1. **Find the equation of a graph passing through (0, -1) and (3, 5).** We assume a linear equation $y = mx + c$. 2. Use the points to find $m$ and $c$. At $(0, -1)$: $-1 = m \times 0 + c \Rightarrow c = -1$. At $(3, 5)$: $5 = m \times 3 - 1 \Rightarrow 3m = 6 \Rightarrow m = 2$. 3. Equation is $y = 2x - 1$. 4. **Find magnitude of $x$ where $O$ is center of circle and $x$ is angle at vertex $C$ on circumference.** By circle theorem, angle at center is twice angle at circumference. If angle at center is $2x$, and triangle properties or given info used, solve for $x$. (Since no numeric data given, assume $x$ is half the central angle.) 5. **Cylinder problem: height $h=12$ cm, area of one plane surface $=154$ cm², circumference $=44$ cm. Find total surface area.** 6. Circumference $C = 2\pi r = 44 \Rightarrow r = \frac{44}{2\pi} = \frac{22}{\pi}$. 7. Area of one plane surface (circle base) $= \pi r^2 = 154$. Check $\pi r^2 = 154 \Rightarrow r^2 = \frac{154}{\pi}$. Calculate $r$ from circumference and area to confirm consistency. 8. Total surface area $= 2\pi r h + 2\pi r^2$. Calculate: $2\pi r h = 2 \times \pi \times \frac{22}{\pi} \times 12 = 2 \times 22 \times 12 = 528$ cm². $2\pi r^2 = 2 \times 154 = 308$ cm². Total surface area $= 528 + 308 = 836$ cm². 9. **Sector perimeter $=64$ cm, radius $=21$ cm. Find angle of sector.** 10. Perimeter $= 2r + l = 64$ where $l$ is arc length. $2 \times 21 + l = 64 \Rightarrow l = 64 - 42 = 22$ cm. 11. Arc length $l = r \theta$ (in radians), so $\theta = \frac{l}{r} = \frac{22}{21} \approx 1.0476$ radians. Convert to degrees: $\theta = 1.0476 \times \frac{180}{\pi} \approx 60^\circ$. 12. **Tree location equidistant from walls AB and BC, 5m from B.** 13. The locus of points equidistant from two intersecting lines is the angle bisector. 14. The tree lies on the angle bisector of angle at B, at distance 5m from B. 15. **Square based right pyramid: height $=6$ cm, base length $=16$ cm. Find area of four triangular faces.** 16. Slant height $l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm (half base is 8 cm). 17. Area of one triangular face $= \frac{1}{2} \times base \times slant height = \frac{1}{2} \times 16 \times 10 = 80$ cm². 18. Total area of four triangular faces $= 4 \times 80 = 320$ cm². 19. **Events A and B independent, $P(A \cap B) = \frac{1}{5}$, $P(B) = \frac{3}{5}$. Find $P(A')$.** 20. Independence implies $P(A \cap B) = P(A)P(B)$. $\frac{1}{5} = P(A) \times \frac{3}{5} \Rightarrow P(A) = \frac{1/5}{3/5} = \frac{1}{3}$. 21. $P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$. --- **Part B:** 1. Auditorium with semicircular stage and trapezium seating. I. $AB$ is half of $DC = 56$ m, so $AB = 28$ m. Length of arc $AB = \pi r$ where $r = AB/2 = 14$ m. Arc length $= \pi \times 14 = 44$ m. II. Seating area trapezium $= 420$ m². Area trapezium $= \frac{1}{2} (AB + DC) \times h = 420$. $\frac{1}{2} (28 + 56) h = 420 \Rightarrow 42h = 420 \Rightarrow h = 10$ m. III. Total area = semicircle area + trapezium area. Semicircle area $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi 14^2 = 98\pi \approx 307.88$ m². Total area $= 307.88 + 420 = 727.88$ m². IV. Bulbs along arc AB spaced 2m apart including A and B. Number of bulbs $= \frac{44}{2} + 1 = 22 + 1 = 23$ bulbs. V. Bulbs along diameter AB (28 m) spaced 2m apart including ends. Number of bulbs $= \frac{28}{2} + 1 = 14 + 1 = 15$ bulbs. 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 took $\frac{3}{4}$ of remaining. Fraction distributed among main players excluding Amaan $= \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$.