Difference between revisions of "1953 AHSME Problems/Problem 9"
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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 == | + | == 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>. | ||
| + | |||
| + | == Solution 2== | ||
| + | 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> | ||
| + | <math>45=27+3n</math> | ||
| + | <math>18=3n</math> | ||
| + | <math>6=n \imples \textbf{(6)}6}</math> | ||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=8|num-a=10}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 19:45, 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 ![\[9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.\]](http://latex.artofproblemsolving.com/e/4/5/e45cf0114195f1d2081529ea0e7225b5ea47f593.png)
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 
$6=n \imples \textbf{(6)}6}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
