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

m (Problem)
m (Solution: Changed equation outcomes to correct values (ex. 2075 → 20.75))
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
7p + 8n = 415 &\Longrightarrow  35p + 40n = 2075\\
+
7p + 8n = 4.15 &\Longrightarrow  35p + 40n = 20.75\\
5p + 3n = 177 &\Longrightarrow  35p + 21n = 1239
+
5p + 3n = 1.77 &\Longrightarrow  35p + 21n = 12.39
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
Subtracting these equations yields <math>19n = 836 \Longrightarrow n = 44</math>. Backwards solving gives <math>p = 9</math>. Thus the answer is <math>16p + 10n = 584\ \mathrm{(B)}</math>.
+
Subtracting these equations yields <math>19n = 8.36 \Longrightarrow n = .44</math>. Backwards solving gives <math>p = .90</math>. Thus the answer is <math>16p + 10n = 5.84\ \mathrm{(B)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:28, 20 January 2021

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

Subtracting these equations yields $19n = 8.36 \Longrightarrow n = .44$. Backwards solving gives $p = .90$. Thus the answer is $16p + 10n = 5.84\ \mathrm{(B)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
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

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