SOLUTION: If solution A contains 50% nitrogen and solution B contains 29% nitrogen. How much of each solution should be mixed in order to create 30 liters of a solution that contains 36% nit
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Question 1145289: If solution A contains 50% nitrogen and solution B contains 29% nitrogen. How much of each solution should be mixed in order to create 30 liters of a solution that contains 36% nitrogen Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! If solution A contains 50% nitrogen and solution B contains 29% nitrogen.
How much of each solution should be mixed in order to create 30 liters of a solution that contains 36% nitrogen.
:
let x = amt of 50% solution
since the resulting amt is 30 liter:
(30-x) = 29% solution
:
.50x + .29(30-x) = .36(30)
.50x + 8.7 - .29x = 10.8
.50x - .29x = 10.8 - 8.7
.21x = 2.1
x = 2.1/.21
x = 10 liters of 50% liters nitrogen
then
30 - 10 = 20 liters of the 29% liters
:
:
Check
.50(10) = 5
.29(20) = 5.8
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total = 10.8
