Paint Calculations Dad0A1

1. **Problem 2.2:** Calculate the amount of paint required to paint the shipping container with two coats. - Given: Spread rate = $8\text{ m}^2$ per litre, two coats means the total area to be painted is doubled. - Let the surface area of the container be $A$ (in $\text{m}^2$). - Total area to paint = $2 \times A$ (because of two coats). - Amount of paint required $= \frac{2A}{8} = \frac{A}{4}$ litres. 2. **Problem 2.3:** Calculate the cost to paint storage containers with the marine finish paint. - Paint price per 5 litre can = 1154.71 - Price per litre $= \frac{1154.71}{5} = 230.942$ (rounded to 3 decimals). - Amount of paint required from 2.2 is $\frac{A}{4}$ litres. - Number of cans needed $= \frac{\frac{A}{4}}{5} = \frac{A}{20}$ cans. - Since cans must be whole, round up to nearest whole number: $\lceil \frac{A}{20} \rceil$. - Total cost $= 1154.71 \times \lceil \frac{A}{20} \rceil$. 3. **Problem 2.4:** Determine the number of tanks that can be placed inside the tank. - Let the volume of the large tank be $V_L$ and the volume of one small tank be $V_S$. - Number of small tanks $= \left\lfloor \frac{V_L}{V_S} \right\rfloor$ (integer division). **Note:** The exact values of $A$, $V_L$, and $V_S$ are not provided in the prompt, so the answers are expressed in terms of these variables. --- **Final answers:** - Amount of paint required $= \frac{A}{4}$ litres. - Cost to paint $= 1154.71 \times \lceil \frac{A}{20} \rceil$. - Number of tanks inside $= \left\lfloor \frac{V_L}{V_S} \right\rfloor$.