SOLUTION: I had this mixture problem on a quiz and its really hard. If i'd pulled out 10 hairs for each time i tried again, i'd be bald. Can someone help me with it? It is a mixture problem

Algebra ->  Test -> SOLUTION: I had this mixture problem on a quiz and its really hard. If i'd pulled out 10 hairs for each time i tried again, i'd be bald. Can someone help me with it? It is a mixture problem       Log On


   

Question 281845: I had this mixture problem on a quiz and its really hard. If i'd pulled out 10 hairs for each time i tried again, i'd be bald. Can someone help me with it? It is a mixture problem which is:
One solution is 80% acid and another one is 30% acid. How much of each solution is needed to make a 200L solution that is 62% acid?
I really need help with this so someone please answer it please! Thanks in advance!
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
One solution is 80% acid and another one is 30% acid. How much of each solution is needed to make a 200L solution that is 62% acid?
.
Let x = amount of 80% acid
then
200-x = amount of 30% acid
.
Our equation:
.80x + .30(200-x) = .62(200)
.80x + 60 - .30x = 124
.50x + 60 = 124
.50x = 64
x = 128 L (80% acid)
.
30% acid:
200-x = 200-128 = 72 L (30% acid)