SOLUTION: How many liters of a 20% alcohol solution must be miced with 50 liters of a 70% solution to get a 50 % solution?
I can solve this just by using the equation of .20x + 35 = .50(5
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I can solve this just by using the equation of .20x + 35 = .50(5
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Question 230075: How many liters of a 20% alcohol solution must be miced with 50 liters of a 70% solution to get a 50 % solution?
I can solve this just by using the equation of .20x + 35 = .50(50+x)
But this has to be done in a system of equations. I am stumped on how to get the 2nd equation. There is no usual x+y = a certain total amount here.
Not looking for the answer but just the 2nd. equation
Thanks in advance :0) Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! All I can figure is that they want you to say liters is the total liters of solution
and is the other equation (very trivial)
Let = liters of 20% alcohol solution needed
In words:
(final liters of alcohol)/(final total solution) = 50%
given:
Alcohol in 70% solution:
----------------------
(1)
and
(2)
