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

(Created page with "== Problem == The number of ounces of water needed to reduce <math>9</math> ounces of shaving lotion containing <math>50</math> % alcohol to a lotion containing <math>30</mat...")
 
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<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
 
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7</math>
  
== Solution ==
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== Solution 1 ==
  
 
Say we add <math>N</math> ounces of water to the shaving lotion. Since half of a <math>9</math> ounce bottle of shaving lotion is alcohol, we know that we have <math>\frac{9}{2}</math> ounces of alcohol. We want <math>\frac{9}{2}=0.3(9+N)</math> (because we want the amount of alcohol, <math>\frac{9}{2}</math>, to be <math>30\%</math>, or <math>0.3</math>, of the total amount of shaving lotion, <math>9+N</math>). Solving, we find that <cmath>9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.</cmath> So, the total amount of water we need to add is <math>\boxed{\textbf{(D) } 6}</math>.
 
Say we add <math>N</math> ounces of water to the shaving lotion. Since half of a <math>9</math> ounce bottle of shaving lotion is alcohol, we know that we have <math>\frac{9}{2}</math> ounces of alcohol. We want <math>\frac{9}{2}=0.3(9+N)</math> (because we want the amount of alcohol, <math>\frac{9}{2}</math>, to be <math>30\%</math>, or <math>0.3</math>, of the total amount of shaving lotion, <math>9+N</math>). Solving, we find that <cmath>9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.</cmath> So, the total amount of water we need to add is <math>\boxed{\textbf{(D) } 6}</math>.
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== 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>. To get a solution of 30% alcohol, we solve <math>\frac{4.5}{9+n}=\frac{3}{10}</math>
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<math>45=27+3n</math>
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<math>18=3n</math>
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<math>6=n \imples \textbf{(6)}6}</math>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=8|num-a=10}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:45, 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 $\frac{4.5}{9+n}=\frac{3}{10}$ $45=27+3n$ $18=3n$ $6=n \imples \textbf{(6)}6}$ (Error compiling LaTeX. ! Undefined control sequence.)

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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