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>

Revision as of 10:55, 27 May 2014

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

Assume that the constants are positive, as well as $n.$

WLOG let $e,$ $R,$ and $r$ all equal $1.$ Then $C=\frac{n}{1+n}.$ We can see that as $n$ increases from $0,$ it slowly approaches $1.$ Therefore, $C$ $\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}$

If $r$ and $R$ were positive and $e$ was negative, then $C$ would decrease, for example.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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