6pq$, matches original. Answer: A. --- 5. Interior angles of pentagon are $(x + 5)^6$, $(x + 15)^6$, and $(x + 20)^6$. Find $x$. Step 1: Sum of interior angles of pentagon: $$180^6 \times (5 - 2) = 540^6$$ Step 2: Sum given angles + 2 unknown (assume 2 unknown angles sum to $S$): $$ (x + 5) + (x + 15) + (x + 20) + S = 540 $$ Step 3: Sum simplified: $$ 3x + 40 + S = 540 \\ 3x + S = 500 $$ If only three angles are given as interior angles and rest are known or zero, need more info. Assuming these three are all interior angles: $$3x + 40 = 540 \\ 3x = 500 \\ x = \frac{500}{3} = 166.67$$ incorrect. Possibility question incomplete. Using answer options, closest is 68. --- 6. Find value of $\angle RPT$ given $\angle R = 40^\circ$ in pentagon with arrows. Given options include 40°, 23°, 65°, 73°. Assuming $\angle RPT$ equals $40^\circ$ as per given angle at R. Answer: C. --- 7. Evaluate $$ \frac{49^{\frac{2}{3}} + 49^{\frac{1}{6}}}{(\frac{1}{7})^{\frac{7}{6}} \times (\frac{1}{7})^{\frac{1}{6}}} $$ Step 1: Simplify numerator: $$49^{\frac{2}{3}} = (7^2)^{\frac{2}{3}} = 7^{\frac{4}{3}}$$ $$49^{\frac{1}{6}} = (7^2)^{\frac{1}{6}} = 7^{\frac{1}{3}}$$ Step 2: Sum numerator: $$7^{\frac{4}{3}} + 7^{\frac{1}{3}} = 7^{\frac{1}{3}}(7^{1} + 1) = 7^{\frac{1}{3}}(7 + 1) = 8 \times 7^{\frac{1}{3}}$$ Step 3: Simplify denominator: $$(\frac{1}{7})^{\frac{7}{6}} \times (\frac{1}{7})^{\frac{1}{6}} = (\frac{1}{7})^{\frac{8}{6}} = 7^{-\frac{8}{6}} = 7^{-\frac{4}{3}}$$ Step 4: Division becomes multiplication by reciprocal: $$ \frac{8 \times 7^{\frac{1}{3}}}{7^{-\frac{4}{3}}} = 8 \times 7^{\frac{1}{3} + \frac{4}{3}} = 8 \times 7^{\frac{5}{3}} $$ Final answer: $$8 \times 7^{\frac{5}{3}}$$ ---