Difference between revisions of "2007 AMC 10A Problems/Problem 5"
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m (→Solution: Changed equation outcomes to correct values (ex. 2075 → 20.75)) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 7p + 8n = | + | 7p + 8n = 4.15 &\Longrightarrow 35p + 40n = 20.75\\ |
− | 5p + 3n = | + | 5p + 3n = 1.77 &\Longrightarrow 35p + 21n = 12.39 |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Subtracting these equations yields <math>19n = | + | 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 . It also sells 5 pencils and 3 notebooks for . How much do 16 pencils and 10 notebooks cost?
Solution
We let cost of pencils in cents, number of notebooks in cents. Then
Subtracting these equations yields . Backwards solving 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 |
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