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 "1979 AHSME Problems/Problem 15"

(Created page with "== Problem 15 == Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being <math>p : 1</math> in one jar and <m...")
 
m (→‎Solution)
Line 14: Line 14:
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
−
The amount of alcohol in the jars are <math>\frac{p}{p+1}</math> and <math>\frac{q}{q+1}</math>, and the amount of water in the jars are <math>\frac{1}{p+1}</math> and <math>\frac{1}{q+1}</math>. Then the total amount of alcohol is <math>\frac{p}{p+1} + \frac{q}{q+1} = \frac{p+q+2pq}{(p+1)(q+1)}</math>, and the total amount of water is <math>\frac{1}{p+1} and \frac{1}{q+1} = \frac{p+q+2}{(p+1)(q+1)}</math>. The ratio of the volume of alcohol to the volume of water in the mixture is <math>\frac{\frac{p+q+2pq}{(p+1)(q+1}}{\frac{p+q+2}{(p+1)(q+1)}} = \boxed{\textbf{(E) } \frac{p+q+2pq}{p+q+2}</math>.
+
The amount of alcohol in the jars are <math>\frac{p}{p+1}</math> and <math>\frac{q}{q+1}</math>, and the amount of water in the jars are <math>\frac{1}{p+1}</math> and <math>\frac{1}{q+1}</math>. Then the total amount of alcohol is <math>\frac{p}{p+1} + \frac{q}{q+1} = \frac{p+q+2pq}{(p+1)(q+1)}</math>, and the total amount of water is <math>\frac{1}{p+1} and \frac{1}{q+1} = \frac{p+q+2}{(p+1)(q+1)}</math>. The ratio of the volume of alcohol to the volume of water in the mixture is <math>\frac{\frac{p+q+2pq}{(p+1)(q+1}}{\frac{p+q+2}{(p+1)(q+1)}} = \boxed{\textbf{(E) } \frac{p+q+2pq}{p+q+2}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 12:21, 6 January 2017

Problem 15

Two identical jars are filled with alcohol solutions, the ratio of the volume of alcohol to the volume of water being $p : 1$ in one jar and $q : 1$ in the other jar. If the entire contents of the two jars are mixed together, the ratio of the volume of alcohol to the volume of water in the mixture is

$\textbf{(A) }\frac{p+q}{2}\qquad \textbf{(B) }\frac{p^2+q^2}{p+q}\qquad \textbf{(C) }\frac{2pq}{p+q}\qquad \textbf{(D) }\frac{2(p^2+pq+q^2)}{3(p+q)}\qquad \textbf{(E) }\frac{p+q+2pq}{p+q+2}$

Solution

Solution by e_power_pi_times_i

The amount of alcohol in the jars are $\frac{p}{p+1}$ and $\frac{q}{q+1}$, and the amount of water in the jars are $\frac{1}{p+1}$ and $\frac{1}{q+1}$. Then the total amount of alcohol is $\frac{p}{p+1} + \frac{q}{q+1} = \frac{p+q+2pq}{(p+1)(q+1)}$, and the total amount of water is $\frac{1}{p+1} and \frac{1}{q+1} = \frac{p+q+2}{(p+1)(q+1)}$. The ratio of the volume of alcohol to the volume of water in the mixture is $\frac{\frac{p+q+2pq}{(p+1)(q+1}}{\frac{p+q+2}{(p+1)(q+1)}} = \boxed{\textbf{(E) } \frac{p+q+2pq}{p+q+2}}$.

See also

1979 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
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=1979_AHSME_Problems/Problem_15&oldid=82214"