From an Idea to a Principle
In the narrative introduction, mixture problems were viewed as conservation problems. A mixture may change in size or concentration, but the component being tracked must remain accounted for unless some of it is deliberately added or removed.
To turn that idea into algebra, we need a relationship connecting three quantities:
- Concentration: the fraction or percentage of the mixture belonging to the component being studied.
- Total quantity: the complete amount of mixture.
- Component amount: the actual amount of the component being tracked.
These quantities form the foundation of the Mixture Principle.
The Mixture Principle
Let
- \(A\) represent the amount of the component being measured,
- \(C\) represent the concentration of that component, and
- \(T\) represent the total amount of mixture.
Then
\[ \boxed{A=CT}. \]This is the basic Mixture Principle.
It states that the amount of a component is found by multiplying the concentration of that component by the total quantity of the mixture.
Understanding Concentration
Concentration is usually written as a percentage in ordinary language, but percentages should normally be converted to decimal form before they are used in mixture equations.
For example,
\[ 25\%=0.25, \] \[ 40\%=0.40, \]and
\[ 7\%=0.07. \]Therefore, if \(12\) liters of solution are \(25\%\) acid, the amount of acid is
\[ A=(0.25)(12). \]So
\[ A=3. \]The mixture contains
\[ \boxed{3\text{ liters of acid}}. \]Three Forms of the Mixture Principle
Starting from
\[ A=CT, \]we can solve for any one of the three quantities when the other two are known.
To find the component amount:
\[ \boxed{A=CT}. \]To find the concentration:
\[ \boxed{C=\frac{A}{T}}. \]To find the total quantity:
\[ \boxed{T=\frac{A}{C}}. \]These three forms describe the same relationship from different perspectives.
Example 1: Finding Component Amount
A container holds \(20\) liters of a solution that is \(35\%\) alcohol. How many liters of alcohol are present?
Use
\[ A=CT. \]Here,
\[ C=0.35 \]and
\[ T=20. \]Therefore,
\[ A=(0.35)(20)=7. \]The container contains
\[ \boxed{7\text{ liters of alcohol}}. \]Example 2: Finding Concentration
A \(15\)-liter mixture contains \(3\) liters of a particular chemical. What is the concentration of the chemical?
Use
\[ C=\frac{A}{T}. \]Substitute:
\[ C=\frac{3}{15}. \]Then
\[ C=0.20. \]Convert the decimal to a percentage:
\[ 0.20=20\%. \]The mixture is therefore
\[ \boxed{20\%}. \]Example 3: Finding Total Quantity
A mixture contains \(8\) kilograms of copper and is \(40\%\) copper. What is the total mass of the mixture?
Use
\[ T=\frac{A}{C}. \]Substitute:
\[ T=\frac{8}{0.40}. \]Therefore,
\[ T=20. \]The total mass is
\[ \boxed{20\text{ kilograms}}. \]Conservation When Mixtures Are Combined
The relationship \(A=CT\) describes a single mixture. When two or more mixtures are combined, conservation connects them.
Suppose two mixtures are poured together. The first contains component amount
\[ A_1=C_1T_1, \]and the second contains
\[ A_2=C_2T_2. \]If nothing is lost during mixing, then the amount of component in the final mixture must be
\[ A_f=A_1+A_2. \]Since
\[ A_f=C_fT_f, \]we obtain
\[ \boxed{C_1T_1+C_2T_2=C_fT_f}. \]This is the standard conservation equation for combining two mixtures.
The Final Total Quantity
When two ordinary mixtures are simply combined, the final total quantity is usually the sum of the original quantities:
\[ T_f=T_1+T_2. \]The conservation equation can therefore often be written as
\[ \boxed{C_1T_1+C_2T_2=C_f(T_1+T_2)}. \]This form appears frequently in concentration problems.
Example 4: Combining Two Known Mixtures
Suppose \(8\) liters of a \(25\%\) solution are mixed with \(12\) liters of a \(50\%\) solution. Find the concentration of the final mixture.
The first mixture contributes
\[ 0.25(8)=2 \]liters of the component.
The second contributes
\[ 0.50(12)=6. \]The total component amount is
\[ 2+6=8. \]The total mixture is
\[ 8+12=20. \]Therefore,
\[ C_f=\frac{8}{20}=0.40. \]The final mixture is
\[ \boxed{40\%}. \]The Same Example as One Equation
The previous example can also be written directly using the conservation equation:
\[ 0.25(8)+0.50(12)=C_f(20). \]Simplify:
\[ 2+6=20C_f. \] \[ 8=20C_f. \]Divide by \(20\):
\[ C_f=0.40. \]Again,
\[ \boxed{C_f=40\%}. \]Both approaches express the same reasoning. One calculates the contributions separately; the other places them immediately into a single equation.
When One Quantity Is Unknown
Mixture problems become algebra problems when one of the quantities is unknown.
Suppose \(10\) liters of a \(30\%\) solution are mixed with \(x\) liters of a \(60\%\) solution to produce a \(40\%\) solution.
The first mixture contributes
\[ 0.30(10). \]The second contributes
\[ 0.60x. \]The final amount of mixture is
\[ 10+x. \]At \(40\%\) concentration, its component amount is
\[ 0.40(10+x). \]Conservation gives
\[ 0.30(10)+0.60x=0.40(10+x). \]Simplify:
\[ 3+0.60x=4+0.40x. \]Subtract \(0.40x\) from both sides:
\[ 3+0.20x=4. \]Subtract \(3\):
\[ 0.20x=1. \]Therefore,
\[ x=5. \]So
\[ \boxed{5\text{ liters}} \]of the \(60\%\) solution must be added.
Think in Component Amounts
One of the best habits in mixture problems is to stop thinking temporarily about percentages and ask:
How much of the actual component does each mixture contribute?
In the previous example:
- The \(30\%\) solution contributes \(3\) liters of the component.
- The \(60\%\) solution contributes \(0.60x\) liters.
- The final mixture contains \(0.40(10+x)\) liters of the component.
Once those quantities are identified, the equation becomes a statement of conservation rather than a formula to memorize.
The Mixture Table
A table can help organize the three quantities used in the Mixture Principle.
| Mixture | Concentration | Total Quantity | Component Amount |
|---|---|---|---|
| Mixture 1 | \(C_1\) | \(T_1\) | \(C_1T_1\) |
| Mixture 2 | \(C_2\) | \(T_2\) | \(C_2T_2\) |
| Final Mixture | \(C_f\) | \(T_f\) | \(C_fT_f\) |
The component amounts in the first two rows are then connected to the final row through
\[ C_1T_1+C_2T_2=C_fT_f. \]Pure Substances and Zero-Concentration Substances
Two special concentrations appear frequently in mixture problems.
A pure component has concentration
\[ 100\%=1. \]Therefore, \(x\) units of a pure component contain
\[ 1(x)=x \]units of that component.
A substance containing none of the measured component has concentration
\[ 0\%=0. \]Therefore, \(x\) units of that substance contribute
\[ 0(x)=0. \]These are not separate rules. They are ordinary cases of the Mixture Principle.
Dilution as a Mixture Problem
Suppose \(10\) liters of a \(40\%\) solution are diluted by adding \(x\) liters of water until the concentration becomes \(25\%\).
The original amount of component is
\[ 0.40(10)=4. \]Water contributes none of the component:
\[ 0(x)=0. \]The final mixture contains
\[ 10+x \]liters.
So conservation gives
\[ 4=0.25(10+x). \]Expand:
\[ 4=2.5+0.25x. \]Subtract \(2.5\):
\[ 1.5=0.25x. \]Therefore,
\[ x=6. \]So
\[ \boxed{6\text{ liters of water}} \]must be added.
The General Mixture Model
For several mixtures contributing to one final mixture, the same principle extends naturally:
\[ \boxed{ C_1T_1+C_2T_2+C_3T_3+\cdots=C_fT_f }. \]Each term on the left represents the amount of the tracked component contributed by one source.
The right side represents the total amount of that component in the resulting mixture.
If some material is removed instead of added, its contribution may need to be subtracted:
\[ \boxed{ \text{initial component} +\text{component added} -\text{component removed} = \text{final component} }. \]This broader conservation form becomes especially important in replacement problems.
Pay Attention to Units
The quantities in a mixture equation must use compatible units.
For example, if one volume is measured in liters and another in milliliters, convert them to the same unit before combining them.
Likewise, a concentration written as a percentage must be interpreted correctly:
\[ 35\%=0.35. \]Writing
\[ 35(10) \]instead of
\[ 0.35(10) \]would make the component amount one hundred times too large.
Where Should the Final Concentration Fall?
When two positive quantities are combined without adding a pure component or removing material, the final concentration should lie between the original concentrations.
If a \(20\%\) solution is mixed with a \(50\%\) solution, then the final concentration should satisfy
\[ 20\%This provides an important reasonableness check.
Mixtures and Weighted Averages
When two mixtures are combined, the final concentration can be written as
\[ C_f= \frac{C_1T_1+C_2T_2}{T_1+T_2}. \]This is a weighted average of the original concentrations.
The concentrations are not weighted equally unless the quantities are equal. Instead, each concentration is weighted by the amount of mixture to which it applies.
This explains why simply averaging two percentages often gives the wrong answer.
Common Errors
- Averaging concentrations directly. The concentrations should only be averaged directly when the quantities being mixed are equal.
- Using percentages as whole numbers. Convert \(30\%\) to \(0.30\) before multiplying.
- Confusing concentration with component amount. A \(40\%\) mixture does not contain \(40\) units of the component unless the total quantity happens to make that true.
- Forgetting the final total. If \(x\) units are added to \(10\) units, the final amount is usually \(10+x\), not merely \(x\).
- Tracking the wrong component. Choose one component and use it consistently throughout the equation.
- Ignoring material that is removed. In replacement problems, removing part of a mixture also removes a proportional amount of the tracked component.
A General Method
- Identify the component being tracked. Decide whether the problem concerns salt, acid, alcohol, copper, value, or another measurable quantity.
- Convert concentrations to decimals. For example, \(35\%=0.35\).
- Determine the total quantity of each mixture. Introduce a variable for any unknown quantity.
- Compute or represent each component amount. Use \[ A=CT. \]
- Apply conservation. Set the total component amount before or during mixing equal to the final component amount.
- Solve the resulting equation. Use the appropriate algebraic techniques.
- Check the answer in context. Verify the units, concentration range, and physical reasonableness of the result.
Summary of the Mixture Principle
The central relationship is
\[ \boxed{A=CT}. \]For a single mixture:
\[ A=CT, \] \[ C=\frac{A}{T}, \]and
\[ T=\frac{A}{C}. \]For two mixtures combined:
\[ \boxed{C_1T_1+C_2T_2=C_fT_f}. \]When the total quantities simply add,
\[ \boxed{C_1T_1+C_2T_2=C_f(T_1+T_2)}. \]The formulas are useful, but the deeper idea is conservation:
\[ \boxed{\text{component contributed}=\text{component accounted for}}. \]Once that principle is understood, concentration, dilution, replacement, alloy, value, and other mixture problems become different applications of the same mathematical structure.
Continue the Exploration
The next lesson applies the Mixture Principle to ordinary concentration problems, where two or more mixtures with different concentrations are combined and one quantity must be determined.
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Continue to: Concentration Problems
Apply \(A=CT\) and conservation to solve basic mixture equations involving known and unknown concentrations and quantities.