1. **Problem statement:** Lily wants to paint her bedroom walls and ceiling. The room measures 4 m (length) × 3.5 m (width) × 2.5 m (height). There is a door (0.8 m × 2 m) and a window (0.96 m × 0.84 m) which will not be painted. (a)(i) Find the area of the ceiling. (a)(ii) Find the total area of the walls to be painted, excluding the door and window. (a)(iii) Calculate how many tins of paint Lily needs if she applies two coats on the ceiling and three coats on the walls. Each tin covers between 50 m² and 60 m². (a)(iv) Determine the cheapest way for Lily to buy at least 15 litres of paint given prices and offers. --- 2. **Step 1: Area of the ceiling** The ceiling is a rectangle with length 4 m and width 3.5 m. $$\text{Ceiling area} = 4 \times 3.5 = 14 \text{ m}^2$$ 3. **Step 2: Area of the walls** There are 4 walls: two walls of size 4 m × 2.5 m and two walls of size 3.5 m × 2.5 m. Calculate total wall area: $$2 \times (4 \times 2.5) + 2 \times (3.5 \times 2.5) = 2 \times 10 + 2 \times 8.75 = 20 + 17.5 = 37.5 \text{ m}^2$$ Subtract door and window areas: Door area: $$0.8 \times 2 = 1.6 \text{ m}^2$$ Window area: $$0.96 \times 0.84 = 0.8064 \text{ m}^2$$ Total area to paint on walls: $$37.5 - (1.6 + 0.8064) = 37.5 - 2.4064 = 35.0936 \text{ m}^2$$ Rounded to nearest m²: $$35 \text{ m}^2$$ 4. **Step 3: Total paint area with coats** Ceiling needs 2 coats: $$14 \times 2 = 28 \text{ m}^2$$ Walls need 3 coats: $$35.0936 \times 3 = 105.2808 \text{ m}^2$$ Total paint area: $$28 + 105.2808 = 133.2808 \text{ m}^2$$ 5. **Step 4: Number of tins needed** Each tin covers between 50 and 60 m². Use the worst case (50 m²) to ensure enough paint: $$\frac{133.2808}{50} = 2.6656$$ So Lily needs 3 tins of paint. 6. **Step 5: Cheapest option to buy at least 15 litres** Options: - 5-litre tins at 29 each - 10-litre tins at 53 each with "buy one get second half price" offer Calculate cost for 15 litres: - Using only 5-litre tins: $$3 \times 29 = 87$$ - Using 10-litre tins: Buy 1 at 53, second at half price 26.5, total for 20 litres: $$53 + 26.5 = 79.5$$ Since 20 litres > 15 litres, this satisfies the requirement. - Using 1 ten-litre + 1 five-litre tin: $$53 + 29 = 82$$ Cheapest is buying two 10-litre tins with the offer for 79.5. **Final answers:** - Ceiling area: $14$ m² - Walls area to paint: $35$ m² - Number of tins needed: $3$ - Cheapest purchase option: two 10-litre tins with offer costing 79.5