1. **Problem statement:** Find the thickness of the wall of a box with outer dimensions 9 cm, 7 cm, and 5 cm, given that the inner surface area is 142 cm² and the volume of a cylinder equals the outer volume of the box. 2. **Step 1: Calculate the outer volume of the box.** $$V_{outer} = 9 \times 7 \times 5 = 315 \text{ cm}^3$$ 3. **Step 2: Define the thickness of the wall as $t$.** The inner dimensions are: $$(9 - 2t), (7 - 2t), (5 - 2t)$$ 4. **Step 3: Calculate the inner surface area.** The inner surface area is given as 142 cm², and the formula for surface area of a box is: $$S = 2(lw + lh + wh)$$ So, $$2[(9 - 2t)(7 - 2t) + (7 - 2t)(5 - 2t) + (9 - 2t)(5 - 2t)] = 142$$ 5. **Step 4: Expand and simplify the expression inside the brackets:** $$(9 - 2t)(7 - 2t) = 63 - 18t - 14t + 4t^2 = 63 - 32t + 4t^2$$ $$(7 - 2t)(5 - 2t) = 35 - 14t - 10t + 4t^2 = 35 - 24t + 4t^2$$ $$(9 - 2t)(5 - 2t) = 45 - 18t - 10t + 4t^2 = 45 - 28t + 4t^2$$ 6. **Step 5: Sum these:** $$63 - 32t + 4t^2 + 35 - 24t + 4t^2 + 45 - 28t + 4t^2 = (63 + 35 + 45) - (32t + 24t + 28t) + (4t^2 + 4t^2 + 4t^2)$$ $$= 143 - 84t + 12t^2$$ 7. **Step 6: Multiply by 2 and set equal to 142:** $$2(143 - 84t + 12t^2) = 142$$ $$286 - 168t + 24t^2 = 142$$ 8. **Step 7: Rearrange to form a quadratic equation:** $$24t^2 - 168t + 286 - 142 = 0$$ $$24t^2 - 168t + 144 = 0$$ 9. **Step 8: Simplify by dividing entire equation by 24:** $$\cancel{24}t^2 - \cancel{168}t + \cancel{144} = 0 \Rightarrow t^2 - 7t + 6 = 0$$ 10. **Step 9: Factorize the quadratic:** $$(t - 6)(t - 1) = 0$$ 11. **Step 10: Solve for $t$:** $$t = 6 \text{ or } t = 1$$ 12. **Step 11: Check for physical validity:** Thickness cannot be 6 cm because it would make inner dimensions negative or zero. So, thickness $t = 1$ cm. **Final answer:** The thickness of the wall is **1 cm**.