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

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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>
 
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>
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but what if <math>n\leq 0</math>
  
 
==See Also==
 
==See Also==

Revision as of 14:36, 3 July 2022

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 $C$ $\boxed{\mathrm{(A)}\text{ }\mathrm{ Increases}.}$

but what if $n\leq 0$

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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