Overview
Not every mixture problem involves percentages or chemical concentrations. The same mathematical structure appears whenever products with different unit values are combined into a single blend.
A coffee merchant may blend beans costing different amounts per pound. A grocer may combine varieties of nuts or grain. A manufacturer may mix materials with different costs per kilogram. In each case, every portion contributes some amount of value to the final mixture.
The central relationship is
\[ \boxed{\text{Total Value}=\text{Unit Value}\times\text{Quantity}}. \]If we let \(V\) represent total value, \(P\) represent price or value per unit, and \(Q\) represent quantity, then
\[ \boxed{V=PQ}. \]This has the same mathematical structure as the Mixture Principle
\[ A=CT. \]Only the interpretation of the quantities has changed.
From Concentration to Value
In concentration problems, a mixture contributes an amount of a component according to
\[ \text{Component Amount} = \text{Concentration} \times \text{Quantity}. \]In price problems, a product contributes value according to
\[ \text{Total Value} = \text{Unit Price} \times \text{Quantity}. \]The parallel is direct:
| Concentration Mixture | Price or Value Mixture |
|---|---|
| Concentration \(C\) | Unit price \(P\) |
| Quantity \(T\) | Quantity \(Q\) |
| Component amount \(A=CT\) | Total value \(V=PQ\) |
This is why price-and-value problems belong naturally within the same mixture family.
Conservation of Total Value
Suppose two products are combined.
The first has unit price \(P_1\) and quantity \(Q_1\), so its value is
\[ P_1Q_1. \]The second has unit price \(P_2\) and quantity \(Q_2\), so its value is
\[ P_2Q_2. \]If the final blend has unit value \(P_f\) and total quantity \(Q_f\), then its total value is
\[ P_fQ_f. \]Assuming no value is added or lost through the blending process,
\[ \boxed{ P_1Q_1+P_2Q_2=P_fQ_f }. \]If the quantities simply combine,
\[ Q_f=Q_1+Q_2, \]so
\[ \boxed{ P_1Q_1+P_2Q_2=P_f(Q_1+Q_2) }. \]Example 1: Finding the Value of a Blend
A coffee merchant combines \(6\) pounds of coffee worth \(\$8\) per pound with \(4\) pounds of coffee worth \(\$13\) per pound. What is the average value per pound of the blend?
The first coffee contributes
\[ 8(6)=48 \]dollars of value.
The second contributes
\[ 13(4)=52. \]The total value is
\[ 48+52=100. \]The total quantity is
\[ 6+4=10 \]pounds.
Therefore, the value per pound of the blend is
\[ P_f=\frac{100}{10}. \] \[ P_f=10. \]The final blend is worth
\[ \boxed{\$10\text{ per pound}}. \]The Final Price Is a Weighted Average
The final unit price can be written as
\[ \boxed{ P_f= \frac{P_1Q_1+P_2Q_2}{Q_1+Q_2} }. \]This is a weighted average of the two original prices.
In the coffee example,
\[ P_f= \frac{8(6)+13(4)}{10}. \] \[ P_f= \frac{100}{10}=10. \]The prices \(\$8\) and \(\$13\) are not simply averaged unless the quantities are equal.
The amount of each product determines how strongly its price influences the final average.
When Does a Simple Average Work?
Suppose \(5\) pounds of a product worth \(\$6\) per pound are mixed with \(5\) pounds of a product worth \(\$10\) per pound.
Because the quantities are equal, the average value is
\[ \frac{6+10}{2}=8. \]So the blend is worth
\[ \boxed{\$8\text{ per pound}}. \]A simple arithmetic average works here because both prices receive equal weight.
If the quantities differ, the weighted-average model must be used.
Example 2: Finding an Unknown Quantity
A merchant has coffee worth \(\$6\) per pound and coffee worth \(\$10\) per pound. How many pounds of the \(\$10\) coffee should be mixed with \(12\) pounds of the \(\$6\) coffee to produce a blend worth \(\$7.50\) per pound?
Let \(x\) represent the pounds of \(\$10\) coffee.
The first coffee contributes
\[ 6(12)=72 \]dollars of value.
The second contributes
\[ 10x. \]The final quantity is
\[ 12+x. \]At \(\$7.50\) per pound, the final blend has value
\[ 7.50(12+x). \]Apply conservation of value:
\[ 72+10x=7.50(12+x). \]Expand:
\[ 72+10x=90+7.50x. \]Subtract \(7.50x\):
\[ 72+2.50x=90. \]Subtract \(72\):
\[ 2.50x=18. \]Therefore,
\[ x=7.2. \]So the merchant should add
\[ \boxed{7.2\text{ pounds}} \]of the \(\$10\)-per-pound coffee.
Organizing a Value Mixture in a Table
| Product | Unit Value | Quantity | Total Value |
|---|---|---|---|
| Lower-priced product | \(\$6\) | \(12\) | \(\$72\) |
| Higher-priced product | \(\$10\) | \(x\) | \(\$10x\) |
| Final blend | \(\$7.50\) | \(12+x\) | \(\$7.50(12+x)\) |
Reading across the total-value column gives
\[ 72+10x=7.50(12+x). \]As with concentration tables, the purpose of the table is simply to keep the important quantities aligned.
Example 3: A Fixed Final Quantity
A grocer wants to make \(30\) pounds of a nut mixture worth \(\$9\) per pound using nuts worth \(\$6\) per pound and nuts worth \(\$11\) per pound. How many pounds of each should be used?
Let \(x\) be the amount of the \(\$6\) nuts.
Then the amount of the \(\$11\) nuts is
\[ 30-x. \]The total value equation is
\[ 6x+11(30-x)=9(30). \]Expand:
\[ 6x+330-11x=270. \]Combine like terms:
\[ 330-5x=270. \]Subtract \(330\):
\[ -5x=-60. \]Therefore,
\[ x=12. \]So the grocer should use
\[ 12\text{ pounds} \]of the \(\$6\) nuts and
\[ 30-12=18 \]pounds of the \(\$11\) nuts.
Thus,
\[ \boxed{12\text{ pounds at }\$6\text{ and }18\text{ pounds at }\$11}. \]Checking the Value
The lower-priced nuts contribute
\[ 6(12)=72 \]dollars.
The higher-priced nuts contribute
\[ 11(18)=198 \]dollars.
The total value is
\[ 72+198=270. \]The final mixture contains \(30\) pounds, so
\[ \frac{270}{30}=9. \]The blend is therefore worth
\[ \boxed{\$9\text{ per pound}}, \]as required.
The Target Price Must Usually Lie Between the Source Prices
If two products worth \(\$6\) and \(\$11\) per unit are simply blended, their average value must lie between
\[ \$6 \]and
\[ \$11. \]A blend made entirely from those two products cannot ordinarily have an average value of \(\$14\) per unit.
Likewise, it cannot have an average value of \(\$4\) per unit.
For positive quantities,
\[ \boxed{ P_{\min}\leq P_f\leq P_{\max} }. \]This provides the same kind of reasonableness check used in concentration problems.
The Price-Difference Pattern
Just as concentration problems produce a useful difference relationship, price mixtures do the same.
Suppose two products have prices \(P_1\) and \(P_2\), with
\[ P_1Rearranging gives
\[ (P_f-P_1)Q_1=(P_2-P_f)Q_2. \]Therefore,
\[ \boxed{ \frac{Q_1}{Q_2} = \frac{P_2-P_f}{P_f-P_1} }. \]The amount of one product is proportional to the price difference on the opposite side of the target.
Example 4: Using the Difference Pattern
A blend is to be made from products worth \(\$4\) and \(\$10\) per pound so that the final blend is worth \(\$6\) per pound.
The target is
\[ 6-4=2 \]dollars above the lower price and
\[ 10-6=4 \]dollars below the higher price.
Therefore,
\[ \frac{Q_{\$4}}{Q_{\$10}} = \frac{4}{2} = 2. \]So the quantities must be in the ratio
\[ \boxed{2:1}. \]For every \(2\) pounds of the \(\$4\) product, \(1\) pound of the \(\$10\) product should be used.
For example,
\[ 2(4)+1(10)=18 \]dollars of value across \(3\) pounds:
\[ \frac{18}{3}=6. \]The Same Algebra in Different Clothing
Compare these two equations.
A concentration mixture:
\[ 0.30x+0.60(10)=0.40(x+10). \]A price mixture:
\[ 6x+10(10)=8(x+10). \]The numbers have different interpretations, but the mathematical structure is identical.
Both equations say:
\[ \boxed{ \text{contribution from source 1} + \text{contribution from source 2} = \text{total in the final blend} }. \]Recognizing this common structure is more useful than memorizing separate procedures for every type of mixture problem.
Example 5: Finding an Unknown Unit Price
A merchant combines \(8\) pounds of one product with \(12\) pounds of another. The first product is worth \(\$5\) per pound, and the final blend is worth \(\$8\) per pound. What is the price per pound of the second product?
Let \(x\) represent the unknown unit price.
The first product contributes
\[ 5(8)=40 \]dollars of value.
The second contributes
\[ 12x. \]The final blend contains
\[ 8+12=20 \]pounds and is worth
\[ 8(20)=160 \]dollars.
Therefore,
\[ 40+12x=160. \]Subtract \(40\):
\[ 12x=120. \]Thus,
\[ x=10. \]The second product is worth
\[ \boxed{\$10\text{ per pound}}. \]Three or More Products
The same principle extends to any number of products.
For three products,
\[ \boxed{ P_1Q_1+P_2Q_2+P_3Q_3=P_fQ_f }. \]If all quantities combine,
\[ Q_f=Q_1+Q_2+Q_3. \]So
\[ P_1Q_1+P_2Q_2+P_3Q_3 = P_f(Q_1+Q_2+Q_3). \]Each source simply contributes its own value to the whole.
Example 6: A Three-Product Blend
A coffee blend contains \(5\) pounds worth \(\$6\) per pound, \(10\) pounds worth \(\$8\) per pound, and \(5\) pounds worth \(\$12\) per pound. Find the average value per pound.
The three values are
\[ 6(5)=30, \] \[ 8(10)=80, \]and
\[ 12(5)=60. \]The total value is
\[ 30+80+60=170. \]The total quantity is
\[ 5+10+5=20. \]Therefore,
\[ P_f=\frac{170}{20}=8.5. \]The blend is worth
\[ \boxed{\$8.50\text{ per pound}}. \]Mixture Value Is Not Automatically a Selling Price
A price-mixture calculation usually determines the average underlying value or cost of the combined goods.
It does not automatically include profit, markup, taxes, shipping, labor, or other business costs unless the problem explicitly includes them.
For example, a blend whose ingredients average
\[ \$8 \]per pound may be sold for
\[ \$10 \]per pound.
The \(\$8\) figure describes the blended cost or value under the mixture model. The \(\$10\) figure involves an additional pricing decision.
Carefully identify what the stated “price” represents before constructing the equation.
Example 7: Including a Desired Markup
A merchant produces a \(20\)-pound blend at an average ingredient cost of \(\$7.50\) per pound. If the merchant wants a \(20\%\) markup on ingredient cost, what selling price per pound would correspond to that markup?
The ingredient cost per pound is
\[ \$7.50. \]A \(20\%\) markup is
\[ 0.20(7.50)=1.50. \]Therefore, the selling price is
\[ 7.50+1.50=9.00. \]So the corresponding selling price is
\[ \boxed{\$9.00\text{ per pound}}. \]This markup calculation occurs after the mixture calculation. It is not part of the conservation-of-value equation itself.
Value Mixtures and Functions
A product with fixed unit price \(P\) has total value
\[ V(Q)=PQ. \]This is a linear function of quantity.
For a product worth \(\$6\) per pound,
\[ V(Q)=6Q. \]For another product worth \(\$10\) per pound,
\[ V(Q)=10Q. \]A mixture problem combines these value functions:
\[ V_{\text{total}} = V_1+V_2. \]For quantities \(x\) and \(y\),
\[ V_{\text{total}}=6x+10y. \]This shows how mixture problems connect naturally to linear functions and systems of equations.
Connection to Systems of Equations
When neither quantity is expressed in terms of the other, price mixtures can be represented using a system.
Suppose \(x\) pounds of one product and \(y\) pounds of another must produce \(30\) pounds total.
Then
\[ x+y=30. \]If their prices are \(\$6\) and \(\$11\) per pound and the final blend must be worth \(\$9\) per pound, then
\[ 6x+11y=9(30). \]Thus the problem can be modeled by
\[ \begin{cases} x+y=30,\\ 6x+11y=270. \end{cases} \]Solving the system gives the same result as substituting
\[ y=30-x \]into a single mixture equation.
Pay Attention to Units
A unit value should always be multiplied by a compatible quantity.
For example,
\[ \$8/\text{lb}\times6\text{ lb}=\$48. \]Likewise,
\[ \$3/\text{kg}\times10\text{ kg}=\$30. \]If one quantity is measured in pounds and another in ounces, convert them to compatible units before combining values.
Unit analysis makes the relationship
\[ V=PQ \]especially transparent.
Recognizing Impossible Blends
Suppose a merchant has only products worth \(\$4\) and \(\$9\) per pound.
Can positive quantities of these two products be blended to create an average ingredient value of \(\$12\) per pound?
No.
A weighted average of \(\$4\) and \(\$9\) must lie between those two values.
Likewise, an average of \(\$2\) per pound is impossible using only those ingredients.
If the algebra produces a negative quantity, that often signals that the target value lies outside the possible range.
Method for Price and Value Mixtures
- Identify each unit price or unit value. Determine the cost or value associated with one unit of each product.
- Identify each quantity. Introduce a variable for any unknown amount.
- Find or represent each total value. Use \[ V=PQ. \]
- Determine the final quantity. If products are simply combined, add their quantities.
- Write the final total value. Use \[ P_fQ_f. \]
- Apply conservation of value. Set the sum of the source values equal to the value of the final blend.
- Solve the equation or system. Use the appropriate algebraic method.
- Check the answer. Verify that quantities are nonnegative and that the average value lies within the expected range.
Common Errors
- Averaging prices without considering quantities. Prices must generally be weighted by how much of each product is used.
- Adding unit prices instead of total values. The conserved quantities are \(P_1Q_1\), \(P_2Q_2\), and so forth.
- Forgetting the final total quantity. The final value is \(P_fQ_f\), not merely \(P_f\).
- Confusing cost per unit with total cost. A product costing \(\$8\) per pound does not contribute only \(\$8\) if several pounds are used.
- Accepting a target value outside the source range. A positive blend of two products has an average value between their unit values.
- Mixing incompatible units. Convert pounds, ounces, kilograms, or other units before combining quantities.
- Including markup automatically. Profit or selling price should only be included when the problem states that it is part of the model.
Practice Problems
- \(8\) pounds of coffee worth \(\$6\) per pound are mixed with \(12\) pounds worth \(\$9\) per pound. Find the average value per pound.
- \(5\) pounds of nuts worth \(\$4\) per pound are mixed with \(15\) pounds worth \(\$8\) per pound. Find the average value per pound.
- How many pounds of a product worth \(\$12\) per pound should be mixed with \(10\) pounds of a product worth \(\$6\) per pound to obtain a blend worth \(\$8\) per pound?
- A merchant wants \(30\) pounds of a blend worth \(\$7\) per pound using products worth \(\$5\) and \(\$9\) per pound. How many pounds of each should be used?
- Eight pounds of a product worth \(\$4\) per pound are mixed with \(12\) pounds of another product. The final blend is worth \(\$7\) per pound. Find the unit value of the second product.
- A blend contains \(5\) pounds worth \(\$3\) per pound, \(10\) pounds worth \(\$6\) per pound, and \(5\) pounds worth \(\$11\) per pound. Find the average value per pound.
- Can products worth \(\$5\) and \(\$9\) per pound be blended to produce an average value of \(\$11\) per pound? Explain.
- A merchant has products worth \(\$4\) and \(\$10\) per pound. How many pounds of each are needed to make \(24\) pounds of a blend worth \(\$7\) per pound?
Practice Answers
- \[ P_f= \frac{6(8)+9(12)}{20} \] \[ = \frac{48+108}{20} \] \[ = \frac{156}{20} = \boxed{\$7.80\text{ per pound}}. \]
- \[ P_f= \frac{4(5)+8(15)}{20} \] \[ = \frac{20+120}{20} = \boxed{\$7\text{ per pound}}. \]
- Let \(x\) be the amount of the \(\$12\) product: \[ 6(10)+12x=8(10+x). \] \[ 60+12x=80+8x. \] \[ 4x=20. \] \[ \boxed{x=5\text{ pounds}}. \]
- Let \(x\) be the amount of the \(\$5\) product: \[ 5x+9(30-x)=7(30). \] \[ 5x+270-9x=210. \] \[ -4x=-60. \] \[ x=15. \] Therefore, \[ \boxed{15\text{ pounds of each product}}. \]
- Let \(x\) be the unknown price: \[ 4(8)+12x=7(20). \] \[ 32+12x=140. \] \[ 12x=108. \] \[ \boxed{x=\$9\text{ per pound}}. \]
- \[ P_f= \frac{3(5)+6(10)+11(5)}{20}. \] \[ = \frac{15+60+55}{20}. \] \[ = \frac{130}{20} = \boxed{\$6.50\text{ per pound}}. \]
- \[ \boxed{\text{No}}. \] A positive blend of products worth \(\$5\) and \(\$9\) per pound must have an average value between \(\$5\) and \(\$9\) per pound.
- Let \(x\) be the amount of the \(\$4\) product: \[ 4x+10(24-x)=7(24). \] \[ 4x+240-10x=168. \] \[ -6x=-72. \] \[ x=12. \] Therefore, \[ \boxed{12\text{ pounds at }\$4\text{ and }12\text{ pounds at }\$10}. \]
Summary
Price and value mixtures use the relationship
\[ \boxed{V=PQ}, \]where \(V\) is total value, \(P\) is value per unit, and \(Q\) is quantity.
For two products,
\[ \boxed{ P_1Q_1+P_2Q_2=P_fQ_f }. \]When the quantities simply combine,
\[ \boxed{ P_1Q_1+P_2Q_2=P_f(Q_1+Q_2) }. \]The final unit value is therefore a weighted average:
\[ \boxed{ P_f= \frac{P_1Q_1+P_2Q_2}{Q_1+Q_2} }. \]The deeper lesson is that mixture reasoning is not limited to concentration. Whenever each quantity contributes some measurable property according to a rate per unit, the contributions can often be combined through the same general mathematical structure.
Continue the Exploration
Price mixtures show that the Mixture Principle applies beyond liquids and concentrations. The next lesson returns to physical composition and examines alloys and other materials whose makeup is described by percentages of individual components.
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Continue to: Alloy and Composition Problems
Apply conservation and percentage reasoning to metals, materials, and other mixtures whose composition is measured by component percentages.