# SOLUTION: how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture?

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 Question 473567: how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture?Found 2 solutions by stanbon, oberobic:Answer by stanbon(57967)   (Show Source): You can put this solution on YOUR website!how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture? ------- Eqution: alcohol + alcohol = alcohol x + 0.25*20 = 0.85(x+20) ---- Multiply thru 100 to get: 100x + 25*20 = 85x + 85*20 15x = 60*20 x = 5 liters (amt. of pure alcohol needed) ============================================ Cheers, Stan H. ================== Answer by oberobic(2304)   (Show Source): You can put this solution on YOUR website!When solving mixture problems, try to figure out how "pure" stuff you need. In this case, we're told 20 liters of 25% alcohol, which means there are 5 liters of pure alcohol and 15 liters of water. We're also told we need to obtain an unknown amount of pure alcohol to obtain an 85% solution. Let x be the unknown. So when we add that to the 20 liters, that will be (20+x). . 20*25% + x*100% = 85% * (20+x) 20*.25 + x*1.00 = .85 (20+x) 5 + x = 17 + .85x . subtract .85 x from both sides subtract 5 from both sides . .15x = 12 x = 80 , Does this make sense? The best way to tell is to check your answer. . Adding 80 liters of pure alcohol to the 20 liters of 25% alcohol yields 100 liters. We know from above that 20 liters of 25% alcohol contain 5 liters of pure alcohol. 80+5 = 85 So we have 85 liters of pure alcohol in the 100 liters of solution, which is 85% alcohol. . Answer: Add 80 liters of pure alcohol. . Done.