Difference between revisions of "2007 AMC 10A Problems/Problem 5"
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== Problem == | == Problem == | ||
− | A school store sells 7 pencils and 8 notebooks for <math> | + | A school store sells 7 pencils and 8 notebooks for <math> $ </math>4.15<math>. It also sells 5 pencils and 3 notebooks for </math>$<math>1.77</math>. How much do 16 pencils and 10 notebooks cost? |
− | <math>\text{(A)}\ | + | <math>\text{(A)}\ $ </math>1.76 \qquad \text{(B)}\ $ <math>5.84 \qquad \text{(C)}\ $ </math>6.00 \qquad \text{(D)}\ $ <math>6.16 \qquad \text{(E)}\ $ </math>6.32<math> |
== Solution == | == Solution == | ||
− | We let <math>p =< | + | 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*} | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Subtracting these equations yields <math>19n = 836 \Longrightarrow n = 44< | + | 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)}$. |
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Articles with dollar signs]] | [[Category:Articles with dollar signs]] | ||
+ | {{MAA Notice}} |
Revision as of 10:50, 4 July 2013
Problem
A school store sells 7 pencils and 8 notebooks for 4.15$. How much do 16 pencils and 10 notebooks cost?
1.76 \qquad \text{(B)}\ $ 6.00 \qquad \text{(D)}\ $ 6.32p =n = $number of notebooks in cents. Then
<cmath>
Subtracting these equations yields$ (Error compiling LaTeX. Unknown error_msg)19n = 836 \Longrightarrow n = 44p = 916p + 10n = 584\ \mathrm{(B)}$.
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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