1. **Problem statement:** Two pipes can fill a tank together in 6 hours. The larger pipe works twice as fast as the smaller pipe. We need to find how long each pipe would take to fill the tank working separately. 2. **Define variables:** Let the time taken by the smaller pipe to fill the tank alone be $x$ hours. 3. **Rate of filling:** The smaller pipe's rate is $\frac{1}{x}$ tank per hour. The larger pipe works twice as fast, so its rate is $\frac{2}{x}$ tank per hour. 4. **Combined rate:** When both pipes work together, their rates add up: $$\frac{1}{x} + \frac{2}{x} = \frac{3}{x}$$ 5. **Given combined time:** Together they fill the tank in 6 hours, so their combined rate is $\frac{1}{6}$ tank per hour. 6. **Set up equation:** $$\frac{3}{x} = \frac{1}{6}$$ 7. **Solve for $x$:** Multiply both sides by $x$: $$3 = \frac{x}{6}$$ Multiply both sides by 6: $$18 = x$$ 8. **Interpretation:** The smaller pipe takes 18 hours to fill the tank alone. The larger pipe takes half that time, so: $$\frac{18}{2} = 9$$ hours. **Final answer:** - Smaller pipe: 18 hours - Larger pipe: 9 hours