SOLUTION: how many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50 % solution?
Algebra.Com
Question 32399: how many liters of a 30% alcohol solution must be mixed with 80 liters of a 80% solution to get a 50 % solution?
Answer by mukhopadhyay(490) (Show Source): You can put this solution on YOUR website!
Assume x liters of a 30% alcohol solution is to be mixed with 80 liters of a 80% solution to produce (x+80) liters of a 50% solution.
=> x*(30%) + 80(80%) = (x+80)(50%)
=> .3x + 64 = .5x + 40
=> .2x = 24
=> x = 120
Answer: 120 liters
RELATED QUESTIONS
How many liters of a 10% alcohol solution must be mixed with 50 liters of a 80% solution... (answered by mtmorro)
How many liters of a 10% alcohol solution must be mixed with 80 liters of a 80% solution... (answered by Fombitz,edjones)
How many liters of a 10% alcohol solution must be mixed with 80 liters of a 80% solution... (answered by richwmiller)
How many liters of a 30% alcohol solution must be mixed with 80 liters of a 90% solution... (answered by mananth)
How many liters of a 30% alcohol solution must be mixed with 80 liters of a 90% solution... (answered by checkley77)
How many liters (L) of a 30% alcohol solution must be mixed with 50 L of a 80% solution... (answered by nyc_function)
how many liters of a 30% alcohol solution must be mixed with 30 liters of an 80% solution (answered by Alan3354,richwmiller,ewatrrr)
How many liters of a 10% alcohol solution must be mixed with 50 liters of a 80% solution
(answered by Fombitz,modi.kk91)
How many liters of a 50% alcohol solution must be mixed with 80 liters if a 90% solution (answered by stanbon)