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This Lesson (Mixture: one material quantity unknown) was created by by josgarithmetic(39617)
   : View Source, ShowAbout josgarithmetic: Academic and job experience with beginning & intermediate Algebra. Tutorial help mostly for Basic Math and up through intermediate algebra.
MIXTURE: ONE MATERIAL QUANTITY UNKNOWN
A common type of mixture exercise is to find the amount of one unknown material. Concentrations for the mixture could be as percent (parts per hundred of a total) or price or cost (money per mass, or money per item count). Here are a couple of two-part mixture examples requiring one unknown material quantity and therefore only one unknown variable: ! A bulk foods manager wants to mix 30 pounds of old, stale coffee, priced to sell for $9 per pound with fresh coffee priced to sell for $6 per pound. The manager wants the selling price to be $8.40 per pound. How much of the fresh coffee should be added and mixed with the old stale beans? ! A blending tank holds 2210 pounds of liquid at 7.5 percent nonvolatile matter. How much of a similar but much more concentrated, 35 percent nonvolatile liquid should be added and blended so that the mixture results in 9.0 percent nonvolatile matter? Those two examples and many like them can be imagined as holding the same form, or depending how variables are assigned, slightly different forms. These are the same problem but just different examples. SOLVE SYMBOLICALLY Assign Variables to All Numbers: L = lower strength material T = target strength wanted H = higher strength of material p = quantity of the lower strength material, might be unknown or known q = quantity of the higher strength material, might be known or unknown Note that in the variable assignments just described, THE STARTING EQUATION We want (Some Expression) = (Target Concentration or Price). The pure matter of interest is The amount of blend or mixture is The starting equation is Solve for whichever variable is unknown. The resulting formulas will be different. As well, ways of assigning variables can be different from one person to another, but in any case, the fundamental equations, expressions, and skills are the same. THE UNKNOWN IS p THE UNKNOWN IS q Steps proceed much like above done, but change slightly according to wanting the other material quantity as unknown, Depending on which material quantity variables are known and unknown, the final formulas are somewhat different. The basic starting equation is usually easier to produce. FINAL FORMULA AND FINISHING Once you obtain the final formula for the solved variable, you just substitute the known or given values into all of those variables and compute the value of the variable which you solved. The purpose for assigning variables and solving symbolically is to be able to solve a general problem few times, instead of a general problem several times. THE EXAMPLES GIVEN (!) The old-new coffee example. L = 6 T = 8.4 H = 9 p = unknown pounds of new fresh, inexpensive coffee beans q = 30 The starting equation would fit the form shown way above, and the final formula to for the fresh new pounds of coffee to add would fit (!) The liquid with specified percent nonvolatile example. Variables assigned this way: L=7.5 T=9.0 H=35.0 p= 2210 pounds, corresponding to L q = unknown corresponding to H The starting equation would fit the form shown way above, and the final formula for how much of the high concentrated liquid to add, q, would fit This lesson has been accessed 3799 times. |
