Meshing gears (inverse revolutions): A 6-cog wheel meshes with a larger 14-cog wheel. When the smaller wheel makes 21 revolutions, how many revolutions does the larger wheel make?

Introduction / Context:
For meshed gears, revolutions are inversely proportional to the number of cogs (teeth). Fewer cogs imply more turns to pass the same number of teeth, and vice versa. We apply inverse proportion to map revolutions between the two wheels.

Given Data / Assumptions:

• Smaller wheel: 6 cogs; Larger wheel: 14 cogs.
• Smaller wheel makes 21 revolutions.
• No slipping; perfect meshing; same tooth engagement count.

Concept / Approach:
rev_small * cogs_small = rev_large * cogs_large. Solve for rev_large using the product equality of engaged teeth counts.

Step-by-Step Solution:

Verification / Alternative check:
Tooth counts: small engages 21 * 6 = 126 teeth; for the big gear to see 126 teeth with 14 per revolution, it must make 126/14 = 9 revolutions.

Why Other Options Are Wrong:

• 4 or 7: Undercount the necessary turns to match 126 engaged teeth.
• 14: Would over-rotate the larger gear compared to meshing constraints.

Common Pitfalls:
Using direct proportion (wrong here) instead of inverse proportion. Always remember: more cogs ⇒ fewer revolutions for a fixed tooth transfer.

Final Answer: