Difference between revisions of "2007 AMC 10A Problems/Problem 5"
(slightly less algebra solution?) |
|||
(11 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | + | 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? | |
− | <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 one pencil in dollars, <math>n = </math> cost of one notebook in dollars. Then |
<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 < | + | 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>. |
+ | |||
+ | == 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> . | ||
+ | |||
+ | 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.