Finding a Formula for MixturesDate: 11/10/2003 at 21:41:56 From: Gaye Subject: algebra story problem Alfredo needs to make 250 ml of a 27% alcohol solution, using a 15% solution and a 40% solution. How much of each should he use? Date: 11/11/2003 at 12:57:35 From: Doctor Ian Subject: Re: alegebra story problem Hi Gaye, As a guess, let's suppose that he starts with 125 ml of each kind of solution. The first solution will contain x parts alcohol, where 15 parts alcohol x parts alcohol ------------------ = ------------------ 100 parts solution 125 parts solution (0.15 * 125) = x The second will contain y parts alcohol, where 40 parts alcohol y parts alcohol ------------------ = ------------------ 100 parts solution 125 parts solution (0.40 * 125) = y And the concentration of the resulting solution will be (x + y) parts alcohol 27 parts alcohol -------------------------- = ------------------ (125 + 125) parts solution 100 parts solution Does this make sense? We can substitute the values we just found for x and y: (0.15 * 125 + 0.40 * 125) parts alcohol 27 parts alcohol ---------------------------------------- = ------------------ (125 + 125) parts solution 100 parts solution Now, this probably isn't going to work out to be true! In fact, if we evaluate each side, we get 68.75 27 ----- = --- 250 100 Cross-multiplying we get 100 * 68.75 = 250 * 27 6875 = 3712.5 which is false. So our guess was wrong. But what if we don't split the two solutions evenly? For example, suppose we use 100 parts of the 15% solution to 150 parts of the 40% solution? Then we get (0.15 * 100 + 0.40 * 150) parts alcohol 27 parts alcohol ---------------------------------------- = ------------------ (100 + 150) parts solution 100 parts solution This _still_ isn't right, probably. (Work it out if you'd like.) However, we can start to see some patterns. For one thing, the two amounts have to add up to 250 parts, no matter how we split them up: (0.15 * 100 + 0.40 * 150) parts alcohol 27 parts alcohol ---------------------------------------- = ------------------ 250 parts solution 100 parts solution And as soon as we choose the amount of 15% solution, the amount of 40% solution is fixed. That is, if we use P parts of the 15% solution, we get (0.15*P + 0.40*(250-P)) parts alcohol 27 parts alcohol ------------------------------------- = ------------------ 250 parts solution 100 parts solution We can generalize it a little further, by letting T stand for some arbitrary total amount of solution: (0.15*P + 0.40*(T-P)) parts alcohol 27 parts alcohol ----------------------------------- = ------------------ T parts solution 100 parts solution Does this make sense? Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/