1. **Problem 18:** A composite solid is made by drilling a cylindrical hole of radius 2 cm and height 10 cm through the center of a solid cylinder of radius 5 cm and height 10 cm. Find the volume of the remaining solid. Use \(\pi = 3.14\). 2. **Formula:** Volume of a cylinder is given by \(V = \pi r^2 h\). 3. **Step 1:** Calculate the volume of the large cylinder: $$V_{large} = 3.14 \times 5^2 \times 10 = 3.14 \times 25 \times 10 = 785 \text{ cm}^3$$ 4. **Step 2:** Calculate the volume of the drilled hole (small cylinder): $$V_{hole} = 3.14 \times 2^2 \times 10 = 3.14 \times 4 \times 10 = 125.6 \text{ cm}^3$$ 5. **Step 3:** Subtract the hole volume from the large cylinder volume to get the remaining volume: $$V_{remaining} = 785 - 125.6 = 659.4 \text{ cm}^3$$ --- 6. **Problem 19:** A closed box is a cube of side 20 cm. A smaller cube of side 8 cm is removed from its interior (completely taken out). Find the surface area of the remaining hollow box (assume the smaller cube is removed centrally and its faces are aligned with the larger cube faces) — include all exposed faces. 7. **Formula:** Surface area of a cube is \(6a^2\), where \(a\) is the side length. 8. **Step 1:** Calculate surface area of the large cube: $$SA_{large} = 6 \times 20^2 = 6 \times 400 = 2400 \text{ cm}^2$$ 9. **Step 2:** Calculate surface area of the smaller cube: $$SA_{small} = 6 \times 8^2 = 6 \times 64 = 384 \text{ cm}^2$$ 10. **Step 3:** When the smaller cube is removed centrally, the faces of the smaller cube become internal surfaces exposed inside the hollow. The outer surface area remains the same except the area where the smaller cube was removed is now open. 11. **Step 4:** The area of the hole on the large cube is the face of the smaller cube removed, which is \(8 \times 8 = 64 \text{ cm}^2\). 12. **Step 5:** The total surface area of the hollow box is the outer surface area plus the internal surface area exposed by the removed cube: $$SA_{hollow} = SA_{large} - 64 + SA_{small} = 2400 - 64 + 384 = 2720 \text{ cm}^2$$ --- 13. **Problem 20:** A cylindrical can has radius 4 cm and height 12 cm. If a label (a single rectangular sticker) wraps around the side (covers lateral surface) with a 1 cm overlap in width, what are the dimensions (height × width) of the smallest rectangular label that will cover the lateral surface? (Assume label height equals can height.) 14. **Formula:** Lateral surface area of a cylinder is \(A = 2 \pi r h\). 15. **Step 1:** Calculate the circumference of the base (width needed without overlap): $$C = 2 \times 3.14 \times 4 = 25.12 \text{ cm}$$ 16. **Step 2:** Add 1 cm overlap to the width: $$Width = 25.12 + 1 = 26.12 \text{ cm}$$ 17. **Step 3:** The height of the label equals the height of the can: $$Height = 12 \text{ cm}$$ 18. **Final answer:** The smallest rectangular label dimensions are \(12 \text{ cm} \times 26.12 \text{ cm}\).