2007 AMC 10A Problems/Problem 5

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Problem

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

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

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 $16p + 10n = 584\ \mathrm{(B)}$ .

2007 AMC 10A (Problems)
Preceded by Problem 8	Followed by Problem 10
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2007 AMC 10A Problems/Problem 5

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Solution

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