1. **State the problem:** Find the shortest route from Kingwood to Richmond using the given paths and distances. 2. **List the paths and distances:** - Kingwood to Silvergrove: $4 \frac{2}{5}$ miles - Silvergrove to Arlington: $10 \frac{11}{16}$ miles - Arlington to Richmond (longer path): $14 \frac{9}{16}$ miles - Arlington to Richmond (shorter path): $6 \frac{3}{5}$ miles 3. **Convert mixed numbers to improper fractions for easier calculation:** - $4 \frac{2}{5} = \frac{22}{5}$ - $10 \frac{11}{16} = \frac{171}{16}$ - $14 \frac{9}{16} = \frac{233}{16}$ - $6 \frac{3}{5} = \frac{33}{5}$ 4. **Calculate total distances for possible routes:** - Route 1: Kingwood → Silvergrove → Arlington → Richmond (longer path) $$\frac{22}{5} + \frac{171}{16} + \frac{233}{16}$$ - Route 2: Kingwood → Silvergrove → Arlington → Richmond (shorter path) $$\frac{22}{5} + \frac{171}{16} + \frac{33}{5}$$ 5. **Find common denominators and add fractions:** - For Route 1: Common denominator for 5 and 16 is 80. $$\frac{22}{5} = \frac{22 \times 16}{80} = \frac{352}{80}$$ $$\frac{171}{16} = \frac{171 \times 5}{80} = \frac{855}{80}$$ $$\frac{233}{16} = \frac{233 \times 5}{80} = \frac{1165}{80}$$ Sum: $$\frac{352}{80} + \frac{855}{80} + \frac{1165}{80} = \frac{2372}{80}$$ - For Route 2: $$\frac{22}{5} = \frac{352}{80}$$ $$\frac{171}{16} = \frac{855}{80}$$ $$\frac{33}{5} = \frac{33 \times 16}{80} = \frac{528}{80}$$ Sum: $$\frac{352}{80} + \frac{855}{80} + \frac{528}{80} = \frac{1735}{80}$$ 6. **Compare the sums:** - Route 1 total distance: $\frac{2372}{80} = 29 \frac{27}{40}$ miles - Route 2 total distance: $\frac{1735}{80} = 21 \frac{35}{80} = 21 \frac{7}{16}$ miles 7. **Conclusion:** The shortest route from Kingwood to Richmond is via Silvergrove and Arlington using the shorter Arlington to Richmond path, totaling $21 \frac{7}{16}$ miles. **Final answer:** $21 \frac{7}{16}$ mi