1961 AHSME Problems/Problem 15

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Problem

If $x$ men working $x$ hours a day for $x$ days produce $x$ articles, then the number of articles (not necessarily an integer) produced by $y$ men working $y$ hours a day for $y$ days is:

$\textbf{(A)}\ \frac{x^3}{y^2}\qquad \textbf{(B)}\ \frac{y^3}{x^2}\qquad \textbf{(C)}\ \frac{x^2}{y^3}\qquad \textbf{(D)}\ \frac{y^2}{x^3}\qquad \textbf{(E)}\ y$

Solution 1

Let $k$ be the number of articles produced per hour per person. By using dimensional analysis, \[\frac{x \text{ hours}}{\text{day}} \cdot x \text{ days} \cdot \frac{k \text{ articles}}{\text{hours} \cdot \text{person}} \cdot x \text{ people} = x \text{ articles}\] Solving this yields $k = \frac{1}{x^2}$. Using dimensional analysis again, the number of articles produced by $y$ men working $y$ hours a day for $y$ days is \[\frac{y \text{ hours}}{\text{day}} \cdot y \text{ days} \cdot \frac{\frac{1}{x^2} \text{ articles}}{\text{hours} \cdot \text{person}} \cdot y \text{ people} = \frac{y^3}{x^2} \text{ articles}\] The answer is $\boxed{\textbf{(B)}}$.

Solution 2 (Simple logic)

The question is based on the assumption that each person, each hour, each day, will be produce a constant number of items (maybe fractional).

So it takes $x$ men $x$ hours to produce $\frac{x}{x}=1$ item in a day.

In a similar manner, 1 man, 1 hour, for a day, can produce $\frac{1}{x^2}$ items. So $y$ men, $y$ hours a day, for $y$ days produce $\frac{y^3}{x^2}$ items. Therefore, the answer is $\boxed{B}$.

Dimensional analysis is definitely the most rigid, but if you know the ending units (e.g. you know that density is measured in $g/cm^2$ or something like that, you can just treat is as simple proportions and equations.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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