# SOLUTION: Hi i had 2 questions. My first question is this. How many gallons of pure alcohol should be mixed with 20 gal of a 15% alcohol solution to obtain a mixture that is 25% alcohol? I

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Hi i had 2 questions. My first question is this. How many gallons of pure alcohol should be mixed with 20 gal of a 15% alcohol solution to obtain a mixture that is 25% alcohol? I       Log On

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 Click here to see ALL problems on Mixture Word Problems Question 257068: Hi i had 2 questions. My first question is this. How many gallons of pure alcohol should be mixed with 20 gal of a 15% alcohol solution to obtain a mixture that is 25% alcohol? I have done this problem several times and i get as far as .15x + 3 = .25x + 5 but i keep getting a negative answer. Could you please show me how to solve this correctly. My other question is this Marin Caswell needs 10% hydrocholoric acid for a chemistry experiment. How much 5% acid should mix with 60 mL of 20% acid to get a 10% solution? I really want to know how to solve these and set them up so i can do them on the test. Thank youAnswer by ankor@dixie-net.com(15745)   (Show Source): You can put this solution on YOUR website!How many gallons of pure alcohol should be mixed with 20 gal of a 15% alcohol solution to obtain a mixture that is 25% alcohol? ; Let x = amt of pure alcohol required : A per cent equation .15(20) + 1x = .25(x+20) 3 + 1x = .25x + 5 1x - .25x = 5 - 3 .75x = 2 x = : x = 2 gal of pure alcohol required : : Marin Caswell needs 10% hydrocholoric acid for a chemistry experiment. How much 5% acid should mix with 60 mL of 20% acid to get a 10% solution? : Let x = amt of 5% mixture required : .05x + .20(60) = .10(x+60) .05x + 12 = .10x + 6 12 - 6 = .10x - .05x 6 = .05x x = x = 120 ml of 5% stuff : Prove this .05(120) + .20(60) = .10(120 + 60) 6 + 12 = .10(180) 18 = 18 : : You can prove the 1st one, like we did this one.