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Revision as of 13:12, 5 July 2013

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 $\boxed{D}$.

See also

1997 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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All AHSME Problems and Solutions

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