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

(Solution)
(Solution)
 
Line 8: Line 8:
  
 
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>
 
but what if <math>n\leq 0</math>
 
  
 
==See Also==
 
==See Also==

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png