1. **Find the equation of a graph passing through (0, -1) and (3, 5).** We assume the graph is a straight line, so use the slope-intercept form: $$y = mx + c$$ 2. Calculate the slope $m$: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - (-1)}{3 - 0} = \frac{6}{3} = 2$$ 3. Use point (0, -1) to find $c$: $$-1 = 2 \times 0 + c \Rightarrow c = -1$$ 4. Equation of the line: $$y = 2x - 1$$ --- 20. **Find the magnitude of angle $x$ at center $O$ of the circle with triangle $OAB$ inscribed.** Since $O$ is the center, angle $x$ is the central angle subtended by chord $AB$. If $OAB$ is isosceles with $OA = OB$ (radii), then angles at $A$ and $B$ are equal. Using circle properties, the central angle $x$ equals twice the inscribed angle subtending the same arc. Without numeric values, the magnitude of $x$ depends on given lengths or angles; more info needed. --- 21. **Find total surface area of a right cylinder with height 12 cm, one plane surface area 154 cm², and base circumference 44 cm.** 1. The plane surface area given is likely the base area: $$\text{Area} = 154 = \pi r^2$$ 2. Circumference: $$44 = 2 \pi r \Rightarrow r = \frac{44}{2\pi} = \frac{22}{\pi}$$ 3. Calculate $r^2$: $$r^2 = \left(\frac{22}{\pi}\right)^2 = \frac{484}{\pi^2}$$ 4. Check base area: $$\pi r^2 = \pi \times \frac{484}{\pi^2} = \frac{484}{\pi} \approx 154$$ (consistent) 5. Lateral surface area: $$2 \pi r h = 2 \pi \times \frac{22}{\pi} \times 12 = 2 \times 22 \times 12 = 528$$ 6. Total surface area: $$2 \times \text{base area} + \text{lateral area} = 2 \times 154 + 528 = 308 + 528 = 836$$ --- 22. **Find the angle of a sector with perimeter 64 cm and radius 21 cm.** 1. Perimeter of sector: $$P = 2r + l = 64$$ 2. Arc length $l$: $$l = r \theta$$ where $\theta$ in radians. 3. Substitute: $$64 = 2 \times 21 + 21 \theta \Rightarrow 64 = 42 + 21 \theta$$ 4. Solve for $\theta$: $$21 \theta = 22 \Rightarrow \theta = \frac{22}{21} \approx 1.0476 \text{ radians}$$ 5. Convert to degrees: $$\theta = 1.0476 \times \frac{180}{\pi} \approx 60^\circ$$ --- 23. **Locate point $T$ equidistant from walls $AB$ and $BC$ and 5m from $B$.** 1. The locus of points equidistant from two intersecting lines is the angle bisector. 2. Draw angle bisector of angle $ABC$. 3. Mark point $T$ on this bisector such that distance $BT = 5$ m. --- 24. **Find area of four triangular faces of a square-based right pyramid with height 6 cm and base length 16 cm.** 1. Base side $a = 16$ cm. 2. Height $h = 6$ cm. 3. Slant height $l$ of triangular face: $$l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ 4. Area of one triangular face: $$\frac{1}{2} \times a \times l = \frac{1}{2} \times 16 \times 10 = 80$$ 5. Total area of four triangular faces: $$4 \times 80 = 320$$ --- 25. **Given $P(A \cap B) = \frac{1}{5}$ and $P(B) = \frac{3}{5}$, find $P(A^c)$.** 1. Since $A$ and $B$ are independent: $$P(A \cap B) = P(A) P(B)$$ 2. Substitute: $$\frac{1}{5} = P(A) \times \frac{3}{5} \Rightarrow P(A) = \frac{1/5}{3/5} = \frac{1}{3}$$ 3. Complement: $$P(A^c) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$$ --- **Part B** 1. Auditorium with semicircular stage and trapezium seating area: I. Given $AB = \frac{1}{2} DC$, $DC = 56$ m, so $AB = 28$ m. Length of arc $AB$ (semicircle with diameter $AB$): $$\text{Arc length} = \pi r = \pi \times \frac{AB}{2} = \pi \times 14 = 44$$ II. Seating area trapezium = 420 m², bases $AB=28$ m, $DC=56$ m. Height $h$: $$\text{Area} = \frac{1}{2} (AB + DC) h \Rightarrow 420 = \frac{1}{2} (28 + 56) h = 42 h$$ $$h = \frac{420}{42} = 10$$ III. Total area = semicircle area + trapezium area: Semicircle area: $$\frac{1}{2} \pi r^2 = \frac{1}{2} \pi 14^2 = 98 \pi \approx 308.4$$ Total area: $$308.4 + 420 = 728.4$$ IV. Bulbs along arc $AB$ every 2 m including ends: Number of bulbs: $$\frac{44}{2} + 1 = 22 + 1 = 23$$ V. Bulbs along diameter $AB = 28$ m every 2 m including ends: Number of bulbs: $$\frac{28}{2} + 1 = 14 + 1 = 15$$ Additional bulbs needed: 15 --- 2. Cricket team prize money distribution: I. Amaan took $\frac{1}{5}$, remaining: $$1 - \frac{1}{5} = \frac{4}{5}$$ II. Remaining main players (excluding Amaan) took $\frac{3}{4}$ of remaining: Fraction distributed among main players excluding Amaan: $$\frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$$ --- Final answers: 1. $y = 2x - 1$ 21. Total surface area = 836 cm² 22. Sector angle = $60^\circ$ 24. Area of four triangular faces = 320 cm² 25. $P(A^c) = \frac{2}{3}$ Part B1.I. Arc length = 44 m Part B1.II. Height = 10 m Part B1.III. Total area = 728.4 m² Part B1.IV. Bulbs on arc = 23 Part B1.V. Additional bulbs on diameter = 15 Part B2.I. Remaining fraction = $\frac{4}{5}$ Part B2.II. Fraction distributed among main players excluding Amaan = $\frac{3}{5}$