Difference between revisions of "2007 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | 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}}</math>. | + | Since <math>30\%</math> of the original <math>30</math> liters of paint was yellow, and <math>5</math> 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 <math>5</math> liters of paint were added to the original <math>30</math>, there are a total of <math>35</math> 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}}</math>. |
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Latest revision as of 11:50, 24 December 2024
Problem
A mixture of 
liters of paint is 
red tint, 
yellow
tint and 
water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?
Solution
Since of the original liters of paint was yellow, and 
liters of yellow paint were added to make the new mixture, there are 
liters of yellow tint in the new mixture. Since only liters of paint were added to the original , there are a total of 
liters of paint in the new mixture. This gives 
of yellow tint in the new mixture, which is 
.
Video Solution by WhyMath
~savannahsolver
Video Solution
https://youtu.be/StRqfN7yYE8 Soo, DRMS, NM
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
