Difference between revisions of "1953 AHSME Problems/Problem 9"
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The concentration of alcohol after adding <math>n</math> ounces of water is <math>\frac{4.5}{9+n}</math>. | The concentration of alcohol after adding <math>n</math> ounces of water is <math>\frac{4.5}{9+n}</math>. | ||
To get a solution of 30% alcohol, we solve <math>\frac{4.5}{9+n}=\frac{3}{10}</math> | To get a solution of 30% alcohol, we solve <math>\frac{4.5}{9+n}=\frac{3}{10}</math> | ||
+ | |||
<math>45=27+3n</math> | <math>45=27+3n</math> | ||
+ | |||
<math>18=3n</math> | <math>18=3n</math> | ||
− | <math>6=n \imples \textbf{( | + | |
+ | <math>6=n \imples \boxed{\textbf{(D) } 6}</math> | ||
==See Also== | ==See Also== |
Revision as of 20:45, 1 April 2017
Contents
Problem
The number of ounces of water needed to reduce ounces of shaving lotion containing % alcohol to a lotion containing % alcohol is:
Solution 1
Say we add ounces of water to the shaving lotion. Since half of a ounce bottle of shaving lotion is alcohol, we know that we have ounces of alcohol. We want (because we want the amount of alcohol, , to be , or , of the total amount of shaving lotion, ). Solving, we find that So, the total amount of water we need to add is .
Solution 2
The concentration of alcohol after adding ounces of water is . To get a solution of 30% alcohol, we solve
$6=n \imples \boxed{\textbf{(D) } 6}$ (Error compiling LaTeX. ! Undefined control sequence.)
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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All AHSME Problems and Solutions |
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