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Difference between revisions of "1961 AHSME Problems/Problem 15"

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Dimensional analysis is definitely the most rigid, but if you know the ending units (e.g. you know that density is measured in <math>g/cm^2</math> or something like that, you can just treat is as simple proportions and equations.
 
Dimensional analysis is definitely the most rigid, but if you know the ending units (e.g. you know that density is measured in <math>g/cm^2</math> or something like that, you can just treat is as simple proportions and equations.
 +
 +
~hastapasta
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 +
==What's happening here? Why isn't the answer <math>y</math>?==
 +
 +
Notice that if we change the problem to <math>x</math> men produce <math>x</math> items a day, <math>y</math> men produces how many items a day, then the answer would be <math>y</math>. In this case, it would be a direct variation. However, notice that direct variations only have two factors --- an independent and dependent variable each (cause-effect, <math>x</math>-<math>y</math>). However, there are 3 factors, not 1, that are contributing to how many items are produced in the original problem. This is a combined variation problem, not a direct variation problem. This is the reason why the answer is <math>\boxed{B}</math> (also see that the base unit (1 man/1 hour/1 day) is <math>\frac{1}{x^2}</math>.
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Hope that solves your confusions.
  
 
~hastapasta
 
~hastapasta

Revision as of 14:20, 9 March 2022

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.

~hastapasta

What's happening here? Why isn't the answer $y$?

Notice that if we change the problem to $x$ men produce $x$ items a day, $y$ men produces how many items a day, then the answer would be $y$. In this case, it would be a direct variation. However, notice that direct variations only have two factors --- an independent and dependent variable each (cause-effect, $x$-$y$). However, there are 3 factors, not 1, that are contributing to how many items are produced in the original problem. This is a combined variation problem, not a direct variation problem. This is the reason why the answer is $\boxed{B}$ (also see that the base unit (1 man/1 hour/1 day) is $\frac{1}{x^2}$.

Hope that solves your confusions.

~hastapasta

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions


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