Difference between revisions of "2007 AMC 10A Problems/Problem 5"

Revision as of 15:00, 27 December 2008

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 • 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

@@ Line 15: / Line 15: @@
 == See also ==
-{{AMC10 box|year=2007|ab=A|num-b=8|num-a=10}}
+{{AMC10 box|year=2007|ab=A|num-b=4|num-a=6}}
 [[Category:Introductory Algebra Problems]]
 [[Category:Articles with dollar signs]]

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Difference between revisions of "2007 AMC 10A Problems/Problem 5"

Revision as of 15:00, 27 December 2008

Problem

Solution

See also