Mixture Problems. Controlling Equations Approach

Transcription

1 Mixture Problems One class of frequently encountered situations is what has been loosely called mixture problems. These arise when two constituent sources are being combined -- or are considered to -- yield some mixed result. It may be mixing a strong solution and a weak solution to get a medium strength solution; combining an expensive product with a low-cost product to get a medium priced mixture; or investing money into two accounts, one a high-yield account and one a low-yield account, to collect the combined interest from these accounts. There are two approaches to mixture problems, with some variation, they are: A controlling equation approach wherein the information in the problem is viewed through the lens of some equation the controlling equation -- relating the parameters, and the data is frequently managed by use of a table. The columns of the table may represent the parameters of the controlling equation; the rows would then represent the different solutions, materials, or accounts under consideration. A weighted averages approach that views the information in the problem from the point of view of weighted averages. This method emphasizes ratios and proportions. Though it is computationally easier, many more concepts and constructs need to be introduced before this approach is used. Fortunately, since the solution to a mixture problem is independent of route taken. We can get to the computational ease of this approach by sneaking in a side door. There are two side doors. Be patient. Controlling Equations Approach There are several controlling equations that may be loosely categorized as some sort of rate times a parameter yields another parameter. (Keep in mind that a rate is the ratio between two distinct parameters! That is why rates always include their units.) Some examples of these controlling equations are: st nd RATE 1 PARAMETER 2 PARAMETER speed time distance speed of work time number of jobs done price weight value concentration volume of solution volume of pure chemical concentration weight of alloy weight of pure metal annual simple interest rate principal annual dividend

2 The above equations can also be expressed in different forms. For example, the last relation can also be found as: simple interest rate principal years = total dividend. Consider this standard mixture problem: Example 1: How many gallons of a 60% saline solution must be added to 30 gallons of an 85% saline solution to get a 80% solution? The controlling equation is Concentration times Volume of solution equals Volume of pure chemical, cv p. Creating a table, based on the equation and the information, we can let each column stand for a parameter of the equation and each row represent one of the solutions under consideration. Concentration Vol of Sol = Vol of Salt = = = We can further organize the data by listing the solution that are to be USED first, and the solution being MADE last. In this case we get something in the nature of: Concentration Vol of Sol = Vol of Salt 60% = 85% 30 gal = 80% = We will use the variable v to represent the volume of the 60% solution measured in gallons. Keeping in mind that liquids add directly, that is 2cups of water added to 3 cups of water yields 5 cups of water, we can add the 30 gal and the v in like manner. Performing the arithmetic, we get: Concentration Vol of Sol = Vol of Salt 60% v = 0.6v 85% 30 gal = % v+30 = 0.8(v+30) The following equation may be harvested from the right most column of the table: 0.6v v 30 Solving this equation for v leads to the conclusion that7.5 gallons of the 60% solution should be used. [Note: we could also have harvested v30v 30 from the Vol of Sol column, but this equation would not add anything distinct to our knowledge. In other cases, the Vol of Sol column may yield useful equations.]

3 Let us try that again with another problem: Example 2: How much 80% acid solution and how much 30% acid solution should be mixed together to get a 400mL of a 40% acid solution? Just like last problem we identify the controlling equation as: Concentration Volume = Volume of pure chemical. We again utilize a table to organize the information: Concentration Volume = Volume of acid Starting solution = Starting solution = Final solution = Harvest the information and insert it into the table Concentration Volume = Volume of acid Starting solution 30% = Starting solution 80% = Final solution 40% 400 ml = Notice that the task statement in our problem reads How much 80% acid solution and how much 30% acid solution should be mixed. That means we need two variables for the volumes. I will use h and v as my variables. I will also convert the percentage values to the decimal format: Concentration Volume = Volume of acid Starting solution 0.3 v = Starting solution 0.8 h = Final solution ml = Multiplying within the rows, we can fill in the last column. Concentration Volume = Volume of acid Starting solution 0.3 v = 0.3v Starting solution 0.8 h = 0.8h Final solution ml = 160 Following the logic that both the total volume of solution and the volume acid must add up, we get two equations: Volumes of the starting solutions: v h400 Volume of pure acid contained: 0.3v 0.8h 160

4 Focus on the first equation, the one about volume of solutions. We can rearrange that equation by subtracting h from both sides of the equation: vh400v400 h This means that wherever we see a v we can replace it with 400 h. Substituting into the second equation we get: 0.3v 0.8h h 0.8h 160 Solving this last equation, we can find the volume of the 80% solution: h 0.8h h0.8h h h40 h80 We conclude that we need 80mL of the 80% solution. We can use this information to figure out how much 30% solution we need. Since the solution must add up to 400mL, we conclude we need 320mL of the 30% solution. When the mixture problem is couched in terms of price and value, the logic and flow is the same. Consider the following: Example 3: Mr McGregor wishes create a mix of sultanas and peanuts. How many pounds of sultanas, selling for $2.25 per pound, should be mixed with peanuts, selling at $1.50 per pound, to get 40 pounds of a mixture valued at $2.00 per pound? The controlling equation is Price of ingredient times Weight of ingredient equals Value of the ingredient, pw v. Creating a table, based on the equation and the information, we can let each column stand for a parameter of the equation and each row represent one of the ingredients under consideration. We can further organize the information by listing first the ingredients to be USED, and the mixture being MADE last. In this case we get something in the nature of: Price Weight = Value Sultanas $2.25 = Peanuts $1.50 = Mixture $ lbs = We will use the variable n to represent the weight of peanuts measured in pounds; and the variable r to represent the weight of sultanas measured in pounds. We can flesh out table using the variables.

5 Price Weight = Value Sultanas $2.25 r = Peanuts $1.50 n = Mixture $ lbs = Performing the indicated arithmetic we can further fill-in the table: Price Weight = Value Sultanas $2.25 r = 2.25r Peanuts $1.50 n = 1.50n Mixture $ lbs = 80 We can harvest two equations from the weight and value columns of the table: Weight: rn 40 Value: 2.25r1.50n 80 We can solve the weight equation for the variable n: n40 r and substitute for n in the value equation to get: 2.25r r 80 2 This last equation can be solved to yield r 26. This leads to the conclusion that 3 1 n 13. (What leads to this conclusion?) Therefore, Mr McGregor should use 26 lbs of sultanas and 13 lbs of peanuts. 3 3 This same method can be adapted for mixed investment problems: Oleg Oospenskiya invests $29,000 some at 2.0% simple interest and the rest at 2.5% simple interest. If the total amount of interest from the two accounts at the end of the first year is $640, how much was invested in each account? As before, we will utilize a table to organize the information. Unlike before, we only have two explicit accounts. Earlier we had three explicit solutions, or three explicit prices. We will have to adapt. Our data rows will list only two accounts and the resulting totals: Interest Rate Amount Invested = Interest Earned High yield account = Low yield account = Totals Harvesting the information we get:

6 Interest Rate Amount Invested = Interest Earned High yield account 2.5% = Low yield account 2.0% = Totals -- $29,000 $640 We need to choose some variables, so let s assign h to be the amount invested in the high yield account and let y be the amount in the low yield account. We get: Interest Rate Amount Invested = Interest Earned High yield account 2.5% h = 0.025h Low yield account 2.0% y = 0.020y Totals -- $29,000 $640 We can now see the implied equations: h y h0.020y640 Rearrange the first equation to get: hy29000y29000 h Substitute h in the second equation with that to which h is equal, and then solve for y: 0.025h0.020y h h h h h h60 h12000 Since h and the total amount invested was $29,000, we conclude that y So Oleg invested $12,000 at 2.5% and $17,000 at 2.0% Those Side Doors I Told You About THE FIRST SIDE DOOR: The general mixture problem has a high concentration solution or a high priced material or high yield account that will be mixed with a low concentration solution, a low priced material, or low yield account. The mixture will yield a medium concentration solution or a medium priced mix or a medium yield account. So the pattern in High mixed with Low yields Medium. Sometimes what we know is how much solution we want to make, what weight of mixed material we want, or what kind of yield we want. The problem will read something like:

7 Lobelia Demente has on hand a High-% belladonna potion and a Low-% belladonna potion. For a special friend, she wishes to make V ounces of a Medium-% potion. How much High-% belladonna and how much Low-% belladonna should be mixed to get the desired Medium-% potion? Of course, the problem would have numbers instead of High, Low, Medium, and V. But I chose those words to make choice of variable obvious If we proceed with this problem as previously described, we would organize the information into a table that would look something like: Concentration Volume = Volume of pure belladonna Starting potion H = Starting potion L = Final potion M V = I simply used H and L for the values of the high concentration and low concentration potions and V for the volume. Further, we need variables for the volumes of the High and Low concentration potions. We could use the lower case letters for the volumes, if we are careful not to confuse upper and lower cases. I will use c for the volume of the high (concentrated) potion and w for the volume of the low (weak) potion. We get: Concentration Volume = Volume of pure belladonna Starting potion H c = Hc Starting potion L w = Lw Final potion M V = MV The implied equations are: HcLwMV cwv Let us multiply both side of the second equation by the negative of L, L : LcLw LV Let us add Lc Lw to the left hand side of the first equation and LV to the right hand side and simplify. This is shown below:

8 HcLwMV HcLwLcLwMVLV HcLcMVLV Add to each side. c H L V M L Combine like terms Factor/Reverse distribute Divide both side of equation by HL ML cv H L Look at that last equation!!!!!! What does the variable c represent? It is the volume of the High concentration potion! The fraction M L is simply the ratio of the difference between the HL Medium concentration and the Low concentration and the difference between the High concentration and the Low concentration! Why all the exclamation points? If I keep this in mind, I get to avoid work. (I always relish the opportunity to avoid work. If you can figure out how to do a job quicker or more easily and still get the same paycheck, you just gave yourself a pay raise. Just do not tell your boss, she or he will just give you more work to do.) Let us try it. Lobelia Demente has on hand an 80% belladonna potion and a 30% belladonna potion. For a special friend, she wishes to make V ounces of a 65% potion. How much 80% belladonna and how much 30% belladonna should be mixed to get the desired 65% potion? Using the reasoning above, to figure out how much of the 80% potion to use I take: ML cv H L So, Lobelia should use 14 ozs of the 80% potion and 6 oz of the 30% potion. (Where did the 6 oz come from? Think about it.) FOR DISCUSSION/GROUP WORK: In developing the short cut method, the amount of the high concentration solution was found. Could you alter the method to solve for the low concentration solution?

9 FOR DISCUSSION/GROUP WORK: Try working that problem out the other way. Do you get the same answer? How long did it take? This method can also be applied to the investing problem we saw earlier: Oleg Oospenskiya invests $29,000 some at 2.0% simple interest and the rest at 2.5% simple interest. If the total amount of interest from the two accounts at the end of the first year is $640, how much was invested in each account? In order to analyze this problem we need to identify the high yield account the low yield account and the medium yield account. The high is 2.5%, the low is 2.0%. How we find the medium yield account? We use the amount of total interest divided by the initial investment: $ % $29, From our previous discussion, c is the amount invested in the high yield account and V is the total $29,000 invested. We use our short cut formula to get ML cv H L So Oleg invested $12,000 at 2.5%, and $17,000 at 2.0% THE SECOND DOOR What if the information in the problem is altered as to which values are provided? What if we are not told the total volume, but instead are given the volume of one of the solutions to be used? Consider the following example: Example XX: How many fluid ounces of 50% vinegar should be mixed with70 fl oz 5% vinegar to get a solution which is 15% vinegar? We could go back to the tried and true Controlling Equation Method with which we started out. There is nothing wrong with that approach, but where is the fun in that? Let us think about this. The resulting mixture will be 15% in strength. This will be made up of the 50% solution and the 5% solution. To which starting solution is the 15% more similar? Since the difference between the 15% and the 5% is only 10%, whereas the difference between the 50% and the 15% is 35%, we can conclude the 15% is more like the 5% than the 50%. We can also conclude that we will use more of the 5% solution than the 50% solution.

10 It turns out that the amounts of each solution in the mixture are in proportion to these differences, so that these differences may be thought of as a sort of volume terms. The differences are NOT the volumes, but are proportional to the volumes. KEY IDEA: The larger difference is associated with the dominant solution. Let s see how this plays out in our current example. We will is a general proportionality multiplier p in to get the volume terms: Difference factor of Mix Low:15% 5% 10% 10p Difference factor of High Mix:50% 15% 35% 35p Which of those volume terms is larger? That would be the 35p, right? So, the portion of the mixture that is from the 5% is35p. Another way to put that is: 35p is the volume of the 5%. We can write: 35p 70 So, p 2. We now take that p2and apply it to the volume factor that represents the 50% to get p We conclude we require 20 fl oz of the 50% solution. ML 15% 5% 10% 2 cv H L 50% 5% 45% 9 So 20 fl oz is the volume of the 50% vinegar required and the remaining 70 fl oz must be from the 5% vinegar. There are alternate graphic mnemonics for this approach. One such is constructed as a rectangle: High Concentration Med Difference between Medium and Low Low Concentration Difference between High and Medium

11 Using this visual on Example 2 we would have: 50% 15% 5% = 10% 15% 5% This rectangular visualization results in the volume terms we saw earlier: 10x for the High volume and 35x for the Low concentration.