Overview
An alloy is a material formed by combining metals or other elements in specific proportions. Brass, for example, is primarily composed of copper and zinc, while many forms of steel contain iron combined with carbon and other elements.
Mathematically, alloy problems are concentration problems expressed in terms of mass or composition rather than liquid volume.
If an alloy is \(30\%\) copper and has a total mass of \(20\) kilograms, then its copper mass is
\[ 0.30(20)=6 \]kilograms.
The same Mixture Principle applies:
\[ \boxed{A=CT}, \]where \(A\) is the amount of a particular component, \(C\) is its fractional composition, and \(T\) is the total mass or quantity.
Composition Is a Form of Concentration
The word concentration is often associated with liquids, but mathematically it simply describes how much of a whole belongs to a particular component.
An alloy that is \(40\%\) copper has copper concentration
\[ C=0.40. \]If the alloy weighs \(25\) kilograms, then the amount of copper is
\[ A=0.40(25). \]Therefore,
\[ A=10. \]So the alloy contains
\[ \boxed{10\text{ kilograms of copper}}. \]The remaining \(15\) kilograms consist of other material.
Two-Component Alloys
Suppose an alloy contains only copper and zinc.
If it is \(35\%\) copper, then the remaining percentage must be
\[ 100\%-35\%=65\%. \]So the alloy is \(65\%\) zinc.
In decimal form,
\[ 0.35+0.65=1. \]This simple relationship is useful when a problem gives the percentage of only one component.
For a two-component mixture,
\[ \boxed{C_1+C_2=1}. \]Conservation of a Component
When two alloys are melted together, the total amount of a particular metal in the final alloy equals the sum of the amounts contributed by the original alloys.
If one alloy contributes
\[ C_1T_1 \]units of copper and another contributes
\[ C_2T_2, \]then the final alloy contains
\[ C_1T_1+C_2T_2. \]If the final alloy has concentration \(C_f\) and total mass \(T_f\), then
\[ \boxed{ C_1T_1+C_2T_2=C_fT_f }. \]When the original masses simply combine,
\[ T_f=T_1+T_2. \]Therefore,
\[ \boxed{ C_1T_1+C_2T_2=C_f(T_1+T_2) }. \]Example 1: Finding the Final Alloy Composition
A metalworker combines \(10\) kilograms of an alloy that is \(20\%\) copper with \(15\) kilograms of an alloy that is \(40\%\) copper. What percentage of the final alloy is copper?
The first alloy contains
\[ 0.20(10)=2 \]kilograms of copper.
The second contains
\[ 0.40(15)=6 \]kilograms of copper.
The final alloy therefore contains
\[ 2+6=8 \]kilograms of copper.
Its total mass is
\[ 10+15=25 \]kilograms.
So the final copper concentration is
\[ C_f=\frac{8}{25}. \] \[ C_f=0.32. \]Thus the final alloy is
\[ \boxed{32\%\text{ copper}}. \]Checking the Final Composition
The original alloys were \(20\%\) and \(40\%\) copper.
Because both masses are positive and no pure copper or other material was added, the final percentage should lie between the original percentages:
\[ 20\%<32\%<40\%. \]The second alloy also had greater mass and greater copper concentration, so the final result should be pulled somewhat toward \(40\%\).
The answer \(32\%\) is therefore reasonable.
Organizing an Alloy Problem
| Material | Component Percentage | Total Mass | Component Mass |
|---|---|---|---|
| Alloy 1 | \(0.20\) | \(10\) | \(2\) |
| Alloy 2 | \(0.40\) | \(15\) | \(6\) |
| Final alloy | \(C_f\) | \(25\) | \(25C_f\) |
The component-mass column gives
\[ 2+6=25C_f. \]Solving gives
\[ C_f=0.32. \]Example 2: Finding an Unknown Alloy Quantity
How many kilograms of a \(60\%\) copper alloy should be mixed with \(20\) kilograms of a \(30\%\) copper alloy to obtain a \(40\%\) copper alloy?
Let \(x\) be the mass of the \(60\%\) alloy.
The \(30\%\) alloy contains
\[ 0.30(20)=6 \]kilograms of copper.
The \(60\%\) alloy contributes
\[ 0.60x. \]The final mass is
\[ 20+x. \]At \(40\%\) copper, the final alloy contains
\[ 0.40(20+x) \]kilograms of copper.
Apply conservation:
\[ 6+0.60x=0.40(20+x). \]Expand:
\[ 6+0.60x=8+0.40x. \]Subtract \(0.40x\):
\[ 6+0.20x=8. \]Subtract \(6\):
\[ 0.20x=2. \]Therefore,
\[ x=10. \]So
\[ \boxed{10\text{ kilograms}} \]of the \(60\%\) alloy should be used.
Problems with a Fixed Final Mass
Sometimes the final mass is given rather than one of the starting masses.
Suppose \(x\) kilograms of one alloy and \(y\) kilograms of another are combined to produce \(T\) kilograms total.
Then
\[ x+y=T. \]If we use one variable,
\[ y=T-x. \]This allows the composition equation to be written with a single unknown.
Example 3: Making a Fixed Amount of Alloy
A manufacturer needs \(40\) kilograms of an alloy that is \(35\%\) copper. Available alloys are \(20\%\) copper and \(50\%\) copper. How much of each should be used?
Let \(x\) be the kilograms of the \(20\%\) alloy.
Then the amount of the \(50\%\) alloy is
\[ 40-x. \]The copper equation is
\[ 0.20x+0.50(40-x)=0.35(40). \]Expand:
\[ 0.20x+20-0.50x=14. \]Combine like terms:
\[ 20-0.30x=14. \]Subtract \(20\):
\[ -0.30x=-6. \]Therefore,
\[ x=20. \]The amount of the \(50\%\) alloy is
\[ 40-20=20. \]Thus,
\[ \boxed{20\text{ kg of each alloy}}. \]Why Equal Amounts Worked
The target concentration in the previous example was
\[ 35\%. \]This lies exactly halfway between
\[ 20\% \]and
\[ 50\%. \]Since the target is the arithmetic midpoint,
\[ \frac{20\%+50\%}{2}=35\%, \]equal masses of the two alloys produce the desired composition.
If the target were closer to one of the two source percentages, more of that source alloy would be required.
Adding a Pure Metal
Pure copper has copper concentration
\[ 100\%=1. \]So if \(x\) kilograms of pure copper are added, they contribute
\[ x \]kilograms of copper.
The general model becomes
\[ CT+x=C_f(T+x). \]This is exactly the same structure developed earlier for adding a pure substance.
Example 4: Strengthening an Alloy with Pure Metal
A \(30\)-kilogram alloy is \(40\%\) copper. How much pure copper must be added to make the alloy \(50\%\) copper?
The original amount of copper is
\[ 0.40(30)=12. \]Let \(x\) be the kilograms of pure copper added.
Then
\[ 12+x=0.50(30+x). \]Expand:
\[ 12+x=15+0.50x. \]Subtract \(0.50x\):
\[ 12+0.50x=15. \]Subtract \(12\):
\[ 0.50x=3. \]Therefore,
\[ x=6. \]Thus,
\[ \boxed{6\text{ kilograms of pure copper}} \]must be added.
Adding Material with None of the Tracked Component
Suppose a copper alloy is mixed with a material containing no copper.
Relative to copper, the added material has concentration
\[ 0\%=0. \]This lowers the percentage of copper by increasing total mass while leaving copper mass unchanged.
The model is
\[ CT=C_f(T+x), \]the same structure used for dilution.
Example 5: Lowering the Percentage of a Component
A \(20\)-kilogram alloy is \(45\%\) copper. How much copper-free material must be added to reduce the copper percentage to \(30\%\)?
The original copper mass is
\[ 0.45(20)=9. \]Let \(x\) be the mass of the copper-free material.
Then
\[ 9=0.30(20+x). \]Expand:
\[ 9=6+0.30x. \]Subtract \(6\):
\[ 3=0.30x. \]Therefore,
\[ x=10. \]So
\[ \boxed{10\text{ kilograms}} \]of copper-free material must be added.
Tracking More Than One Component
An alloy may contain several metals.
For example, suppose an alloy is
- \(60\%\) copper,
- \(25\%\) zinc, and
- \(15\%\) tin.
The percentages must add to
\[ 60\%+25\%+15\%=100\%. \]For a \(20\)-kilogram alloy, the component masses are
\[ 0.60(20)=12 \]kilograms of copper,
\[ 0.25(20)=5 \]kilograms of zinc, and
\[ 0.15(20)=3 \]kilograms of tin.
Together,
\[ 12+5+3=20. \]Each component can be tracked independently using the Mixture Principle.
The Multi-Component View
Suppose a mixture contains several components with fractions
\[ C_1,C_2,\ldots,C_n. \]Because they describe the whole mixture,
\[ \boxed{ C_1+C_2+\cdots+C_n=1 }. \]If the total quantity is \(T\), then the component amounts are
\[ A_1=C_1T, \] \[ A_2=C_2T, \]and so forth.
The total of all component amounts must satisfy
\[ A_1+A_2+\cdots+A_n=T. \]This is the mathematical foundation of composition problems with several materials.
Example 6: Finding a Missing Composition Percentage
An alloy is \(55\%\) iron, \(18\%\) chromium, and the remainder nickel. What percentage of the alloy is nickel?
The known percentages total
\[ 55\%+18\%=73\%. \]Therefore, the nickel percentage is
\[ 100\%-73\%=27\%. \]So the alloy is
\[ \boxed{27\%\text{ nickel}}. \]Example 7: Finding Component Masses
A \(50\)-kilogram alloy is \(64\%\) iron, \(20\%\) chromium, and \(16\%\) nickel. Find the mass of each metal.
Iron:
\[ 0.64(50)=32. \]Chromium:
\[ 0.20(50)=10. \]Nickel:
\[ 0.16(50)=8. \]Thus the alloy contains
\[ \boxed{32\text{ kg iron},\ 10\text{ kg chromium},\ 8\text{ kg nickel}}. \]As a check,
\[ 32+10+8=50. \]Purity Problems
Composition problems also appear in the language of purity.
A metal described as \(90\%\) pure gold contains
\[ 0.90T \]units of gold in a total quantity \(T\).
The remaining
\[ 10\% \]consists of other material.
Purity problems therefore use exactly the same mathematical structure as alloy and concentration problems.
Example 8: Mixing Different Purities
A jeweler combines \(8\) grams of gold that is \(75\%\) pure with \(12\) grams of gold that is \(90\%\) pure. What is the purity of the resulting mixture?
The first sample contains
\[ 0.75(8)=6 \]grams of pure gold.
The second contains
\[ 0.90(12)=10.8 \]grams.
The total pure gold is
\[ 6+10.8=16.8. \]The total mass is
\[ 8+12=20. \]Therefore,
\[ C_f=\frac{16.8}{20}=0.84. \]The resulting mixture is
\[ \boxed{84\%\text{ pure gold}}. \]Composition May Be Expressed in Other Scales
Not every composition problem uses percentages directly.
Gold purity, for example, may be expressed in karats, where \(24\)-karat gold represents pure gold.
The gold fraction corresponding to \(k\) karats is
\[ \frac{k}{24}. \]For example, \(18\)-karat gold has gold fraction
\[ \frac{18}{24}=0.75. \]So \(18\)-karat gold is
\[ \boxed{75\%\text{ gold}}. \]Once the purity scale is converted to a fraction or decimal, ordinary mixture methods can be used.
Connection to Systems of Equations
Alloy problems can also be modeled using systems of equations.
Suppose \(x\) kilograms of a \(20\%\) alloy and \(y\) kilograms of a \(50\%\) alloy must produce \(40\) kilograms of a \(35\%\) alloy.
The total-mass equation is
\[ x+y=40. \]The copper equation is
\[ 0.20x+0.50y=0.35(40). \]So the system is
\[ \begin{cases} x+y=40,\\ 0.20x+0.50y=14. \end{cases} \]This expresses two conservation laws at once:
- total mass is conserved, and
- copper mass is conserved.
Composition as a Weighted Average
When two alloys are combined, the final composition is
\[ \boxed{ C_f= \frac{C_1T_1+C_2T_2}{T_1+T_2} }. \]This is a weighted average of the original compositions.
The larger the mass of one alloy, the more strongly its composition influences the final result.
This same weighted-average structure appeared in concentration problems and price-and-value mixtures.
The Composition-Difference Pattern
Suppose two alloys with compositions \(C_1\) and \(C_2\) are combined to obtain target composition \(C_f\), where
\[ C_1This is the same difference relationship seen in ordinary concentration and price-mixture problems.
A target closer to one source composition requires more of that source.
Recognizing Impossible Compositions
Suppose only \(20\%\) copper alloy and \(50\%\) copper alloy are available.
Can they be combined to make a \(70\%\) copper alloy?
No.
A weighted average of \(20\%\) and \(50\%\) must lie between those values.
Therefore,
\[ 20\%\leq C_f\leq50\%. \]To obtain \(70\%\), a source containing more than \(70\%\) copper, such as pure copper, would need to be introduced.
Method for Alloy and Composition Problems
- Identify the component being tracked. Choose copper, gold, iron, nickel, or another material.
- Convert percentages to decimals. For example, \[ 35\%=0.35. \]
- Identify each total quantity. Use mass, volume, or another compatible measurement.
- Find each component amount. Use \[ A=CT. \]
- Determine the final total. If materials are simply combined, add their quantities.
- Apply conservation. Set the sum of the component amounts before mixing equal to the component amount afterward.
- Solve the resulting equation or system. Use the algebraic method appropriate to the problem.
- Check the result. Verify that the percentages are between \(0\%\) and \(100\%\), quantities are nonnegative, and the final composition is physically reasonable.
Common Errors
- Adding percentages directly. The component amounts add, not the percentages themselves.
- Forgetting to convert percentages to decimals. Use \(0.40\), not \(40\), when multiplying by mass.
- Confusing component mass with total alloy mass. A \(30\%\) copper alloy is not \(30\) kilograms of copper unless the total mass makes that true.
- Ignoring the remainder of the composition. If an alloy has only two components and one is \(35\%\), the other must be \(65\%\).
- Assuming every mixture contains only two components. Some composition problems require several percentages to be tracked separately.
- Accepting impossible target percentages. A blend of two ordinary alloys cannot exceed the stronger composition or fall below the weaker one.
- Mixing incompatible units. Convert grams, kilograms, pounds, or other measurements before combining quantities.
Practice Problems
- A \(10\)-kilogram alloy is \(20\%\) copper and a \(15\)-kilogram alloy is \(50\%\) copper. Find the copper percentage of the final alloy.
- How many kilograms of a \(60\%\) copper alloy should be mixed with \(12\) kilograms of a \(30\%\) copper alloy to produce a \(40\%\) alloy?
- A manufacturer needs \(30\) kilograms of a \(45\%\) alloy using \(30\%\) and \(60\%\) alloys. How much of each should be used?
- A \(20\)-kilogram alloy is \(35\%\) copper. How much pure copper must be added to make the alloy \(50\%\) copper?
- A \(24\)-kilogram alloy is \(50\%\) copper. How much copper-free material must be added to reduce the composition to \(30\%\) copper?
- An alloy is \(62\%\) iron and \(18\%\) nickel, with the remainder chromium. What percentage is chromium?
- A \(40\)-kilogram alloy is \(55\%\) iron, \(25\%\) nickel, and \(20\%\) chromium. Find the mass of each component.
- Ten grams of gold that is \(70\%\) pure are mixed with \(15\) grams of gold that is \(90\%\) pure. Find the purity of the final mixture.
Practice Answers
- Copper amount: \[ 0.20(10)+0.50(15)=2+7.5=9.5. \] Total mass: \[ 10+15=25. \] Thus, \[ C_f=\frac{9.5}{25}=0.38. \] \[ \boxed{38\%\text{ copper}}. \]
- Let \(x\) be the mass of the \(60\%\) alloy: \[ 0.30(12)+0.60x=0.40(12+x). \] \[ 3.6+0.60x=4.8+0.40x. \] \[ 0.20x=1.2. \] \[ \boxed{x=6\text{ kg}}. \]
- Let \(x\) be the amount of the \(30\%\) alloy: \[ 0.30x+0.60(30-x)=0.45(30). \] \[ 0.30x+18-0.60x=13.5. \] \[ -0.30x=-4.5. \] \[ x=15. \] Therefore, \[ \boxed{15\text{ kg of each alloy}}. \]
- \[ 0.35(20)+x=0.50(20+x). \] \[ 7+x=10+0.50x. \] \[ 0.50x=3. \] \[ \boxed{x=6\text{ kg}}. \]
- \[ 0.50(24)=0.30(24+x). \] \[ 12=7.2+0.30x. \] \[ 4.8=0.30x. \] \[ \boxed{x=16\text{ kg}}. \]
- \[ 100\%-62\%-18\%=20\%. \] \[ \boxed{20\%\text{ chromium}}. \]
- Iron: \[ 0.55(40)=22\text{ kg}. \] Nickel: \[ 0.25(40)=10\text{ kg}. \] Chromium: \[ 0.20(40)=8\text{ kg}. \] Therefore, \[ \boxed{22\text{ kg iron},\ 10\text{ kg nickel},\ 8\text{ kg chromium}}. \]
- Pure gold: \[ 0.70(10)+0.90(15) = 7+13.5 = 20.5. \] Total mass: \[ 10+15=25. \] Therefore, \[ C_f=\frac{20.5}{25}=0.82. \] \[ \boxed{82\%\text{ pure gold}}. \]
Summary
Alloy and composition problems are ordinary mixture problems expressed in terms of material composition.
The fundamental relationship remains
\[ \boxed{A=CT}. \]For two materials combined,
\[ \boxed{ C_1T_1+C_2T_2=C_fT_f }. \]When their quantities simply add,
\[ \boxed{ C_1T_1+C_2T_2=C_f(T_1+T_2) }. \]For a mixture containing several components,
\[ \boxed{ C_1+C_2+\cdots+C_n=1 }. \]The same reasoning applies to alloys, purity, mineral composition, manufactured materials, and many other situations where the parts of a whole are described by fractions or percentages.
The deeper idea remains conservation: each component of the final material must come from the component amounts contributed by the materials used to create it.
Continue the Exploration
The mixture problems studied so far have generally provided the concentrations of the source materials. The next lesson reverses the situation and focuses on problems in which a concentration itself is unknown and must be discovered from the quantities and final mixture.
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Continue to: Unknown Concentration Problems
Learn how to represent an unknown concentration algebraically and recover it from conservation equations involving known quantities and final composition.