Difference between revisions of "1980 AHSME Problems/Problem 10"

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== Solution ==
 
== Solution ==
The distance that each of the gears rotate is constant. Let us have the number of teeth per minute equal to <math>k</math>. The revolutions per minute are in ratio of:
+
Since the gears are meshed, they must all rotate the same number of teeth per minute.  Let <math>k</math> be this rate. Then gears <math>A</math>, <math>B</math>, and <math>C</math> must have angular speeds of <math>\frac{k}{x}</math>, <math>\frac{k}{y}</math>, and <math>\frac{k}{z}</math>, respectively.  Thus the ratio of the three gears' angular speeds is <math>\frac{k}{x}:\frac{k}{y}:\frac{k}{z}</math>.  Multiplying each term of this ratio by <math>\frac{xyz}{k}</math> yields <math>\boxed{(\textbf{D})\ yz:xz:xy}</math>.
<cmath>\frac{k}{x}:\frac{k}{y}:\frac{k}{z}</cmath>
 
<cmath>yz:xz:xy.</cmath>
 
Therefore, the answer is <math>\fbox{D: yz:xz:xy}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 20:36, 25 July 2025

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 $\boxed{(\textbf{D})\ yz:xz:xy}$.

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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