SOLUTION: A Chemist has a solution that is 20 % acid and 80 % water. He adds 5 liters of water to make a solution that is 15 % acid. What is the volume of the original solution. answer

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A Chemist has a solution that is 20 % acid and 80 % water. He adds 5 liters of water to make a solution that is 15 % acid. What is the volume of the original solution. answer      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 164484: A Chemist has a solution that is 20 % acid and 80 % water. He adds 5 liters of water to make a solution that is 15 % acid. What is the volume of the original solution.

answer:The original
solution had 15 liters of water

need a break down of how my text book got: the original solution had 15 liters of water
Found 3 solutions by nerdybill, ptaylor, themathprof:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = volume of original solution
.
"amt of acid in original solution" = "amt of acid new solution"
.20x = .15(x+5)
.20x = .15x + .75
.05x = .75
x = 15 liters (volume of original solution)

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=volume of the original solution
Now we know that the amount of pure acid in the original solution (0.20x) plus the amount of pure acid added (0) has to equal the amount of pure acid in the final solution (0.15(5+x)). So our equation to solve is:
0.20x=0.15(5+x) get rid of parens (distributive law)
0.20x=0.75+0.15x subtract 0.15x from each side
0.20x-0.15x==0.75+0.15x-0.15x
0.05x=0.75 divide each side by 0.05
x=15 liters-----------------------------------volume of original solution
CK
0.20*15=0.15*20
3=3

Hope this helps---ptaylor

Answer by themathprof(21) About Me  (Show Source):
You can put this solution on YOUR website!
A Chemist has a solution that is 20 % acid and 80 % water. He adds 5 liters of water to make a solution that is 15 % acid. What is the volume of the original solution.
Let x=volume of original. From info in problem we have this table:
percent acid * amount = amt acid
water .00 .80x .00
acid 1.00 .20x .20x
solution x .20x
Now add 5 liters of water. New table looks like this:
water .00 .80x+5 .00
acid 1.00 .20x .20x
solution .15 x+5 .15(x+5)
From Column three:
.00+.20x=.15(x+5)
20x=15(x+5)
20x=15x+75
5x=75
Professor Martin Weissman
More Word problems worked out
in my Math911 software. It's free.
download now and tell your friends!
www.math911.com
x=15