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>\mathdollar 4.15</math>. It also sells 5 pencils and 3 notebooks for <math>\mathdollar 1.77</math>. How much do 16 pencils and 10 notebooks cost? |
| − | <math>\text{(A)}\ | + | <math>\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</math> |
== Solution == | == Solution == | ||
| − | We let < | + | 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 < | + | 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>. |
== See also == | == See also == | ||
Revision as of 19:45, 11 July 2013
Problem
A 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
