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 one pencil in dollars, $n =$ cost of one notebook in dollars. Then

$\begin{align*} 7p + 8n = 4.15 &\Longrightarrow 35p + 40n = 20.75\\ 5p + 3n = 1.77 &\Longrightarrow 35p + 21n = 12.39 \end{align*}$

Subtracting these equations yields $19n = 8.36 \Longrightarrow n = 0.44$ . Solving backwards gives $p = 0.09$ . Thus the answer is $16p + 10n = 5.84\ \mathrm{(B)}$ .

Solution 2

Since 5 pencils and 3 notebooks cost 1.77 dollars, then 3 times that or 15 pencils and 9 notebooks costs 5.31 dollars which is 1 pencil and 1 notebook off. Looking at answer choices, it can only be 5.84 so $\mathrm{(B)}$ .

Note: 6.00 dollars would imply that 1 pencil and 1 notebook would cost more than 30% of 5 pencils and 3 notebooks, which is incorrect.

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2007 AMC 10A Problems/Problem 5

Contents

Problem

Solution

Solution 2

See also