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Difference between revisions of "1997 AHSME Problems/Problem 8"

(Created page with "==Problem== Mientka Publishing Company prices its bestseller Where's Walter? as follows: <math> C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le...")
 
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Since <math>C(25) = 11\cdot 25 = 275</math>, we want to find the least value of <math>n</math> for which <math>C(n) > 275</math>.
 
Since <math>C(25) = 11\cdot 25 = 275</math>, we want to find the least value of <math>n</math> for which <math>C(n) > 275</math>.
  
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If <math>n \le 24</math>, then <math>C(n) = 12n</math>, so for <math>C(n) > 275</math>, <math>12n > 275</math>, which is equivalent to <math>n > 22.91</math>  Thus, both <math>n=23</math> and <math>n=24</math> will be more expensive than <math>n=25</math>.
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If <math>n \le 24</math>, then <math>C(n) = 12n</math>, so for <math>C(n) > 275</math>, <math>12n > 275</math>, which is equivalent to <math>n > 22.91</math>. Thus, both <math>n=23</math> and <math>n=24</math> will be more expensive than <math>n=25</math>.
  
 
Since <math>C(49) = 10\cdot 49 = 490</math>, we want to find the least value of <math>n</math> for which <math>C(n) > 490</math>.
 
Since <math>C(49) = 10\cdot 49 = 490</math>, we want to find the least value of <math>n</math> for which <math>C(n) > 490</math>.
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Thus, there are <math>2 + 4 = \boxed{6}</math> values of <math>n</math> where it's cheaper to buy more books, and the answer is <math>\boxed{D}</math>.
 
Thus, there are <math>2 + 4 = \boxed{6}</math> values of <math>n</math> where it's cheaper to buy more books, and the answer is <math>\boxed{D}</math>.
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== See also ==
 +
{{AHSME box|year=1997|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 21:58, 15 February 2018

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
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

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