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Difference between revisions of "1954 AHSME Problems/Problem 23"

(Created page with "== Problem 23== If the margin made on an article costing <math>C</math> dollars and selling for <math>S</math> dollars is <math>M=\frac{1}{n}C</math>, then the margin is give...")
 
m (→‎Solution)
 
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Doing some algebraic manipulation to get <math>M</math> on one side,
 
Doing some algebraic manipulation to get <math>M</math> on one side,
  
−
<math>M=\frac{1}{n}{S-M}</math>
+
<math>M=\frac{1}{n}(S-M)</math>
  
 
<math>M=\frac{S}{n}-\frac{M}{n}</math>
 
<math>M=\frac{S}{n}-\frac{M}{n}</math>

Latest revision as of 17:26, 15 April 2017

Problem 23

If the margin made on an article costing $C$ dollars and selling for $S$ dollars is $M=\frac{1}{n}C$, then the margin is given by:

$\textbf{(A)}\ M=\frac{1}{n-1}S\qquad\textbf{(B)}\ M=\frac{1}{n}S\qquad\textbf{(C)}\ M=\frac{n}{n+1}S\\ \textbf{(D)}\ M=\frac{1}{n+1}S\qquad\textbf{(E)}\ M=\frac{n}{n-1}S$

Solution

We are given the margin in terms of the cost of the article. Looking at the answers, it appears we need to find the margin in terms of the selling price. The relationship between cost and selling price is that selling price minus the margin is the cost, $S-M=C$.

Since $M=\frac{1}{n}C$ and $C=S-M$, $M=\frac{1}{n}(S-M)$.

Doing some algebraic manipulation to get $M$ on one side,

$M=\frac{1}{n}(S-M)$

$M=\frac{S}{n}-\frac{M}{n}$

$M+\frac{M}{n}=\frac{S}{n}$

$M(1+\frac{1}{n})=\frac{S}{n}$

$M=\frac{S}{n}\cdot \frac{1}{1+\frac{1}{n}}$

$M=\frac{S}{n+1}$

$M=\frac{1}{n+1}S \implies \boxed{\textbf{(D)} M=\frac{1}{n+1}S}$.

1954 AHSC (Problems • Answer Key • Resources)
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