Difference between revisions of "2007 AMC 10A Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | We let <math>p =</math> cost of | + | We let <math>p =</math> cost of one pencil in dollars, <math>n = </math> cost of one notebook in dollars. Then |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| Line 12: | Line 12: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | Subtracting these equations yields <math>19n = 8.36 \Longrightarrow n = .44</math>. | + | Subtracting these equations yields <math>19n = 8.36 \Longrightarrow n = 0.44</math>. Solving backwards gives <math>p = 0.09</math>. Thus the answer is <math>16p + 10n = 5.84\ \mathrm{(B)}</math>. |
== See also == | == See also == | ||
Latest revision as of 12:10, 3 June 2021
Problem
The school store sells 7 pencils and 8 notebooks for 
. It also sells 5 pencils and 3 notebooks for 
. How much do 16 pencils and 10 notebooks cost?
Solution
We let 
cost of one pencil in dollars, 
cost of one notebook in dollars. Then
Subtracting these equations yields 
. Solving backwards gives 
. Thus the answer is 
.
See also
| 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. 
