Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1953 AHSME Problems/Problem 9"
(→‎Solution 2)
Line 10: Line 10:
  
 
== Solution 2==
 
== 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>
+
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>45=27+3n</math>
 
<math>18=3n</math>
 
<math>18=3n</math>
−
<math>6=n \imples \textbf{(6)}6}</math>
+
<math>6=n \imples \textbf{(6)}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 $\frac{4.5}{9+n}=\frac{3}{10}$ $45=27+3n$ $18=3n$ $6=n \imples \textbf{(6)}6$ (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_9&oldid=85120"