1. **State the problem:** A couple traveled a total distance $D$ during a camping trip. They went one-third of the way by boat, 10 miles by foot, and one-sixth of the way by horse. We need to find the total distance $D$. 2. **Set up the equation:** Let the total distance be $D$ miles. They traveled: - By boat: $\frac{1}{3}D$ - By foot: 10 miles - By horse: $\frac{1}{6}D$ The sum of these parts equals the total distance: $$\frac{1}{3}D + 10 + \frac{1}{6}D = D$$ 3. **Combine like terms:** $$\frac{1}{3}D + \frac{1}{6}D = \frac{2}{6}D + \frac{1}{6}D = \frac{3}{6}D = \frac{1}{2}D$$ So the equation becomes: $$\frac{1}{2}D + 10 = D$$ 4. **Isolate $D$:** Subtract $\frac{1}{2}D$ from both sides: $$\cancel{\frac{1}{2}D} + 10 = \cancel{\frac{1}{2}D} + D - \frac{1}{2}D$$ $$10 = D - \frac{1}{2}D = \frac{1}{2}D$$ 5. **Solve for $D$:** Multiply both sides by 2: $$2 \times 10 = 2 \times \frac{1}{2}D$$ $$20 = D$$ 6. **Answer:** The total distance of the trip is **20 miles**. **Final answer: Option B, 20 miles.