SOLUTION: How do you solve a mixture problem like this?: One solution is 80% acid and another one is 30% acid. How many of each solution is needed to make a 200L solution that is 62% acid?

Question 281844: How do you solve a mixture problem like this?:
One solution is 80% acid and another one is 30% acid. How many of each solution is needed to make a 200L solution that is 62% acid?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Let 80% acid required be x
30% acid will be 200 -x
80%-----------------------30%---------------------62%
x-------------------------200-x-------------------200 quantity
0.8x+0.3(200-x)=0.62*200
0.8x+60-0.3x=124
0.5x=124-60
0.5x=64
x=64/0.5
=128 liters ---- 80% acid
The balance will be 30 % acid solution=72 liters

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