1. **Problem statement:** Two warehouses A and B have the same workload. Alone, A takes 10 hours to move goods in warehouse A, B takes 12 hours in warehouse B, and C takes 15 hours alone. A starts moving goods in warehouse A, B starts moving goods in warehouse B simultaneously, and C helps A first, then switches to help B. We need to find how many hours C helped A and B. 2. **Define variables:** Let $x$ be the hours C helps A, and $y$ be the hours C helps B. 3. **Work rates:** - A's rate: $\frac{1}{10}$ workload/hour - B's rate: $\frac{1}{12}$ workload/hour - C's rate: $\frac{1}{15}$ workload/hour 4. **Work done in warehouse A:** A works alone for $(x+y)$ hours minus the time C helps A, so total time for A is $x + (T - x)$ where $T$ is total time to finish. But since C helps A for $x$ hours, the combined rate during those $x$ hours is $\frac{1}{10} + \frac{1}{15} = \frac{1}{6}$ workload/hour. For the remaining time $(T - x)$, A works alone at $\frac{1}{10}$ workload/hour. 5. **Work done in warehouse B:** Similarly, B works alone for $(T - y)$ hours and with C for $y$ hours. Combined rate during $y$ hours is $\frac{1}{12} + \frac{1}{15} = \frac{3}{20}$ workload/hour. 6. **Total work equals 1 for each warehouse:** For warehouse A: $$\frac{1}{6}x + \frac{1}{10}(T - x) = 1$$ For warehouse B: $$\frac{3}{20}y + \frac{1}{12}(T - y) = 1$$ 7. **Total time $T$ is the same for both warehouses:** Since A and B finish simultaneously, total time $T$ is the same. 8. **C's total working time:** C helps A for $x$ hours and B for $y$ hours, so total time C works is $x + y = T$. 9. **Rewrite equations:** From A: $$\frac{1}{6}x + \frac{1}{10}T - \frac{1}{10}x = 1 \implies \left(\frac{1}{6} - \frac{1}{10}\right)x + \frac{1}{10}T = 1$$ Calculate $\frac{1}{6} - \frac{1}{10} = \frac{5}{30} - \frac{3}{30} = \frac{2}{30} = \frac{1}{15}$: $$\frac{1}{15}x + \frac{1}{10}T = 1 \implies \frac{1}{15}x = 1 - \frac{1}{10}T$$ From B: $$\frac{3}{20}y + \frac{1}{12}T - \frac{1}{12}y = 1 \implies \left(\frac{3}{20} - \frac{1}{12}\right)y + \frac{1}{12}T = 1$$ Calculate $\frac{3}{20} - \frac{1}{12} = \frac{9}{60} - \frac{5}{60} = \frac{4}{60} = \frac{1}{15}$: $$\frac{1}{15}y + \frac{1}{12}T = 1 \implies \frac{1}{15}y = 1 - \frac{1}{12}T$$ 10. **Use $x + y = T$:** Multiply both sides of $x + y = T$ by $\frac{1}{15}$: $$\frac{1}{15}x + \frac{1}{15}y = \frac{1}{15}T$$ 11. **Substitute from above:** $$\left(1 - \frac{1}{10}T\right) + \left(1 - \frac{1}{12}T\right) = \frac{1}{15}T$$ Simplify left side: $$2 - \left(\frac{1}{10} + \frac{1}{12}\right)T = \frac{1}{15}T$$ Calculate $\frac{1}{10} + \frac{1}{12} = \frac{6}{60} + \frac{5}{60} = \frac{11}{60}$: $$2 - \frac{11}{60}T = \frac{1}{15}T$$ 12. **Solve for $T$:** Bring terms with $T$ to one side: $$2 = \frac{1}{15}T + \frac{11}{60}T = \left(\frac{4}{60} + \frac{11}{60}\right)T = \frac{15}{60}T = \frac{1}{4}T$$ Multiply both sides by 4: $$T = 8$$ 13. **Find $x$ and $y$:** From step 9: $$\frac{1}{15}x = 1 - \frac{1}{10}T = 1 - \frac{1}{10} \times 8 = 1 - 0.8 = 0.2$$ $$x = 0.2 \times 15 = 3$$ Similarly for $y$: $$\frac{1}{15}y = 1 - \frac{1}{12}T = 1 - \frac{1}{12} \times 8 = 1 - \frac{2}{3} = \frac{1}{3}$$ $$y = \frac{1}{3} \times 15 = 5$$ 14. **Answer:** C helped A for 3 hours and B for 5 hours.