Difference between revisions of "2007 AMC 8 Problems/Problem 17"

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<math>\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50</math>
 
<math>\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50</math>
  
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==Solution==
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Since <math>30\%</math> of the original <math>30</math> liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are <math>9+5=14</math> liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives <math>40\%</math> of yellow tint in the new mixture, which is <math>\boxed{\textbf{(C) 40}\6}</math>.
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2007|num-b=16|num-a=18}}
 
{{AMC8 box|year=2007|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:02, 26 July 2013

Problem

A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow tint and $45\%$ water. Five liters of yellow tint are added to the original mixture. What is the percent of yellow tint in the new mixture?

$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$

Solution

Since $30\%$ of the original $30$ liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives $40\%$ of yellow tint in the new mixture, which is $\boxed{\textbf{(C) 40}\6}$ (Error compiling LaTeX. Unknown error_msg).

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AJHSME/AMC 8 Problems and Solutions

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