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

(See also)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
A school store sells 7 pencils and 8 notebooks for <math> \</math> <math>4.15</math>. It also sells 5 pencils and 3 notebooks for <math>\</math><math>1.77</math>. How much do 16 pencils and 10 notebooks cost?
+
A school store sells 7 pencils and 8 notebooks for <math> &#036; </math>4.15<math>. It also sells 5 pencils and 3 notebooks for </math>&#036;<math>1.77</math>. How much do 16 pencils and 10 notebooks cost?
  
<math>\text{(A)}\ \</math> <math>1.76 \qquad \text{(B)}\ \</math> <math>5.84 \qquad \text{(C)}\ \</math> <math>6.00 \qquad \text{(D)}\ \</math> <math>6.16 \qquad \text{(E)}\ \</math> <math>6.32</math>
+
<math>\text{(A)}\ &#036; </math>1.76 \qquad \text{(B)}\ &#036; <math>5.84 \qquad \text{(C)}\ &#036; </math>6.00 \qquad \text{(D)}\ &#036; <math>6.16 \qquad \text{(E)}\ &#036; </math>6.32<math>
  
 
== Solution ==
 
== Solution ==
We let <math>p =</math> cost of pencils in cents, <math>n = </math> number of notebooks in cents. Then
+
We let </math>p =<math> cost of pencils in cents, </math>n = <math> number of notebooks in cents. Then
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 12: Line 12:
 
\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 = 836 \Longrightarrow n = 44<math>. Backwards solving gives </math>p = 9<math>. Thus the answer is </math>16p + 10n = 584\ \mathrm{(B)}$.
  
 
== See also ==
 
== See also ==
Line 19: Line 19:
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Articles with dollar signs]]
 
[[Category:Articles with dollar signs]]
 +
{{MAA Notice}}

Revision as of 10:50, 4 July 2013

Problem

A school store sells 7 pencils and 8 notebooks for $&#036;$4.15$. It also sells 5 pencils and 3 notebooks for$$$1.77$. How much do 16 pencils and 10 notebooks cost?

$\text{(A)}\ &#036;$1.76 \qquad \text{(B)}\ $ $5.84 \qquad \text{(C)}\ &#036;$6.00 \qquad \text{(D)}\ $ $6.16 \qquad \text{(E)}\ &#036;$6.32$== Solution == We let$p =$cost of pencils in cents,$n = $number of notebooks in cents. Then

<cmath>7p+8n=41535p+40n=20755p+3n=17735p+21n=1239</cmath>

Subtracting these equations yields$ (Error compiling LaTeX. Unknown error_msg)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
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. AMC logo.png