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Difference between revisions of "2007 AMC 10A Problems/Problem 5"

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== Problem ==
 
== Problem ==
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A school store sells 7 pencils and 8 notebooks for <math> \</math> <math>4.15</math>. It also sells 5 pencils and 3 notebooks for <math>\</math><math>1.77</math>. How much do 16 pencils and 10 notebooks cost?
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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?
  
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<math>\text{(A)}\ \</math> <math>1.76 \qquad \text{(B)}\ \</math> <math>5.84 \qquad \text{(C)}\ \</math> <math>6.00 \qquad \text{(D)}\ \</math> <math>6.16 \qquad \text{(E)}\ \</math> <math>6.32</math>
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<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 ==
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We let <math>p =</math> cost of pencils in cents, <math>n = </math> number of notebooks in cents. Then
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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*}
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7p + 8n = 415 &\Longrightarrow  35p + 40n = 2075\\
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7p + 8n = 4.15 &\Longrightarrow  35p + 40n = 20.75\\
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5p + 3n = 177 &\Longrightarrow  35p + 21n = 1239
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5p + 3n = 1.77 &\Longrightarrow  35p + 21n = 12.39
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
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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>.
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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 = 5.84\ \mathrm{(B)}</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.  
  
 
== 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]]
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{{MAA Notice}}

Latest revision as of 11:42, 6 August 2024

Problem

The school store sells 7 pencils and 8 notebooks for $\mathdollar 4.15$. It also sells 5 pencils and 3 notebooks for $\mathdollar 1.77$. How much do 16 pencils and 10 notebooks cost?

$\text{(A)}\mathdollar 1.76 \qquad \text{(B)}\mathdollar 5.84 \qquad \text{(C)}\mathdollar 6.00 \qquad \text{(D)}\mathdollar 6.16 \qquad \text{(E)}\mathdollar 6.32$

Solution

We let $p =$ cost of one pencil in dollars, $n =$ cost of one notebook in dollars. Then

\begin{align*} 7p + 8n = 4.15 &\Longrightarrow 35p + 40n = 20.75\\ 5p + 3n = 1.77 &\Longrightarrow 35p + 21n = 12.39 \end{align*}

Subtracting these equations yields $19n = 8.36 \Longrightarrow n = 0.44$. Solving backwards gives $p = 0.09$. Thus the answer is $16p + 10n = 5.84\ \mathrm{(B)}$.

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 $\mathrm{(B)}$ .

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

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