A toothed wheel of diameter 50 cm is attached to a smaller wheel of diameter 30 cm. How many revolutions will the smaller wheel make when the larger one makes 15 revolutions?

### Concept & Logic For meshed or connected wheels/gears, the linear distance moved along the circumference is identical for both wheels: $$n_1 \times C_1 = n_2 \times C_2$$ Since $C = \pi d$, this simplifies to an inverse relationship between revolutions and diameters: $$n_1 \times d_1 = n_2 \times d_2$$ ### Step-by-Step Solution * **Given:** * Diameter of larger wheel ($d_1$) = $50\text{ cm}$ * Revolutions of larger wheel ($n_1$) = $15$ * Diameter of smaller wheel ($d_2$) = $30\text{ cm}$ * Let revolutions of smaller wheel be $n_2$. * **Using the relationship:** $$n_1 \times d_1 = n_2 \times d_2$$ $$15 \times 50 = n_2 \times 30$$ $$750 = 30n_2$$ $$n_2 = \frac{750}{30} = 25$$ ### Exam Strategy & Shortcut Direct ratio method: $$n_2 = n_1 \times \frac{d_1}{d_2} = 15 \times \frac{50}{30} = 15 \times \frac{5}{3} = 25$$ ### Common Pitfall Setting up a direct proportion ($\frac{n_1}{d_1} = \frac{n_2}{d_2}$) instead of an inverse proportion. The smaller wheel must turn more times than the larger wheel. ### Final Answer Therefore, the correct answer is **25**.