1. **Problem statement:** We want to find how many revolutions the smaller gear completes during one full revolution of the larger gear. 2. **Formula and concept:** The number of revolutions a gear makes is inversely proportional to its number of teeth (or size). If the larger gear has $N_L$ teeth and the smaller gear has $N_S$ teeth, then the ratio of their revolutions is given by: $$\text{Revolutions of smaller gear} = \frac{N_L}{N_S} \times \text{Revolutions of larger gear}$$ 3. **Given data:** - Larger gear teeth $N_L = 7$ - Smaller gear teeth $N_S = 3$ - Revolutions of larger gear = 1 4. **Calculate revolutions of smaller gear:** $$\text{Revolutions of smaller gear} = \frac{7}{3} \times 1 = \frac{7}{3}$$ 5. **Simplify the fraction:** $$\frac{7}{3} = 2 \frac{1}{3}$$ 6. **Interpretation:** The smaller gear completes $2 \frac{1}{3}$ revolutions during one revolution of the larger gear. **Final answer:** The smaller gear completes $\frac{7}{3}$ revolutions or approximately 2.33 revolutions during one revolution of the larger gear.