SOLUTION: how many liters of a 50% acid solution must be mixed with a 15% acid solution to get a 210L 40% solution
I came up with .5x+.15(x-210)=.40(210)
but keep getting it wrong
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I came up with .5x+.15(x-210)=.40(210)
but keep getting it wrong
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Question 945961: how many liters of a 50% acid solution must be mixed with a 15% acid solution to get a 210L 40% solution
I came up with .5x+.15(x-210)=.40(210)
but keep getting it wrong Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = liters of 50% acid solution needed = liters of acid in 50% solution
Let = liters of 15% acid solution needed = liter od acid is 15% solution
------------------------------------------
(1)
(2)
--------------------------------
(2)
(2)
(2)
(2)
--------------------------
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)
------------------------
and, since
(1)
(1)
(1)
150 liters of 50% acid solution are needed
60 liters of 15% acid solution are needed
check:
(2)
(2)
(2)
(2)
Ok
