1980 AHSME Problems/Problem 10

Problem

The number of teeth in three meshed gears $A$ , $B$ , and $C$ are $x$ , $y$ , and $z$ , respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of $A$ , $B$ , and $C$ are in the proportion

$\text{(A)} \ x: y: z ~~\text{(B)} \ z: y: x ~~ \text{(C)} \ y: z: x~~ \text{(D)} \ yz: xz: xy ~~ \text{(E)} \ xz: yx: zy$

Solution

Since the gears are meshed, they must all rotate the same number of teeth per minute. Let $k$ be this rate. Then gears $A$ , $B$ , and $C$ must have angular speeds of $\frac{k}{x}$ , $\frac{k}{y}$ , and $\frac{k}{z}$ , respectively. Thus the ratio of the three gears' angular speeds is $\frac{k}{x}:\frac{k}{y}:\frac{k}{z}$ . Multiplying each term of this ratio by $\frac{xyz}{k}$ yields $\boxed{(\textbf{D})\ yz:xz:xy}$ .

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1980 AHSME Problems/Problem 10

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