Difference between revisions of "2007 AMC 10A Problems/Problem 5"
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== Problem == | == Problem == | ||
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The 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? | The 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? | ||
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== Solution == | == Solution == | ||
| − | + | Let <math>p</math> be cost of one pencil in dollars and <math>n</math> the cost of one notebook in dollars. Then | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | 7p + 8n = 4.15 &\ | + | 7p + 8n = 4.15 &\implies 35p + 40n = 20.75\\ |
| − | 5p + 3n = 1.77 &\ | + | 5p + 3n = 1.77 &\implies 35p + 21n = 12.39 |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | 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 = \mathdollar 5.84 | + | 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 = \text{(B)}\mathdollar 5.84</math>. |
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| + | == Solution 2 == | ||
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| + | Since 5 pencils and 3 notebooks cost 1.77 dollars, then 3 times that or 15 pencils and 9 notebooks costs 5.31 dollars which is 1 pencil and 1 notebook off. Looking at answer choices, it can only be 5.84 so <math>\mathrm{(B)}</math> . | ||
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| + | Note: 6.00 dollars would imply that 1 pencil and 1 notebook would cost more than 30% of 5 pencils and 3 notebooks, which is incorrect. | ||
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| + | == See Also == | ||
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{{AMC10 box|year=2007|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2007|ab=A|num-b=4|num-a=6}} | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:06, 22 March 2025
Contents
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
Let 
be cost of one pencil in dollars and 
the cost of one notebook in dollars. Then
Subtracting these equations yields 
. Solving backwards gives 
. Thus the answer is 
.
Solution 2
Since 5 pencils and 3 notebooks cost 1.77 dollars, then 3 times that or 15 pencils and 9 notebooks costs 5.31 dollars which is 1 pencil and 1 notebook off. Looking at answer choices, it can only be 5.84 so 
.
Note: 6.00 dollars would imply that 1 pencil and 1 notebook would cost more than 30% of 5 pencils and 3 notebooks, which is incorrect.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
