1. **State the problem:** Harry needs more than $6 \frac{2}{5}$ cups of flour. He currently has $1 \frac{2}{3}$ cups and borrows flour in scoops of $\frac{1}{3}$ cup each from Dan. We want to find how many scoops $x$ Dan needs to add so that Harry has more than $6 \frac{2}{5}$ cups. 2. **Write the inequality:** $$\frac{1}{3}x + 1 \frac{2}{3} > 6 \frac{2}{5}$$ 3. **Convert mixed numbers to improper fractions:** $$1 \frac{2}{3} = \frac{5}{3}$$ $$6 \frac{2}{5} = \frac{32}{5}$$ 4. **Rewrite the inequality with improper fractions:** $$\frac{1}{3}x + \frac{5}{3} > \frac{32}{5}$$ 5. **Subtract $\frac{5}{3}$ from both sides:** $$\frac{1}{3}x + \frac{5}{3} - \frac{5}{3} > \frac{32}{5} - \frac{5}{3}$$ $$\frac{1}{3}x > \frac{32}{5} - \frac{5}{3}$$ 6. **Find common denominator for right side:** The denominators are 5 and 3, so common denominator is 15. $$\frac{32}{5} = \frac{32 \times 3}{15} = \frac{96}{15}$$ $$\frac{5}{3} = \frac{5 \times 5}{15} = \frac{25}{15}$$ 7. **Subtract fractions:** $$\frac{96}{15} - \frac{25}{15} = \frac{71}{15}$$ 8. **Inequality now:** $$\frac{1}{3}x > \frac{71}{15}$$ 9. **Multiply both sides by 3 to solve for $x$:** $$\cancel{3} \times \frac{1}{\cancel{3}} x > 3 \times \frac{71}{15}$$ $$x > \frac{213}{15}$$ 10. **Simplify fraction:** $$\frac{213}{15} = \frac{71}{5} = 14 \frac{1}{5}$$ **Final answer:** Dan needs to add more than $14 \frac{1}{5}$ scoops of flour. Since scoops must be whole, Dan must add at least 15 scoops.