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Please Help! Gabe Amodeo a nuclear physicists needs 80 liters of a 30% acid solution. He currently has a 20% and a 60% solution How many liters of each does he need to make the needed 80 liters of 30% acid solution?

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IAmAMathAddict
he needs 6 litres of the 20% and 2 litres of the 60% a simple trial and error method will do in this type of equation
mreijepatmat
Let x be the number of liters needed from the first acide concentration (20%)
Let y be the number of liters needed from the first acide concentration (60%)

We know at the end we will have:
x+y = 80 ; we also know that:
20%(x) + 60%(y) = 30%(80) OR 0.2x + 0.6y = 0.3(80) = 2x+6y=240

Hence we have :
x+y = 80
2x+6y = 240
Solving this system of 2 equations give: x = 60 and y = 20
In short you need to mix 60 litres at 20% and 20 liters at 60%