SOLUTION: I need help with this story problem in algebra.
How many liters of a 10% acid solution must be mixed with 40 liters of a 50% acid solution to get a 20% acid solution?
I tried a l
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-> SOLUTION: I need help with this story problem in algebra.
How many liters of a 10% acid solution must be mixed with 40 liters of a 50% acid solution to get a 20% acid solution?
I tried a l
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Question 793184: I need help with this story problem in algebra.
How many liters of a 10% acid solution must be mixed with 40 liters of a 50% acid solution to get a 20% acid solution?
I tried a lot of ways, but the way this is set yp is screwing me up. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! --- percent ---------------- Amount
Solution I 50 ---------------- 40 liters
solution II 10 ---------------- x liters
Mixture 20 ---------------- 40 + x liters
50 * 40 + 10 x = 20 ( 40 + x )
2000 + 10 x = 800 + 20 x
10 x -20 x = -2000 + 800
-10 x = -1200
/ -10
x = 120 liters solution II
