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Difference between revisions of "1953 AHSME Problems/Problem 9"

(→‎Solution 2)
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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{(6)}6</math>
+
 
 +
<math>6=n \imples \boxed{\textbf{(D) } 6}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:45, 1 April 2017

Problem

The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution 1

Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\frac{9}{2}$ ounces of alcohol. We want $\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\frac{9}{2}$, to be $30\%$, or $0.3$, of the total amount of shaving lotion, $9+N$). Solving, we find that \[9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.\] So, the total amount of water we need to add is $\boxed{\textbf{(D) } 6}$.

Solution 2

The concentration of alcohol after adding $n$ ounces of water is $\frac{4.5}{9+n}$. To get a solution of 30% alcohol, we solve $\frac{4.5}{9+n}=\frac{3}{10}$

$45=27+3n$

$18=3n$

$6=n \imples \boxed{\textbf{(D) } 6}$ (Error compiling LaTeX. Unknown error_msg)

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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