1967 AHSME Problems/Problem 30

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Problem

A dealer bought $n$ radios for $d$ dollars, $d$ a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8 on each radio sold. If the overall profit was $72, then the least possible value of $n$ for the given information is:

$\textbf{(A)}\ 18\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 15\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 11$

Solution

The first $2$ radios are sold for $\frac{d}{2}$ dollars each, for income of $2 \cdot \frac{d}{2}$, or simply $d$.

The remaining $n-2$ radios are sold for $d+8$ dollars each, for income of $(n-2)(d+8)$, which expands to $nd +8n -2d - 16$.

The amount invested in buying $n$ radios for $d$ dollars is $nd$.

Thus, the total profit is $(d) + (nd - 2d + 8n - 16) - nd$, which simplifies to $8n - d - 16$.

Setting this equal to $72$ gives $8n - d - 16 = 72$, or $n = 11 + \frac{d}{8}$.

Since $n$ must be an integer, $d$ must be a multiple of $8$. If we want to minimize $n$, we want to minimize $d$. Since $d>0$, setting $d=8$ yields $n=12$, giving the answer of $\fbox{D}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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