Overview
In many mixture problems, the quantity being added is unknown. In this lesson, the situation is reversed: the quantity may already be known, while the concentration of one of the mixtures must be discovered.
The same Mixture Principle still governs the problem:
\[ A=CT. \]If the concentration is unknown, we simply represent it with a variable such as \(x\). A mixture of \(10\) liters with unknown concentration \(x\) contains
\[ 10x \]liters of the component being tracked.
Conservation then allows the unknown concentration to be recovered from the information given about the other mixtures and the final result.
Concentration as the Variable
Suppose a \(12\)-liter solution has an unknown salt concentration.
Let
\[ x=\text{the salt concentration as a decimal}. \]Then the amount of salt in the solution is
\[ A=x(12). \]or
\[ \boxed{A=12x}. \]If solving the problem produces
\[ x=0.35, \]then the concentration is
\[ \boxed{35\%}. \]The variable represents a decimal concentration during the algebra, not a whole-number percentage.
The Basic Unknown-Concentration Model
Suppose a mixture of quantity \(T_1\) has unknown concentration \(x\), while another mixture has known concentration \(C_2\) and quantity \(T_2\).
The first mixture contributes
\[ xT_1. \]The second contributes
\[ C_2T_2. \]If the final concentration is \(C_f\) and the mixtures simply combine, then the final quantity is
\[ T_1+T_2. \]Conservation gives
\[ \boxed{ xT_1+C_2T_2=C_f(T_1+T_2) }. \]The unknown concentration can then be isolated using ordinary algebra.
Example 1: Finding an Unknown Concentration
A \(10\)-liter solution of unknown concentration is mixed with \(10\) liters of a \(60\%\) solution. The resulting \(20\)-liter mixture is \(40\%\). What was the concentration of the first solution?
Let \(x\) represent the unknown concentration.
The first solution contributes
\[ 10x \]liters of the measured component.
The second contributes
\[ 0.60(10)=6. \]The final mixture contains
\[ 0.40(20)=8. \]Therefore,
\[ 10x+6=8. \]Subtract \(6\):
\[ 10x=2. \]Divide by \(10\):
\[ x=0.20. \]Thus the unknown solution was
\[ \boxed{20\%}. \]Checking the Result
A \(10\)-liter \(20\%\) solution contains
\[ 0.20(10)=2 \]liters of component.
The \(10\)-liter \(60\%\) solution contains
\[ 0.60(10)=6. \]Together they contain
\[ 2+6=8 \]liters of component in
\[ 20 \]liters total.
Therefore,
\[ \frac{8}{20}=0.40. \]The final concentration is indeed
\[ \boxed{40\%}. \]Organizing an Unknown Concentration
| Mixture | Concentration | Total Quantity | Component Amount |
|---|---|---|---|
| Unknown solution | \(x\) | \(10\) | \(10x\) |
| Known solution | \(0.60\) | \(10\) | \(6\) |
| Final solution | \(0.40\) | \(20\) | \(8\) |
The final column immediately gives
\[ 10x+6=8. \]The table is especially helpful because it keeps the unknown concentration separate from the component amount it produces.
Example 2: Unequal Quantities
A \(12\)-liter solution of unknown concentration is mixed with \(8\) liters of a \(50\%\) solution. The resulting mixture is \(38\%\). Find the unknown concentration.
Let \(x\) represent the unknown concentration.
The unknown solution contributes
\[ 12x. \]The \(50\%\) solution contributes
\[ 0.50(8)=4. \]The final mixture contains \(20\) liters, so the final component amount is
\[ 0.38(20)=7.6. \]Therefore,
\[ 12x+4=7.6. \]Subtract \(4\):
\[ 12x=3.6. \]Divide by \(12\):
\[ x=0.30. \]Thus the unknown concentration was
\[ \boxed{30\%}. \]Solving the General Equation
Begin with
\[ xT_1+C_2T_2=C_f(T_1+T_2). \]Subtract the known contribution:
\[ xT_1=C_f(T_1+T_2)-C_2T_2. \]Divide by \(T_1\):
\[ \boxed{ x= \frac{C_f(T_1+T_2)-C_2T_2}{T_1} }. \]This formula gives the unknown concentration directly.
As with the other formulas in this exploration, however, the conservation equation is usually more important to understand than the solved form.
What Does the Variable Represent?
It is important to define the variable carefully.
If we write
\[ x=\text{concentration}, \]then \(x\) should normally be represented as a decimal between \(0\) and \(1\).
Thus,
\[ x=0.45 \]means
\[ 45\%. \]Another approach is to let \(p\) represent the percentage itself. Then the decimal concentration would be
\[ \frac{p}{100}. \]For example, the component amount in \(10\) liters would be
\[ \frac{p}{100}(10). \]Both methods are valid, but using a decimal concentration usually produces cleaner algebra.
Example 3: Unknown Concentration After Dilution
A \(15\)-liter solution of unknown concentration is diluted with \(5\) liters of water. The resulting \(20\)-liter mixture is \(30\%\). What was the original concentration?
Let \(x\) represent the original concentration.
The original component amount is
\[ 15x. \]Water contributes
\[ 0. \]The final mixture contains
\[ 0.30(20)=6 \]liters of component.
Therefore,
\[ 15x=6. \]So
\[ x=\frac{6}{15}. \] \[ x=0.40. \]The original solution was
\[ \boxed{40\%}. \]Example 4: Unknown Concentration Before Adding Pure Component
A \(20\)-kilogram mixture has an unknown concentration. Five kilograms of pure component are added, producing a \(25\)-kilogram mixture that is \(52\%\) component. Find the original concentration.
Let \(x\) represent the original concentration.
The original component amount is
\[ 20x. \]The pure component contributes
\[ 5. \]The final mixture contains
\[ 0.52(25)=13 \]kilograms of component.
So
\[ 20x+5=13. \]Subtract \(5\):
\[ 20x=8. \]Therefore,
\[ x=0.40. \]The original mixture was
\[ \boxed{40\%}. \]Example 5: Unknown Concentration in a Replacement Problem
A \(20\)-liter tank contains a solution of unknown concentration. Five liters are removed and replaced with water. The final concentration is \(30\%\). Find the original concentration.
Let \(x\) be the original concentration.
The original component amount is
\[ 20x. \]Because the mixture is well mixed, the \(5\) liters removed contain
\[ 5x \]liters of component.
Water contributes none.
The final mixture still contains \(20\) liters and is \(30\%\) component, so the final component amount is
\[ 0.30(20)=6. \]Therefore,
\[ 20x-5x=6. \]Combine like terms:
\[ 15x=6. \]Thus,
\[ x=0.40. \]The original concentration was
\[ \boxed{40\%}. \]Sometimes the Final Concentration Is the Unknown
An unknown-concentration problem does not necessarily involve an unknown starting concentration. Sometimes the final concentration itself is unknown.
Suppose \(8\) liters of a \(25\%\) solution are mixed with \(12\) liters of a \(45\%\) solution.
Let \(x\) be the final concentration.
Then
\[ 0.25(8)+0.45(12)=20x. \]Calculate the contributions:
\[ 2+5.4=20x. \] \[ 7.4=20x. \]Therefore,
\[ x=0.37. \]The final concentration is
\[ \boxed{37\%}. \]Unknown Concentration with a Fixed Final Total
Suppose two mixtures make a fixed final quantity.
One mixture has an unknown concentration \(x\), while the concentration of the second is known.
The same conservation equation applies:
\[ xT_1+C_2T_2=C_fT_f. \]When
\[ T_f=T_1+T_2, \]the total-quantity relationship provides an additional check that the problem has been modeled correctly.
Example 6: A Fixed Alloy Composition
A manufacturer combines \(15\) kilograms of an alloy with unknown copper concentration and \(25\) kilograms of a \(50\%\) copper alloy. The resulting \(40\)-kilogram alloy is \(35\%\) copper. Find the composition of the unknown alloy.
Let \(x\) represent its copper concentration.
The unknown alloy contributes
\[ 15x. \]The known alloy contributes
\[ 0.50(25)=12.5. \]The final alloy contains
\[ 0.35(40)=14 \]kilograms of copper.
Therefore,
\[ 15x+12.5=14. \]Subtract \(12.5\):
\[ 15x=1.5. \]Thus,
\[ x=0.10. \]The unknown alloy was
\[ \boxed{10\%\text{ copper}}. \]Use Weighted-Average Reasoning to Check the Answer
In the previous example, the final concentration was \(35\%\), while one of the source alloys was \(50\%\).
For the final average to fall below \(35\%\), the other alloy must have a concentration below \(35\%\).
The calculated concentration,
\[ 10\%, \]satisfies that requirement.
This kind of qualitative reasoning can often tell us where the unknown concentration should lie before any equation is solved.
Unknown Unit Value Uses the Same Mathematics
The same structure applies to price-and-value mixtures.
Suppose \(8\) pounds of a product worth \(\$5\) per pound are mixed with \(12\) pounds of a product with unknown price \(x\). The final blend is worth \(\$8\) per pound.
Then
\[ 5(8)+12x=8(20). \]So
\[ 40+12x=160. \] \[ 12x=120. \] \[ x=10. \]The unknown product is worth
\[ \boxed{\$10\text{ per pound}}. \]Mathematically, an unknown concentration and an unknown unit price occupy the same position in their respective mixture models.
What If Two Concentrations Are Unknown?
One conservation equation cannot usually determine two unrelated unknown concentrations.
Suppose
\[ 10x+10y=8. \]There are many possible pairs \((x,y)\) satisfying this equation.
To determine both concentrations uniquely, the problem must provide another independent relationship.
For example, suppose the second solution is twice as concentrated as the first:
\[ y=2x. \]Then the two equations can be solved together.
This turns the mixture problem into a system of equations.
Example 7: Related Unknown Concentrations
Ten liters of one solution are mixed with \(10\) liters of another. The second solution is twice as concentrated as the first. The final mixture is \(45\%\). Find both original concentrations.
Let the first concentration be
\[ x. \]Then the second is
\[ 2x. \]The component equation is
\[ 10x+10(2x)=0.45(20). \]Simplify:
\[ 10x+20x=9. \] \[ 30x=9. \]Therefore,
\[ x=0.30. \]So the first solution is
\[ 30\%, \]and the second is
\[ 2(30\%)=60\%. \]Thus,
\[ \boxed{30\%\text{ and }60\%}. \]Related Concentrations May Be Expressed by Differences
A problem may state that one mixture is \(15\) percentage points more concentrated than another.
If the first concentration is represented by
\[ x, \]then the second is
\[ x+0.15. \]Notice that \(15\) percentage points becomes
\[ 0.15 \]when concentrations are represented as decimals.
This translation step is important before the mixture equation is written.
Example 8: Concentrations Related by a Difference
Eight liters of one solution are mixed with \(12\) liters of another solution that is \(20\) percentage points stronger. The final mixture is \(42\%\). Find both concentrations.
Let the weaker concentration be
\[ x. \]Then the stronger concentration is
\[ x+0.20. \]Apply conservation:
\[ 8x+12(x+0.20)=0.42(20). \]Expand:
\[ 8x+12x+2.4=8.4. \]Combine:
\[ 20x+2.4=8.4. \]Subtract \(2.4\):
\[ 20x=6. \]Therefore,
\[ x=0.30. \]The weaker solution is
\[ 30\%, \]and the stronger solution is
\[ 0.30+0.20=0.50. \]Thus,
\[ \boxed{30\%\text{ and }50\%}. \]Percent and Percentage Points Are Not the Same
Suppose one concentration is \(20\%\) and another is \(30\%\).
The second is
\[ 10 \]percentage points higher.
But relative to \(20\%\), the increase is
\[ \frac{30-20}{20}=0.50, \]or
\[ 50\%. \]Therefore:
- 10 percentage points higher: \(20\%\rightarrow30\%\)
- 50% higher: \(20\%\rightarrow30\%\)
Mixture problems that describe relationships between concentrations must be translated carefully.
Concentration Has Natural Restrictions
A concentration expressed as a decimal normally satisfies
\[ \boxed{0\leq x\leq1}. \]As a percentage,
\[ \boxed{0\%\leq x\leq100\%}. \]If an equation produces
\[ x=-0.20 \]or
\[ x=1.30, \]the result is not a physically meaningful ordinary concentration.
Such a result usually indicates that the target mixture cannot be created under the stated conditions or that the model has been set up incorrectly.
Example 9: An Impossible Unknown Concentration
Ten liters of an unknown solution are mixed with \(10\) liters of a \(40\%\) solution to produce a \(70\%\) mixture. What concentration would the unknown solution need to have?
Let \(x\) be the unknown concentration.
Then
\[ 10x+0.40(10)=0.70(20). \]So
\[ 10x+4=14. \] \[ 10x=10. \] \[ x=1. \]The unknown solution would have to be
\[ \boxed{100\%}. \]This is the maximum possible ordinary concentration. If the target were any higher, the algebra would require a concentration greater than \(100\%\), showing that the proposed mixture is impossible.
Find the Missing Contribution First
A useful alternative strategy is to determine how much component the unknown mixture must contribute before solving for its concentration.
Suppose the final mixture must contain \(8\) liters of component and the known mixture contributes \(6\).
Then the unknown mixture must contribute
\[ 8-6=2. \]If the unknown mixture has total quantity \(10\), its concentration is immediately
\[ \frac{2}{10}=0.20. \]This viewpoint emphasizes the conservation structure:
\[ \boxed{ \text{missing contribution} = \text{required final amount} - \text{known contributions} }. \]Unknown Concentration with Several Sources
The same method extends to three or more mixtures.
If one source has unknown concentration \(x\), then
\[ xT_1+C_2T_2+C_3T_3+\cdots=C_fT_f. \]Solving for \(x\) gives
\[ \boxed{ x= \frac{ C_fT_f- C_2T_2- C_3T_3-\cdots }{T_1} }. \]The unknown source must contribute whatever component amount is still needed after the known contributions have been accounted for.
Example 10: Three Mixtures with One Unknown Concentration
Five liters of an unknown solution are mixed with \(10\) liters of a \(30\%\) solution and \(5\) liters of a \(50\%\) solution. The final \(20\)-liter mixture is \(35\%\). Find the unknown concentration.
Let \(x\) represent the unknown concentration.
The equation is
\[ 5x+0.30(10)+0.50(5)=0.35(20). \]Calculate the known contributions:
\[ 5x+3+2.5=7. \]Combine:
\[ 5x+5.5=7. \]Subtract \(5.5\):
\[ 5x=1.5. \]Therefore,
\[ x=0.30. \]The unknown solution was
\[ \boxed{30\%}. \]Unknown Concentration and Functions
For a fixed quantity \(T\), the component amount depends directly on concentration:
\[ A(C)=CT. \]For example, if a container holds \(12\) liters,
\[ A(C)=12C. \]A \(10\%\) concentration gives
\[ A(0.10)=1.2, \]while a \(50\%\) concentration gives
\[ A(0.50)=6. \]Unknown-concentration problems reverse this relationship. Instead of being given \(C\) and finding \(A\), the component amount is inferred from conservation and then used to recover \(C\).
Forward and Reverse Mixture Reasoning
Many earlier problems followed the direction
\[ \text{concentration} \rightarrow \text{component amount}. \]Unknown-concentration problems often reverse it:
\[ \text{component amount} \rightarrow \text{concentration}. \]Because
\[ A=CT, \]we can solve for concentration:
\[ \boxed{ C=\frac{A}{T} }. \]This ability to move forward and backward through a mathematical relationship is an important algebraic skill that appears throughout functions, formulas, and mathematical modeling.
Method for Unknown Concentration Problems
- Choose the component to track. Use the same component throughout the problem.
- Represent the unknown concentration. Usually let \[ x=\text{unknown concentration as a decimal}. \]
- Multiply the unknown concentration by its quantity. If the quantity is \(T\), its component contribution is \[ xT. \]
- Calculate all known component amounts. Use \[ A=CT. \]
- Calculate the required final component amount. Use the final concentration and final total quantity.
- Apply conservation. Set the sum of all source contributions equal to the final component amount.
- Solve for the concentration. Isolate \(x\).
- Convert to a percentage if necessary. For example, \[ x=0.35 \] means \[ 35\%. \]
- Check the domain and context. The concentration should ordinarily satisfy \[ 0\leq x\leq1. \]
Common Errors
- Treating \(x\) as both a decimal and a percentage. Decide whether \(x=0.30\) or \(x=30\) represents \(30\%\), and use that convention consistently.
- Using the concentration alone as the contribution. An unknown concentration \(x\) in \(10\) liters contributes \(10x\), not merely \(x\).
- Forgetting to multiply the final concentration by the final quantity. The final component amount is \(C_fT_f\).
- Ignoring known contributions. Subtract or account for all known component amounts before solving for the unknown.
- Accepting concentrations below \(0\%\) or above \(100\%\). Such results are not meaningful for ordinary mixture composition.
- Confusing percentage points with percent change. A concentration \(20\) percentage points higher than \(30\%\) is \(50\%\), not \(36\%\).
- Assuming one equation can determine several unrelated unknown concentrations. Additional unknowns require additional independent relationships.
Practice Problems
- A \(10\)-liter solution of unknown concentration is mixed with \(10\) liters of a \(50\%\) solution. The final mixture is \(35\%\). Find the unknown concentration.
- A \(15\)-liter solution of unknown concentration is mixed with \(5\) liters of a \(60\%\) solution. The resulting mixture is \(45\%\). Find the unknown concentration.
- A \(12\)-liter solution of unknown concentration is diluted with \(8\) liters of water. The final mixture is \(24\%\). Find the original concentration.
- A \(20\)-kilogram alloy of unknown copper concentration receives \(5\) kilograms of pure copper. The final alloy is \(44\%\) copper. Find the original copper concentration.
- A \(25\)-liter solution of unknown concentration has \(5\) liters removed and replaced with water. The final concentration is \(32\%\). Find the original concentration.
- Ten liters of one solution are mixed with \(10\) liters of another solution that is three times as concentrated. The final mixture is \(40\%\). Find both concentrations.
- Six liters of one solution are mixed with \(14\) liters of a solution that is \(20\) percentage points stronger. The final concentration is \(44\%\). Find both original concentrations.
- Five liters of an unknown solution are mixed with \(10\) liters of a \(20\%\) solution and \(5\) liters of a \(60\%\) solution. The final mixture is \(35\%\). Find the unknown concentration.
Practice Answers
- \[ 10x+0.50(10)=0.35(20) \] \[ 10x+5=7 \] \[ 10x=2 \] \[ x=0.20. \] \[ \boxed{20\%}. \]
- \[ 15x+0.60(5)=0.45(20) \] \[ 15x+3=9 \] \[ 15x=6 \] \[ x=0.40. \] \[ \boxed{40\%}. \]
- \[ 12x=0.24(20) \] \[ 12x=4.8 \] \[ x=0.40. \] \[ \boxed{40\%}. \]
- \[ 20x+5=0.44(25) \] \[ 20x+5=11 \] \[ 20x=6 \] \[ x=0.30. \] \[ \boxed{30\%}. \]
- The amount remaining after removing \(5\) liters is \[ 25x-5x=20x. \] The final component amount is \[ 0.32(25)=8. \] Therefore, \[ 20x=8. \] \[ x=0.40. \] \[ \boxed{40\%}. \]
- Let the weaker concentration be \(x\). Then the stronger concentration is \(3x\): \[ 10x+10(3x)=0.40(20). \] \[ 40x=8. \] \[ x=0.20. \] Therefore, \[ \boxed{20\%\text{ and }60\%}. \]
- Let the weaker concentration be \(x\). Then the stronger concentration is \(x+0.20\): \[ 6x+14(x+0.20)=0.44(20). \] \[ 6x+14x+2.8=8.8. \] \[ 20x=6. \] \[ x=0.30. \] Therefore, \[ \boxed{30\%\text{ and }50\%}. \]
- \[ 5x+0.20(10)+0.60(5)=0.35(20). \] \[ 5x+2+3=7. \] \[ 5x=2. \] \[ x=0.40. \] \[ \boxed{40\%}. \]
Summary
Unknown-concentration problems reverse the usual mixture question. Instead of using a known concentration to calculate component amount, conservation is used to determine the component amount that an unknown mixture must contain.
The Mixture Principle remains
\[ \boxed{A=CT}. \]If concentration is unknown,
\[ \boxed{A=xT}. \]For two mixtures,
\[ \boxed{ xT_1+C_2T_2=C_f(T_1+T_2) }. \]Solving for the unknown concentration gives
\[ \boxed{ x= \frac{C_f(T_1+T_2)-C_2T_2}{T_1} }. \]The same reasoning works for dilution, adding pure substances, replacement, alloys, price mixtures, and several-source mixtures.
The central question is always:
\[ \boxed{\text{How much must the unknown source contribute?}} \]Once that missing contribution is known, concentration follows from
\[ \boxed{C=\frac{A}{T}}. \]Continue the Exploration
You have now examined the major individual structures of mixture problems: ordinary concentration, pure additions, dilution, replacement, price and value, composition, and unknown concentrations. The final lesson combines these ideas without identifying the required method in advance.
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Continue to: Mixed Mixture Problems
Practice selecting the correct mixture model, tracking conserved quantities, and solving problems that combine several ideas from across the exploration.