1. **Problem Statement:** For three consecutive homework assignments, each assignment has one more question than the previous one. The total number of questions on the three assignments is greater than 51. Let $q$ be the number of questions on the first assignment. 2. **Write the inequality:** - First assignment: $q$ questions - Second assignment: $q + 1$ questions - Third assignment: $q + 2$ questions The total number of questions is greater than 51, so: $$q + (q + 1) + (q + 2) > 51$$ 3. **Simplify the inequality:** $$q + q + 1 + q + 2 > 51$$ $$3q + 3 > 51$$ 4. **Solve the inequality:** Subtract 3 from both sides: $$3q + \cancel{3} - \cancel{3} > 51 - 3$$ $$3q > 48$$ Divide both sides by 3: $$\frac{3q}{\cancel{3}} > \frac{48}{\cancel{3}}$$ $$q > 16$$ 5. **Interpretation:** Since $q$ must be greater than 16, the smallest integer value for $q$ is 17. 6. **Minimum number of questions on the third assignment:** The third assignment has $q + 2$ questions. Substitute $q = 17$: $$17 + 2 = 19$$ **Answer:** The minimum number of questions on the third assignment is 19 because $q$ must be greater than 16 to satisfy the inequality, and the third assignment has two more questions than the first.