Difference between revisions of "2007 AMC 10A Problems/Problem 5"
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| − | 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 = \ | + | 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>. |
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| + | == 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 <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. | ||
== See also == | == See also == | ||
Latest revision as of 11:42, 6 August 2024
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
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
