SOLUTION: " 8 pints of a 20% solution of alcohol in water are mixed with 2 pints of a 40% solution of alcohol in water. What is the percentage of alcohol in the new solution? "
I can't f
Algebra ->
Human-and-algebraic-language
-> SOLUTION: " 8 pints of a 20% solution of alcohol in water are mixed with 2 pints of a 40% solution of alcohol in water. What is the percentage of alcohol in the new solution? "
I can't f
Log On
Question 922842: " 8 pints of a 20% solution of alcohol in water are mixed with 2 pints of a 40% solution of alcohol in water. What is the percentage of alcohol in the new solution? "
I can't figure out the formula. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=percentage of alcohol in new solution
Now we know that the AMOUNT of pure alcohol that exists before the mixture takes place(0.20*8+0.40*2) has to equal the AMOUNT of pure alcohol that exists after the mixture takes place(x*10). Sooo our equation to solve is:
0.20*8+0.40*2=10x
1.6+0.8=10x
10x=2.4
x=0.24 or 24%
CK
0.20*8+0.40*2=0.24*10
1.6+0.8=2.4
2.4=2.4
Hope this helps---ptaylor
