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

(Solution to Problem 15)
 
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\textbf{(E)}\ y</math>
 
\textbf{(E)}\ y</math>
  
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==Solution==
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== Solution 1 ==
  
 
Let <math>k</math> be the number of articles produced per hour per person.  By using dimensional analysis,
 
Let <math>k</math> be the number of articles produced per hour per person.  By using dimensional analysis,
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<cmath>\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}</cmath>
 
<cmath>\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}</cmath>
 
The answer is <math>\boxed{\textbf{(B)}}</math>.
 
The answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 +
== 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 <math>x</math> men <math>x</math> hours to produce <math>\frac{x}{x}=1</math> item in a day.
 +
 +
In a similar manner, 1 man, 1 hour, for a day, can produce <math>\frac{1}{x^2}</math> items. So <math>y</math> men, <math>y</math> hours a day, for <math>y</math> days produce <math>\frac{y^3}{x^2}</math> items. Therefore, the answer is <math>\boxed{B}</math>.
 +
 +
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.
  
 
==See Also==
 
==See Also==

Revision as of 13:04, 4 February 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.

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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