Difference between revisions of "1950 AHSME Problems/Problem 11"

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== Problem==
 
== Problem==
  
If in the formula <math> C =\frac{en}{R+nr} </math>, <math>n</math> is increased while <math>e</math>, <math>R</math> and <math>r</math> are kept constant, then <math>C</math>:
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If in the formula <math> C =\frac{en}{R+nr} </math>, where <math>e</math>, <math>n</math>, <math>R</math> and <math>r</math> are all positive, <math>n</math> is increased while <math>e</math>, <math>R</math> and <math>r</math> are kept constant, then <math>C</math>:
  
 
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\ \textbf{(E)}\ \text{Decreases and then increases} </math>
 
<math> \textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\ \textbf{(E)}\ \text{Decreases and then increases} </math>
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==Solution==
 
==Solution==
  
Assume that the constants are positive, as well as <math>n.</math>
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Divide both the numerator and denominator by <math>n</math>, to get <math>C=\frac{e}{\frac{R}{n}+r}</math>.  If <math>n</math> increases then the denominator decreases; so that <math>C</math> <math>\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}</math>
  
WLOG let <math>e,</math> <math>R,</math> and <math>r</math> all equal <math>1.</math> Then <math>C=\frac{n}{1+n}.</math> We can see that as <math>n</math> increases from <math>0,</math> it slowly approaches <math>1.</math> Therefore, <math>C</math> <math>\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}</math>
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==See Also==
  
If <math>r</math> and <math>R</math> were positive and <math>e</math> was negative, then <math>C</math> would decrease, for example.
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{{AHSME 50p box|year=1950|num-b=10|num-a=12}}
 
 
==See Also==
 
  
{{AHSME box|year=1950|num-b=10|num-a=12}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:28, 18 November 2024

Problem

If in the formula $C =\frac{en}{R+nr}$, where $e$, $n$, $R$ and $r$ are all positive, $n$ is increased while $e$, $R$ and $r$ are kept constant, then $C$:

$\textbf{(A)}\ \text{Increases}\qquad\textbf{(B)}\ \text{Decreases}\qquad\textbf{(C)}\ \text{Remains constant}\qquad\textbf{(D)}\ \text{Increases and then decreases}\qquad\\ \textbf{(E)}\ \text{Decreases and then increases}$

Solution

Divide both the numerator and denominator by $n$, to get $C=\frac{e}{\frac{R}{n}+r}$. If $n$ increases then the denominator decreases; so that $C$ $\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AHSME Problems and Solutions

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