1. **Calculate the length of the rod** The rod forms a right triangle with the wall and the ground. Given: - Distance from foot of rod to wall (base) = 3.6 m - Height of rod above ground = 4.8 m Use Pythagoras theorem: $$\text{Length} = \sqrt{(3.6)^2 + (4.8)^2}$$ $$= \sqrt{12.96 + 23.04} = \sqrt{36} = 6 \text{ m}$$ 2. **Calculate the density of the rod** Given: - Volume = 600 cm³ - Mass = 2.4 kg Density formula: $$\text{Density} = \frac{\text{Mass}}{\text{Volume}} = \frac{2.4}{600} = 0.004 \text{ kg/cm}^3$$ 3. **Determine the local time at town P** Given: - Town Q longitude = 40°E - Town P longitude = 50°W - Local time at Q = 4:00 p.m. Total longitude difference: $$40 + 50 = 90^\circ$$ Each 15° corresponds to 1 hour time difference: $$\frac{90}{15} = 6 \text{ hours}$$ Town P is west of Q, so local time at P is: $$4:00 \text{ p.m.} - 6 \text{ hours} = 10:00 \text{ a.m.}$$ 4. **Calculate import duty, excise duty, and VAT** Customs value = 1,200,000 (a) Import duty at 20%: $$1,200,000 \times 0.20 = 240,000$$ (b) Excise duty at 18%: $$1,200,000 \times 0.18 = 216,000$$ (c) VAT at 16%: VAT is charged on customs value + import duty + excise duty: $$\text{VAT base} = 1,200,000 + 240,000 + 216,000 = 1,656,000$$ $$\text{VAT} = 1,656,000 \times 0.16 = 264,960$$ 5. **Thanksgiving ceremony attendance** Ratio men:women:children = 5:7:3 Children = 60 Find the multiplier: $$\frac{60}{3} = 20$$ (a) Number of men: $$5 \times 20 = 100$$ (b) Number of women: $$7 \times 20 = 140$$ Difference women - men: $$140 - 100 = 40$$ 6. **Packing flour problem** 5 men work 6 hours/day for 12 days: Total man-hours: $$5 \times 6 \times 12 = 360$$ 3 men work 8 hours/day, let days = $d$: $$3 \times 8 \times d = 360$$ $$24d = 360 \Rightarrow d = \frac{360}{24} = 15$$ More days taken: $$15 - 12 = 3 \text{ days}$$ 7. **Express decimal 0.17 as fraction** $$0.17 = \frac{17}{100}$$ 8. **Name of quadrilateral with given properties** - Opposite sides equal - Diagonals bisect at 90° - Opposite angles equal This describes a rhombus. 9. **Length of wall on drawing** Scale 1:100, actual length = 3 m = 300 cm Drawing length: $$\frac{300}{100} = 3 \text{ cm}$$ 10. **Gradient of line L₂ perpendicular to L₁** Given L₁: $2y = x + 3 \Rightarrow y = \frac{1}{2}x + \frac{3}{2}$ Gradient of L₁ = $\frac{1}{2}$ Gradient of L₂ (perpendicular) = negative reciprocal: $$-2$$ 11. **New length of photograph** Original length = 16 cm Enlarged to twice size: $$16 \times 2 = 32 \text{ cm}$$