1. **Problem (a):** Luc uses $\frac{3}{4}$ of a tin per door and needs to paint 7 doors. Find the least number of tins needed. 2. **Formula:** Total paint needed $= 7 \times \frac{3}{4} = \frac{21}{4} = 5.25$ tins. 3. Since Luc cannot buy a fraction of a tin, he needs to buy at least 6 tins. --- 1. **Problem (b):** Jan buys tins for 17.16 each and sells at a 25% profit. Find the selling price. 2. **Formula:** Selling price $= \text{Cost price} \times (1 + \text{Profit rate})$ 3. Calculate: $17.16 \times 1.25 = 21.45$ --- 1. **Problem (c):** The cost 17.16 is 4% more than last year. Find last year's cost. 2. **Formula:** Current cost $= \text{Previous cost} \times 1.04$ 3. Calculate previous cost: $\frac{17.16}{1.04} = 16.5$ --- 1. **Problem (d):** Tin costs 17.16 dollars in America and 13.32 euros in Italy. Exchange rate $1 = 0.72$ euros. Find difference in dollars. 2. Convert euros to dollars: $\frac{13.32}{0.72} = 18.5$ dollars. 3. Difference: $18.5 - 17.16 = 1.34$ dollars. --- 1. **Problem (e)(i):** Convert 750 ml to cm$^3$. 2. Since 1 ml = 1 cm$^3$, 750 ml = 750 cm$^3$. --- 1. **Problem (e)(ii):** Calculate radius of cylindrical tin with height 11 cm and volume 750 cm$^3$. 2. Volume formula: $$V = \pi r^2 h$$ 3. Rearrange for radius: $$r = \sqrt{\frac{V}{\pi h}} = \sqrt{\frac{750}{\pi \times 11}}$$ 4. Calculate: $$r = \sqrt{\frac{750}{34.5575}} = \sqrt{21.7} = 4.7 \text{ cm (to 1 decimal place)}$$ --- 1. **Problem (e)(iii):** Similar tin with height 22 cm. Find volume in litres. 2. Since tins are similar, volume scales by cube of scale factor. 3. Scale factor for height: $\frac{22}{11} = 2$ 4. Volume scales by $2^3 = 8$ 5. New volume: $750 \times 8 = 6000$ cm$^3 = 6$ litres. --- 1. **Problem (f):** Mass of one tin is 890 g to nearest 10 g. Find upper bound for 10 tins. 2. Upper bound for one tin: $890 + 5 = 895$ g 3. Total upper bound: $895 \times 10 = 8950$ g --- 1. **Problem (g):** Probability a tin is dented is 0.07. Out of 3000 tins, expected dented tins? 2. Expected number: $3000 \times 0.07 = 210$ --- 1. **Problem (h):** Filling rate is 2 m$^3$ per minute. How many 750 ml tins filled in 1 hour? 2. Convert 1 hour to minutes: $60$ minutes. 3. Total volume filled: $2 \times 60 = 120$ m$^3$ 4. Convert 750 ml to m$^3$: $750$ ml = $750 \times 10^{-6} = 0.00075$ m$^3$ 5. Number of tins: $\frac{120}{0.00075} = 160000$ tins. **Final answers:** (a) 6 tins (b) 21.45 (c) 16.5 (d) 1.34 (e)(i) 750 cm$^3$ (e)(ii) 4.7 cm (e)(iii) 6 litres (f) 8950 g (g) 210 (h) 160000