1967 AHSME Problems/Problem 30

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 $\fbox{D}$

1967 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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
All AHSME Problems and Solutions

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

Page

Toolbox

Search

1967 AHSME Problems/Problem 30

Problem

Solution

See also