1997 AHSME Problems/Problem 8

Problem

Mientka Publishing Company prices its bestseller Where's Walter? as follows:

$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$

where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ books cost less than $24$ books. For how many values of $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution

Clearly, the areas of concern are where the piecewise function shifts value.

Since $C(25) = 11\cdot 25 = 275$ , we want to find the least value of $n$ for which $C(n) > 275$ .

If $n \le 24$ , then $C(n) = 12n$ , so for $C(n) > 275$ , $12n > 275$ , which is equivalent to $n > 22.91$ Thus, both $n=23$ and $n=24$ will be more expensive than $n=25$ .

Since $C(49) = 10\cdot 49 = 490$ , we want to find the least value of $n$ for which $C(n) > 490$ .

If $25 \le n \le 48$ , then $C(n) = 11n$ , so for $C(n) > 490$ , we have $11n > 490$ , leading to $n > 44.5$ . Thus, $n=45, 46, 47, 48$ will be more expensive than $n=49$ .

Thus, there are $2 + 4 = \boxed{6}$ values of $n$ where it's cheaper to buy more books, and the answer is $\boxed{D}$ .

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1997 AHSME Problems/Problem 8

Problem

Solution