Container Capacity A9Baa7

1. **State the problem:** We need to find the total capacity of the container in gallons. 2. **Define variables:** Let $C$ be the total capacity of the container in gallons. 3. **Translate the problem into equations:** - Initially, the container is $\frac{1}{3}$ full, so it contains $\frac{1}{3}C$ gallons of water. - After adding 13 gallons, the container is $\frac{1}{6}$ full, so it contains $\frac{1}{6}C$ gallons. 4. **Set up the equation:** $$\frac{1}{3}C + 13 = \frac{1}{6}C$$ 5. **Solve for $C$:** $$\frac{1}{3}C + 13 = \frac{1}{6}C$$ Subtract $\frac{1}{6}C$ from both sides: $$\frac{1}{3}C - \frac{1}{6}C + 13 = 0$$ Find common denominator and subtract: $$\frac{2}{6}C - \frac{1}{6}C + 13 = 0$$ $$\frac{1}{6}C + 13 = 0$$ Subtract 13 from both sides: $$\frac{1}{6}C = -13$$ This result is negative, which contradicts the problem context, so let's re-examine the problem statement carefully. **Re-examining:** The problem states the container is $\frac{1}{3}$ full initially, and after adding 13 gallons, it is $\frac{1}{6}$ full. Since $\frac{1}{6}$ is less than $\frac{1}{3}$, adding water cannot reduce the water level. Possibly the problem meant the container was $\frac{1}{6}$ full after pouring out 13 gallons, or the fractions are reversed. Assuming the problem meant the container was $\frac{1}{6}$ full initially and $\frac{1}{3}$ full after adding 13 gallons, then: $$\frac{1}{6}C + 13 = \frac{1}{3}C$$ 6. **Solve the corrected equation:** $$\frac{1}{6}C + 13 = \frac{1}{3}C$$ Subtract $\frac{1}{6}C$ from both sides: $$13 = \frac{1}{3}C - \frac{1}{6}C$$ $$13 = \frac{2}{6}C - \frac{1}{6}C$$ $$13 = \frac{1}{6}C$$ Multiply both sides by 6: $$6 \times 13 = C$$ $$78 = C$$ 7. **Answer:** The container can hold **78 gallons** of water.