SOLUTION: Not in book A mixture of 10% acid and 90% water is added to 5 liters of pure acid. The final mixture is 40% water. How many liters of water are in the final mixture?

Question 79617: Not in book
A mixture of 10% acid and 90% water is added to 5 liters of pure acid. The final mixture is 40% water. How many liters of water are in the final mixture?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

Let x=number of liters of acid/water mixture added to the 5 liters of pure acid.
Then0.90x=amount of pure water added to the 5 liters of pure acid
Now we know that the amount of pure water added to the 5 liters of pure acid equals the amount of pure water in the final mixture. So our equation to solve is:
0.90x=0.40(5+x) get rid of parens
0.90x=2+0.40x subtract 0.40x from each side
0.90x-0.40x=2+0.40x-0.40x collect like terms
0.50x=2 divide both sides by 0.50
x=4 liters------------------amount of mixture added to the pure acid
0.90x=0.90*4=3.6 liters---------------------amount of pure water in final mixture
CK.
0.40(5+4)=3.6
3.6=3.6
Hope this helps----ptaylor

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