The Central Idea
Suppose a quantity \(Q\) has a per-unit property \(v\).
Its total contribution is
\[ \boxed{A=vQ}. \]Depending on the problem, \(v\) might represent:
- a component fraction;
- a metal percentage;
- a price per unit;
- a value per kilogram;
- a grade or quality measure that combines linearly.
Mixture problems conserve the total contribution:
\[ \boxed{ v_1Q_1+v_2Q_2 = v_f(Q_1+Q_2) }. \]One Algebraic Structure, Several Applications
The equation
\[ v_1Q_1+v_2Q_2 = v_f(Q_1+Q_2) \]can describe many settings.
For an alloy, \(v\) may be the fraction of copper.
For a product blend, \(v\) may be dollars per pound.
For a material mixture, \(v\) may be the proportion of a particular ingredient.
The context changes, but the weighted-contribution algebra stays the same.
Alloy Composition
An alloy is a mixture containing metals or other materials in specified proportions.
If an alloy has mass \(M\) and is \(p\) percent copper, then its copper mass is
\[ \boxed{ \frac{p}{100}M }. \]Percentages should be converted to decimal or fractional form before multiplication.
Example: Component Mass in an Alloy
A \(24\)-kg alloy is \(35\%\) copper.
Copper mass:
\[ 0.35(24)=8.4. \]Therefore,
\[ \boxed{8.4\text{ kg of copper}}. \]Combining Two Alloys
Suppose alloy 1 has mass \(M_1\) and component fraction \(c_1\), while alloy 2 has mass \(M_2\) and fraction \(c_2\).
The final component fraction \(c_f\) satisfies
\[ \boxed{ c_1M_1+c_2M_2 = c_f(M_1+M_2) }. \]Example: Find the Composition of a Combined Alloy
Combine \(10\) kg of an alloy that is \(20\%\) copper with \(15\) kg of an alloy that is \(40\%\) copper.
Copper contributed:
\[ 0.20(10)+0.40(15) = 2+6 = 8. \]Total mass:
\[ 10+15=25. \]Therefore,
\[ c_f=\frac8{25}=0.32. \]Thus the resulting alloy is
\[ \boxed{32\%\text{ copper}}. \]Unknown Amount of One Alloy
How much \(60\%\) copper alloy should be mixed with \(20\) kg of a \(30\%\) copper alloy to obtain a \(40\%\) copper alloy?
Let \(x\) be the mass of the \(60\%\) alloy.
\[ 0.60x+0.30(20)=0.40(x+20). \]Simplify:
\[ 0.60x+6=0.40x+8. \] \[ 0.20x=2. \] \[ x=10. \]Therefore,
\[ \boxed{10\text{ kg}}. \]Unknown Source Composition
An \(8\)-kg alloy of unknown copper fraction \(c\) is mixed with \(12\) kg of a \(25\%\) copper alloy to obtain a \(40\%\) copper alloy.
Write
\[ 8c+0.25(12)=0.40(20). \] \[ 8c+3=8. \] \[ 8c=5. \] \[ c=0.625. \]Therefore,
\[ \boxed{62.5\%\text{ copper}}. \]Fixed Final Mass
If the final alloy must have mass \(M\), let one source amount be \(x\).
The other source amount is then
\[ \boxed{M-x}. \]This automatically satisfies total-mass conservation.
Example: Fixed-Mass Alloy
How much \(20\%\) and \(50\%\) copper alloy should be combined to make \(30\) kg of a \(35\%\) copper alloy?
Let \(x\) be the mass of the \(20\%\) alloy.
Then \(30-x\) is the mass of the \(50\%\) alloy.
\[ 0.20x+0.50(30-x)=0.35(30). \] \[ 0.20x+15-0.50x=10.5. \] \[ -0.30x=-4.5. \] \[ x=15. \]Therefore,
\[ \boxed{15\text{ kg of each alloy}}. \]Adding Pure Metal
Pure component has fraction
\[ \boxed{1}. \]If \(x\) kg of pure copper is added to an alloy, then the copper amount and total mass both increase by \(x\).
The equation takes the form
\[ \boxed{ A+x=c_f(M+x) }. \]Example: Strengthen an Alloy with Pure Metal
A \(20\)-kg alloy is \(30\%\) copper. How much pure copper must be added to make it \(50\%\) copper?
Initial copper:
\[ 0.30(20)=6. \]Let \(x\) kg of pure copper be added:
\[ 6+x=0.50(20+x). \] \[ 6+x=10+0.5x. \] \[ 0.5x=4. \] \[ x=8. \]Therefore,
\[ \boxed{8\text{ kg of pure copper}}. \]Composition Problems Need Not Be Alloys
The same structure applies to any mixture described by component proportions.
Examples include:
- animal-feed mixtures;
- fertilizer blends;
- food ingredients;
- construction materials;
- manufacturing composites.
The relevant component may be protein, nitrogen, fiber, recycled content, or another measurable constituent.
Example: Material Composition
A \(40\)-kg material is \(25\%\) recycled content. It is mixed with \(20\) kg of material that is \(70\%\) recycled content.
Recycled material:
\[ 0.25(40)+0.70(20) = 10+14 = 24. \]Total material:
\[ 60. \]Therefore,
\[ \frac{24}{60}=0.40. \]Thus,
\[ \boxed{40\%\text{ recycled content}}. \]Price Mixtures
In a price mixture, the conserved quantity is total monetary value.
If \(Q\) units cost \(p\) dollars per unit, then total value is
\[ \boxed{V=pQ}. \]When two products are blended,
\[ \boxed{ p_1Q_1+p_2Q_2 = p_f(Q_1+Q_2) }. \]Example: Find the Price of a Blend
A merchant combines \(10\) lb of a product worth \(\$4\) per pound with \(15\) lb worth \(\$8\) per pound.
Total value:
\[ 4(10)+8(15) = 40+120 = 160. \]Total weight:
\[ 10+15=25. \]Therefore, the value per pound is
\[ \frac{160}{25}=6.4. \]Thus,
\[ \boxed{\$6.40\text{ per lb}}. \]The Final Unit Price Is a Weighted Average
In general,
\[ p_f = \frac{p_1Q_1+p_2Q_2}{Q_1+Q_2}. \]Therefore,
\[ \boxed{ p_f = \frac{p_1Q_1+p_2Q_2}{Q_1+Q_2} }. \]The source with more quantity receives more weight in the final unit price.
Equal Quantities and Average Price
If equal quantities \(Q\) are blended at prices \(p_1\) and \(p_2\), then
\[ p_f = \frac{p_1Q+p_2Q}{2Q}. \]Therefore,
\[ \boxed{ p_f=\frac{p_1+p_2}{2} }. \]The ordinary arithmetic mean works only because the quantities are equal.
Unknown Quantity in a Price Blend
How many pounds of a \(\$10\)-per-pound product should be mixed with \(20\) lb of a \(\$4\)-per-pound product to obtain a blend worth \(\$6\) per pound?
Let \(x\) be the amount of the \(\$10\) product.
\[ 10x+4(20)=6(x+20). \] \[ 10x+80=6x+120. \] \[ 4x=40. \] \[ x=10. \]Therefore,
\[ \boxed{10\text{ lb}}. \]Fixed-Quantity Price Blend
A merchant wants \(50\) lb of a blend worth \(\$7\) per pound using products worth \(\$5\) and \(\$10\) per pound.
Let \(x\) be the pounds of the \(\$5\) product.
Then
\[ 50-x \]pounds come from the \(\$10\) product.
\[ 5x+10(50-x)=7(50). \] \[ 5x+500-10x=350. \] \[ -5x=-150. \] \[ x=30. \]Therefore,
\[ \boxed{30\text{ lb at }\$5} \]and
\[ \boxed{20\text{ lb at }\$10}. \]Value Tables
A value table mirrors a concentration table.
\[ \begin{array}{c|c|c|c} \text{source} & \text{quantity} & \text{unit value} & \text{total value}\\ \hline 1 & Q_1 & p_1 & p_1Q_1\\ 2 & Q_2 & p_2 & p_2Q_2\\ \hline \text{final} & Q_1+Q_2 & p_f & p_f(Q_1+Q_2) \end{array} \]The total-value column creates the conservation equation.
Composition Tables
Alloy and composition problems can be organized similarly:
\[ \begin{array}{c|c|c|c} \text{source} & \text{mass} & \text{component fraction} & \text{component mass}\\ \hline 1 & M_1 & c_1 & c_1M_1\\ 2 & M_2 & c_2 & c_2M_2\\ \hline \text{final} & M_1+M_2 & c_f & c_f(M_1+M_2) \end{array} \]Two Unknown Amounts
If a final mixture has fixed total quantity \(Q\), and two source amounts \(x\) and \(y\) are unknown, use
\[ \boxed{x+y=Q} \]together with the weighted-contribution equation
\[ \boxed{ v_1x+v_2y=v_fQ }. \]This is a system of two linear equations.
Example: Two-Equation Price System
A \(60\)-lb blend is worth \(\$7.50\) per pound. It is made from products worth \(\$6\) and \(\$10\) per pound.
Let \(x\) and \(y\) be the source quantities.
\[ x+y=60. \] \[ 6x+10y=7.5(60). \]Since
\[ 7.5(60)=450, \]substitute
\[ x=60-y. \] \[ 6(60-y)+10y=450. \] \[ 360-6y+10y=450. \] \[ 4y=90. \] \[ y=22.5. \]Therefore,
\[ x=37.5. \]Thus,
\[ \boxed{37.5\text{ lb at }\$6} \]and
\[ \boxed{22.5\text{ lb at }\$10}. \]Target Values Must Lie between Source Values
For positive source quantities, a weighted average lies between the source values.
Therefore, a blend of products worth \(\$4\) and \(\$8\) per unit cannot have a unit value of \(\$10\) without some additional source or adjustment.
In general,
\[ \boxed{ \min(v_1,v_2) \le v_f \le \max(v_1,v_2) }. \]The Ratio Shortcut
Suppose
\[ v_1<v_f<v_2. \]The required source quantities satisfy
\[ \boxed{ \frac{Q_1}{Q_2} = \frac{v_2-v_f}{v_f-v_1} }. \]The source closer to the target generally needs the larger quantity.
Example: Ratio for a Price Blend
In what ratio should products worth \(\$4\) and \(\$10\) per pound be mixed to obtain a blend worth \(\$6\) per pound?
Use the distance from the target:
\[ \frac{Q_4}{Q_{10}} = \frac{10-6}{6-4} = \frac42 = 2. \]Therefore,
\[ \boxed{2:1}. \]Example: Ratio for an Alloy
In what ratio should \(20\%\) and \(60\%\) copper alloys be mixed to obtain a \(30\%\) copper alloy?
\[ \frac{Q_{20}}{Q_{60}} = \frac{60-30}{30-20} = \frac{30}{10} = 3. \]Therefore,
\[ \boxed{3:1}. \]Why the Ratio Is Crossed
The amount of the lower-value source is proportional to the distance from the target to the higher value.
Likewise, the amount of the higher-value source is proportional to the distance from the lower value to the target.
This “crossed difference” comes directly from rearranging the weighted average equation.
Deriving the Ratio Shortcut
Start with
\[ v_1Q_1+v_2Q_2=v_f(Q_1+Q_2). \]Expand:
\[ v_1Q_1+v_2Q_2=v_fQ_1+v_fQ_2. \]Rearrange:
\[ (v_f-v_1)Q_1=(v_2-v_f)Q_2. \]Therefore,
\[ \boxed{ \frac{Q_1}{Q_2} = \frac{v_2-v_f}{v_f-v_1} }. \]Three-Source Value Mixtures
With three sources,
\[ \boxed{ v_1Q_1+v_2Q_2+v_3Q_3 = v_f(Q_1+Q_2+Q_3) }. \]If several source quantities are unknown, additional conditions are needed to determine a unique solution.
Example: Three Known Prices
A blend contains:
- \(5\) lb at \(\$4\) per lb;
- \(10\) lb at \(\$6\) per lb;
- \(15\) lb at \(\$8\) per lb.
Total value:
\[ 4(5)+6(10)+8(15) = 20+60+120 = 200. \]Total quantity:
\[ 30. \]Therefore,
\[ p_f=\frac{200}{30}=\frac{20}{3}. \]Thus,
\[ \boxed{\$6.67\text{ per lb approximately}}. \]Markup Is Not Part of the Mixture Value unless Stated
A price-mixture equation usually tracks the value or cost of the blended goods themselves.
If a later selling price includes markup, tax, commission, or other adjustments, those should be modeled separately unless the problem explicitly includes them in the mixture value.
The dedicated lesson on taxes, discounts, commissions, and markups develops those transformations later.
Cost versus Selling Price
Do not mix cost per unit and selling price per unit in the same weighted conservation equation unless the problem defines them as the same measured property.
A mixture equation must use one consistent kind of value:
\[ \boxed{ \text{same unit property on every term} }. \]Value Conservation
If two goods are blended without gain, loss, or additional processing cost, the total value of the sources is
\[ V_1+V_2. \]Since
\[ V_i=p_iQ_i, \]the final value equation is
\[ \boxed{ p_1Q_1+p_2Q_2=p_fQ_f }. \]When Total Quantity Changes
Most simple blend problems use
\[ Q_f=Q_1+Q_2. \]If the process includes waste, evaporation, removal, or another quantity change, the final total must be modeled separately.
Never assume quantity conservation when the problem explicitly states that material is lost or removed.
Composition with More than One Component
An alloy may contain several components.
If copper and zinc are both tracked, each conserved component can have its own equation.
For copper:
\[ c_{1,\text{Cu}}Q_1+c_{2,\text{Cu}}Q_2 = c_{f,\text{Cu}}Q_f. \]For zinc:
\[ c_{1,\text{Zn}}Q_1+c_{2,\text{Zn}}Q_2 = c_{f,\text{Zn}}Q_f. \]Often one component equation is redundant when the listed component fractions sum to \(1\).
Complementary Components
If an alloy contains only copper and zinc, then
\[ c_{\text{Cu}}+c_{\text{Zn}}=1. \]Therefore, a \(35\%\) copper alloy is automatically
\[ 65\% \]zinc.
Tracking one component may be enough to determine the other.
Example: Complementary Composition
A \(50\)-kg alloy is \(42\%\) copper and contains only copper and zinc.
Zinc percentage:
\[ 100\%-42\%=58\%. \]Zinc mass:
\[ 0.58(50)=29. \]Therefore,
\[ \boxed{29\text{ kg of zinc}}. \]Unknown Price of One Source
Ten pounds of a product with unknown unit price \(p\) are mixed with \(20\) lb worth \(\$5\) per lb. The final \(30\)-lb blend is worth \(\$7\) per lb.
\[ 10p+5(20)=7(30). \] \[ 10p+100=210. \] \[ 10p=110. \] \[ p=11. \]Therefore,
\[ \boxed{\$11\text{ per lb}}. \]Unknown Final Unit Value
If all source quantities and values are known, compute
\[ \boxed{ v_f = \frac{\sum v_iQ_i}{\sum Q_i} }. \]This is the general weighted-average formula.
Weighted Average Notation
For \(n\) sources,
\[ \boxed{ \bar v = \frac{ v_1Q_1+v_2Q_2+\cdots+v_nQ_n }{ Q_1+Q_2+\cdots+Q_n } }. \]The quantities \(Q_i\) act as weights.
Weights as Fractions of the Whole
Define
\[ w_i=\frac{Q_i}{Q_1+\cdots+Q_n}. \]Then
\[ w_1+\cdots+w_n=1. \]The weighted average becomes
\[ \boxed{ \bar v=w_1v_1+\cdots+w_nv_n }. \]Example: Weight Fractions
A blend contains \(20\) kg at value \(v_1\) and \(30\) kg at value \(v_2\).
Total quantity is
\[ 50. \]Therefore, the weights are
\[ w_1=\frac{20}{50}=0.4 \]and
\[ w_2=\frac{30}{50}=0.6. \]Hence
\[ \boxed{ \bar v=0.4v_1+0.6v_2 }. \]Weighted Averages Stay within Bounds
If all weights are nonnegative and sum to \(1\), then the weighted average cannot lie outside the smallest and largest source values.
Therefore,
\[ \boxed{ \min(v_i)\le\bar v\le\max(v_i) }. \]This fact is useful for checking alloy percentages, unit prices, and other blend values.
Closer Targets Require More of the Nearby Source
Suppose a target price is much closer to the cheaper source than to the expensive source.
Then the mixture generally requires more of the cheaper source.
This follows from the ratio formula and gives a fast qualitative check on the solution.
Example: Predict before Solving
A blend uses products worth \(\$4\) and \(\$12\) per pound to obtain a target of \(\$6\) per pound.
Since \(\$6\) is much closer to \(\$4\) than to \(\$12\), we expect more of the \(\$4\) product.
The ratio confirms:
\[ \frac{Q_4}{Q_{12}} = \frac{12-6}{6-4} = \frac62 = 3. \]Therefore,
\[ \boxed{3:1}. \]Checking by Total Mass or Quantity
If the final blend is supposed to contain \(50\) kg and the solved source amounts are \(20\) kg and \(30\) kg, verify
\[ 20+30=50. \]Total-quantity conservation is the first check.
Checking by Total Component
For an alloy, compute the actual component mass from the solved source amounts.
Then divide by the final mass to verify the target percentage.
For example,
\[ \frac{ 0.20(15)+0.50(15) }{30} = \frac{3+7.5}{30} = 0.35. \]The resulting alloy is indeed \(35\%\).
Checking by Total Value
Suppose a \(50\)-lb blend is claimed to be worth \(\$7\) per pound.
Its total value should be
\[ 7(50)=350. \]The source values should add to the same total.
Checking by Bounds
A mixture of \(25\%\) and \(60\%\) alloys cannot produce a standard positive-quantity blend that is \(80\%\).
Likewise, blending products worth \(\$3\) and \(\$9\) per unit cannot produce a value of \(\$12\) per unit without another effect.
Common Error: Averaging Source Values Directly
The arithmetic mean
\[ \frac{v_1+v_2}{2} \]is valid only for equal source quantities.
Otherwise, use
\[ \boxed{ v_f = \frac{v_1Q_1+v_2Q_2}{Q_1+Q_2} }. \]Common Error: Treating Percent as a Whole Number
A \(35\%\) alloy should be represented by
\[ 0.35, \]not
\[ 35. \]Thus \(20\) kg of \(35\%\) alloy contain
\[ 0.35(20)=7 \]kg of the tracked component.
Common Error: Adding Prices Instead of Values
Unit prices do not add directly.
A \(5\)-lb source at \(\$4\) per lb contributes
\[ 4(5)=20 \]dollars of value.
Total values add; unit prices are recovered only after dividing by total quantity.
Common Error: Mixing Cost and Quantity Units
If one price is quoted per kilogram and another per pound, convert the units before combining values.
The same unit of quantity must underlie every unit-value term.
Common Error: Forgetting the Final Quantity
If \(x\) units are added to \(Q\) existing units, the final quantity is
\[ \boxed{Q+x}. \]The final contribution must be based on the actual final amount.
Common Error: Target Outside the Source Range
If the target value is outside the interval formed by the source values, the standard positive two-source mixture has no solution.
A negative source quantity from the algebra is often evidence of this impossibility.
Common Error: Confusing Markup with Mixture Value
A blend's weighted unit value and its later selling price are different concepts.
Apply markup, discount, tax, or commission only after establishing which value the problem asks the mixture equation to conserve.
A Reliable Procedure for Alloy and Composition Problems
- Choose the component to track.
- Convert percentages to decimal fractions.
- Multiply each source fraction by its quantity.
- Add the component amounts.
- Write the final component amount as target fraction times final quantity.
- Solve for the unknown amount or composition.
- Check total quantity and component mass.
- Verify that the target lies within the feasible source range.
A Reliable Procedure for Price and Value Mixtures
- Identify the unit value for each source.
- Multiply unit value by quantity to find each source's total value.
- Add the source values.
- Write the final value as final unit value times final quantity.
- Solve the resulting equation or system.
- Keep cost units and quantity units consistent.
- Check the final unit value against the source-value range.
A Reliable Procedure for Fixed-Total Blends
- Let one source quantity be \(x\).
- Write the other as \(Q-x\).
- Write each weighted contribution.
- Set their sum equal to the target contribution.
- Solve for \(x\).
- Find the other source amount.
- Verify that both quantities are nonnegative and sum to \(Q\).
Practice 1: Alloy Component Mass
A \(32\)-kg alloy is \(37.5\%\) copper. Find the mass of copper.
Practice 2: Combined Alloy Percentage
Mix \(12\) kg of a \(25\%\) copper alloy with \(18\) kg of a \(50\%\) copper alloy. Find the final copper percentage.
Practice 3: Unknown Alloy Amount
How many kilograms of a \(70\%\) copper alloy should be mixed with \(20\) kg of a \(30\%\) alloy to obtain a \(50\%\) alloy?
Practice 4: Unknown Alloy Composition
Ten kilograms of an unknown alloy are mixed with \(15\) kg of a \(20\%\) copper alloy to form a \(40\%\) copper alloy. Find the unknown copper percentage.
Practice 5: Fixed-Mass Alloy
How many kilograms of \(20\%\) and \(60\%\) copper alloys are needed to make \(40\) kg of a \(35\%\) alloy?
Practice 6: Add Pure Copper
A \(24\)-kg alloy is \(25\%\) copper. How much pure copper must be added to produce a \(40\%\) copper alloy?
Practice 7: Final Price of a Blend
Mix \(20\) lb of a product worth \(\$3\) per pound with \(30\) lb worth \(\$7\) per pound. Find the unit value of the final blend.
Practice 8: Unknown Price-Blend Quantity
How many pounds of a \(\$10\)-per-pound product should be mixed with \(30\) lb of a \(\$4\)-per-pound product to make a blend worth \(\$6\) per pound?
Practice 9: Fixed-Quantity Price Blend
A merchant wants \(40\) lb of a blend worth \(\$8\) per pound using products worth \(\$5\) and \(\$11\) per pound. Find the amount of each.
Practice 10: Unknown Source Price
Ten pounds of a product at unknown price \(p\) are mixed with \(20\) lb at \(\$6\) per pound to obtain a \(30\)-lb blend worth \(\$8\) per pound. Find \(p\).
Practice 11: Ratio Shortcut for Alloy
In what ratio should \(10\%\) and \(50\%\) alloys be mixed to obtain a \(20\%\) alloy?
Practice 12: Ratio Shortcut for Price
In what ratio should products worth \(\$5\) and \(\$13\) per unit be mixed to obtain a blend worth \(\$7\) per unit?
Practice 13: Three-Source Value Blend
A blend contains \(5\) kg worth \(\$4\) per kg, \(10\) kg worth \(\$6\) per kg, and \(15\) kg worth \(\$10\) per kg. Find the final value per kilogram.
Practice 14: Feasibility Check
Can a \(75\%\) alloy be made using only \(20\%\) and \(60\%\) alloys with positive source amounts? Explain.
Practice 15: Identify the Error
A merchant mixes \(10\) lb worth \(\$4\) per lb with \(30\) lb worth \(\$8\) per lb and says the blend is worth
\[ \frac{4+8}{2}=\$6\text{ per lb}. \]Explain the error and find the correct unit value.
Practice Answers
Practice 1
\[ 0.375(32)=12. \]Therefore,
\[ \boxed{12\text{ kg of copper}}. \]Practice 2
Copper mass:
\[ 0.25(12)+0.50(18) = 3+9 = 12. \]Total mass:
\[ 12+18=30. \]Therefore,
\[ \frac{12}{30}=0.40. \] \[ \boxed{40\%\text{ copper}}. \]Practice 3
Let \(x\) be the amount of \(70\%\) alloy.
\[ 0.70x+0.30(20)=0.50(x+20). \] \[ 0.70x+6=0.50x+10. \] \[ 0.20x=4. \] \[ x=20. \]Therefore,
\[ \boxed{20\text{ kg}}. \]Practice 4
Let \(c\) be the unknown fraction.
\[ 10c+0.20(15)=0.40(25). \] \[ 10c+3=10. \] \[ 10c=7. \] \[ c=0.70. \]Therefore,
\[ \boxed{70\%}. \]Practice 5
Let \(x\) be the amount of \(20\%\) alloy.
\[ 0.20x+0.60(40-x)=0.35(40). \] \[ 0.20x+24-0.60x=14. \] \[ -0.40x=-10. \] \[ x=25. \]Therefore,
\[ \boxed{25\text{ kg of }20\%\text{ alloy}} \]and
\[ \boxed{15\text{ kg of }60\%\text{ alloy}}. \]Practice 6
Initial copper:
\[ 0.25(24)=6. \]Let \(x\) be pure copper added:
\[ 6+x=0.40(24+x). \] \[ 6+x=9.6+0.4x. \] \[ 0.6x=3.6. \] \[ x=6. \]Therefore,
\[ \boxed{6\text{ kg}}. \]Practice 7
Total value:
\[ 3(20)+7(30) = 60+210 = 270. \]Total quantity:
\[ 50. \]Therefore,
\[ \frac{270}{50}=5.4. \] \[ \boxed{\$5.40\text{ per lb}}. \]Practice 8
Let \(x\) be the amount of the \(\$10\) product.
\[ 10x+4(30)=6(x+30). \] \[ 10x+120=6x+180. \] \[ 4x=60. \] \[ x=15. \]Therefore,
\[ \boxed{15\text{ lb}}. \]Practice 9
Let \(x\) be the amount of the \(\$5\) product.
\[ 5x+11(40-x)=8(40). \] \[ 5x+440-11x=320. \] \[ -6x=-120. \] \[ x=20. \]Therefore,
\[ \boxed{20\text{ lb at }\$5} \]and
\[ \boxed{20\text{ lb at }\$11}. \]Practice 10
\[ 10p+6(20)=8(30). \] \[ 10p+120=240. \] \[ 10p=120. \] \[ p=12. \]Therefore,
\[ \boxed{\$12\text{ per lb}}. \]Practice 11
\[ \frac{Q_{10}}{Q_{50}} = \frac{50-20}{20-10} = \frac{30}{10} = 3. \]Therefore,
\[ \boxed{3:1}. \]Practice 12
\[ \frac{Q_5}{Q_{13}} = \frac{13-7}{7-5} = \frac62 = 3. \]Therefore,
\[ \boxed{3:1}. \]Practice 13
Total value:
\[ 4(5)+6(10)+10(15) = 20+60+150 = 230. \]Total quantity:
\[ 30. \]Therefore,
\[ \frac{230}{30} = \frac{23}{3} \approx7.67. \]Thus,
\[ \boxed{\$7.67\text{ per kg approximately}}. \]Practice 14
No. With positive quantities, the final alloy percentage must lie between \(20\%\) and \(60\%\).
\[ \boxed{\text{Not possible under the stated model}}. \]Practice 15
The quantities are unequal, so the unit prices cannot be averaged directly.
Total value:
\[ 4(10)+8(30) = 40+240 = 280. \]Total quantity:
\[ 40. \]Therefore,
\[ \frac{280}{40}=7. \]Thus,
\[ \boxed{\$7\text{ per lb}}. \]What to Remember
Alloy, composition, price, and value mixtures all use the same weighted contribution structure:
\[ \boxed{ v_1Q_1+v_2Q_2 = v_f(Q_1+Q_2) }. \]For composition, \(v\) is a component fraction.
For price mixtures, \(v\) is a unit price and \(vQ\) is total monetary value.
The final per-unit value is
\[ \boxed{ v_f = \frac{\sum v_iQ_i}{\sum Q_i} }. \]The most important habit is to identify the quantity that is truly being conserved—component mass, total value, or another weighted contribution— before writing the equation.
Where We Go Next
Mixture models track how quantities and values combine. Another major applied algebra family tracks how money moves through a business: costs, sales, revenue, profit, and the point where revenue exactly covers cost.
The next lesson, Cost, Revenue, Profit, and Break-Even, develops those relationships.