1. **Problem Statement:** We have a cube whose side length increases by 10%. We need to find: a) The percentage increase in the total surface area. b) The percentage increase in the volume. 2. **Formulas and Rules:** - Surface area of a cube with side length $s$ is $6s^2$. - Volume of a cube with side length $s$ is $s^3$. - If the side length increases by 10%, the new side length is $s_{new} = s \times 1.10$. 3. **Calculations for Surface Area Increase:** - Original surface area: $6s^2$. - New surface area: $6(s \times 1.10)^2 = 6s^2 \times (1.10)^2 = 6s^2 \times 1.21$. - Percentage increase in surface area: $$\frac{6s^2 \times 1.21 - 6s^2}{6s^2} \times 100 = (1.21 - 1) \times 100 = 21\%$$. 4. **Calculations for Volume Increase:** - Original volume: $s^3$. - New volume: $(s \times 1.10)^3 = s^3 \times (1.10)^3 = s^3 \times 1.331$. - Percentage increase in volume: $$\frac{s^3 \times 1.331 - s^3}{s^3} \times 100 = (1.331 - 1) \times 100 = 33.1\%$$. 5. **Problem Statement for Gift Box Volume:** - Standard box volume: 240 cm³ with length 10 cm. - Large box length: 15 cm. - Find the correct volume of the large box from options: 360, 540, 720, 810 cm³. 6. **Assuming similar shape, volume scales with the cube of the length ratio:** - Length ratio: $\frac{15}{10} = 1.5$. - Volume of large box: $$240 \times (1.5)^3 = 240 \times 3.375 = 810\text{ cm}^3$$. 7. **Answer:** - a) Surface area increases by 21%. - b) Volume increases by 33.1%. - Large box volume is 810 cm³ (option d).