1954 AHSME Problems/Problem 23

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)
Preceded by Problem 22	Followed by Problem 24
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1954 AHSME Problems/Problem 23

Problem 23

Solution