2007 AMC 10A Problems/Problem 5

Problem

The school store sells 7 pencils and 8 notebooks for $\mathdollar 4.15$. It also sells 5 pencils and 3 notebooks for $\mathdollar 1.77$. How much do 16 pencils and 10 notebooks cost?

$\text{(A)}\mathdollar 1.76 \qquad \text{(B)}\mathdollar 5.84 \qquad \text{(C)}\mathdollar 6.00 \qquad \text{(D)}\mathdollar 6.16 \qquad \text{(E)}\mathdollar 6.32$

Solution

We let $p =$ cost of pencils in cents, $n =$ number of notebooks in cents. Then

\begin{align*} 7p + 8n = 415 &\Longrightarrow  35p + 40n = 2075\\ 5p + 3n = 177 &\Longrightarrow  35p + 21n = 1239 \end{align*}

Subtracting these equations yields $19n = 836 \Longrightarrow n = 44$. Backwards solving gives $p = 9$. Thus the answer is $16p + 10n = 584\ \mathrm{(B)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 10 Problems and Solutions

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