The Central Idea
A mixture contains a total quantity and one or more component quantities.
If a mixture has total amount \(Q\) and a component makes up fraction \(c\) of the mixture, then the amount of that component is
\[ \boxed{A=cQ}. \]Here:
- \(Q\) is the total quantity;
- \(c\) is the component fraction or concentration;
- \(A\) is the amount of the component.
Mixture equations are built by conserving total quantity and component quantity.
Percent Concentration as a Decimal
A concentration such as \(30\%\) means
\[ 30\%=0.30. \]If \(20\) liters of solution are \(30\%\) solute, then the solute amount is
\[ 0.30(20)=6. \]Therefore,
\[ \boxed{6\text{ liters of solute}}. \]The Basic Mixture Equation
Suppose two mixtures are combined.
Mixture 1 has quantity \(Q_1\) and concentration \(c_1\).
Mixture 2 has quantity \(Q_2\) and concentration \(c_2\).
The final mixture has total quantity
\[ Q_1+Q_2. \]If its concentration is \(c_f\), then conservation of the component gives
\[ \boxed{ c_1Q_1+c_2Q_2=c_f(Q_1+Q_2) }. \]Why the Equation Works
The first mixture contributes
\[ c_1Q_1 \]units of the component.
The second contributes
\[ c_2Q_2. \]After mixing, the total amount of that component has not disappeared or appeared from nowhere.
Therefore,
\[ \text{component before} = \text{component after}. \]Total Quantity Is Also Conserved
If no material is added or removed except through the stated mixtures, then
\[ \boxed{ Q_f=Q_1+Q_2 }. \]This total-quantity equation often works together with the component-conservation equation.
In simple two-source problems, the final quantity is usually obvious from the context.
Example: Find the Final Concentration
Mix \(10\) liters of a \(20\%\) solution with \(15\) liters of a \(40\%\) solution.
Solute from the first mixture:
\[ 0.20(10)=2. \]Solute from the second:
\[ 0.40(15)=6. \]Total solute:
\[ 2+6=8. \]Total solution:
\[ 10+15=25. \]Therefore, the final concentration is
\[ \frac{8}{25}=0.32. \]Thus,
\[ \boxed{32\%}. \]The Final Concentration Is a Weighted Average
For two mixtures,
\[ c_f = \frac{c_1Q_1+c_2Q_2}{Q_1+Q_2}. \]Therefore,
\[ \boxed{ c_f = \frac{c_1Q_1+c_2Q_2}{Q_1+Q_2} }. \]This is a weighted average of the original concentrations, weighted by the quantities being mixed.
Why the Ordinary Average Can Fail
Mixing a \(20\%\) solution with a \(40\%\) solution does not automatically produce a \(30\%\) solution.
That happens only when equal quantities are mixed.
If \(10\) liters of \(20\%\) solution are mixed with \(30\) liters of \(40\%\) solution, then
\[ c_f = \frac{0.20(10)+0.40(30)}{40}. \] \[ c_f = \frac{2+12}{40} = 0.35. \]Therefore,
\[ \boxed{35\%}. \]Equal Quantities Give the Arithmetic Mean
If equal quantities \(Q\) are mixed at concentrations \(c_1\) and \(c_2\), then
\[ c_f = \frac{c_1Q+c_2Q}{2Q}. \]Factor and simplify:
\[ c_f = \frac{c_1+c_2}{2}. \]Therefore,
\[ \boxed{ c_f=\frac{c_1+c_2}{2} } \]when the mixed quantities are equal.
Mixture Tables
A table helps organize quantity, concentration, and component amount.
\[ \begin{array}{c|c|c|c} \text{source} & \text{quantity} & \text{concentration} & \text{component}\\ \hline 1 & Q_1 & c_1 & c_1Q_1\\ 2 & Q_2 & c_2 & c_2Q_2\\ \hline \text{final} & Q_1+Q_2 & c_f & c_f(Q_1+Q_2) \end{array} \]The component column creates the algebraic equation.
Unknown Quantity of One Mixture
Suppose \(x\) liters of a \(50\%\) solution are mixed with \(20\) liters of a \(20\%\) solution to make a \(30\%\) solution.
The solute equation is
\[ 0.50x+0.20(20)=0.30(x+20). \]Simplify:
\[ 0.50x+4=0.30x+6. \] \[ 0.20x=2. \] \[ x=10. \]Therefore,
\[ \boxed{10\text{ liters}}. \]Check the Unknown-Quantity Example
The \(10\) liters of \(50\%\) solution contain
\[ 0.50(10)=5 \]liters of solute.
The \(20\) liters of \(20\%\) solution contain
\[ 0.20(20)=4. \]Total solute:
\[ 5+4=9. \]Total mixture:
\[ 10+20=30. \]Therefore,
\[ \frac9{30}=0.30. \]The result checks.
Unknown Concentration of One Source
Mix \(8\) liters of a solution with unknown concentration \(c\) with \(12\) liters of a \(25\%\) solution to obtain a \(40\%\) solution.
Write
\[ 8c+0.25(12)=0.40(20). \]Simplify:
\[ 8c+3=8. \] \[ 8c=5. \] \[ c=\frac58=0.625. \]Therefore, the unknown concentration is
\[ \boxed{62.5\%}. \]Unknown Final Concentration
If quantities and starting concentrations are known, the final concentration is found by
\[ \boxed{ c_f = \frac{\text{total component}}{\text{total mixture}} }. \]This is usually the most direct method when the final percentage is the only unknown.
Unknown Amounts with a Fixed Final Total
Suppose a final mixture must contain \(30\) liters.
Let \(x\) liters come from one source.
Then the amount from the second source is
\[ \boxed{30-x}. \]This substitution automatically enforces conservation of total quantity.
Example: Fixed Final Total
How many liters of a \(20\%\) solution and a \(50\%\) solution should be mixed to make \(30\) liters of a \(35\%\) solution?
Let \(x\) be the amount of \(20\%\) solution.
Then the amount of \(50\%\) solution is
\[ 30-x. \]Write the component equation:
\[ 0.20x+0.50(30-x)=0.35(30). \]Expand:
\[ 0.20x+15-0.50x=10.5. \] \[ -0.30x=-4.5. \] \[ x=15. \]Therefore, the mixture requires
\[ \boxed{15\text{ L of each solution}}. \]The Target Must Usually Lie between the Sources
If two positive quantities are mixed without adding or removing the component, the final concentration lies between the original concentrations.
For example, mixing \(20\%\) and \(50\%\) solutions cannot produce a \(70\%\) solution.
In a standard two-source mixture,
\[ \boxed{ \min(c_1,c_2) \le c_f \le \max(c_1,c_2) }. \]This is a powerful reasonableness check.
Strictly between for Positive Amounts
If both source quantities are positive and their concentrations differ, then the final concentration lies strictly between them:
\[ \boxed{ \min(c_1,c_2) < c_f < \max(c_1,c_2) }. \]Equality occurs only when one source contributes zero amount or both source concentrations are already equal.
Mixtures with Pure Material
Pure component has concentration
\[ \boxed{100\%=1}. \]A component-free carrier such as pure water in a salt-water model has concentration
\[ \boxed{0\%=0}. \]These values make many mixture equations especially simple.
Example: Add Pure Component
A \(20\)-liter solution is \(30\%\) solute. How much pure solute must be added to make the final mixture \(50\%\) solute?
Let \(x\) liters of pure solute be added.
Initial solute:
\[ 0.30(20)=6. \]Added solute:
\[ x. \]Final equation:
\[ 6+x=0.50(20+x). \]Solve:
\[ 6+x=10+0.5x. \] \[ 0.5x=4. \] \[ x=8. \]Therefore,
\[ \boxed{8\text{ liters}}. \]Example: Add Component-Free Material
A \(15\)-liter solution is \(40\%\) solute. How much pure water should be added to reduce the concentration to \(30\%\)?
Let \(x\) liters of water be added.
Solute remains
\[ 0.40(15)=6 \]liters.
Final equation:
\[ 6=0.30(15+x). \]Solve:
\[ 6=4.5+0.3x. \] \[ 1.5=0.3x. \] \[ x=5. \]Therefore,
\[ \boxed{5\text{ liters of water}}. \]Mixture Problems Are Conservation Problems
The deepest structure is not the percentage arithmetic itself.
The important principle is
\[ \boxed{ \text{amount before} = \text{amount after} } \]for each conserved component.
This framework extends to concentration, alloy composition, price blends, chemical mixtures, and many physical models.
Two Conservation Equations
When two unknown source quantities \(x\) and \(y\) are mixed to create a final quantity \(Q\), we may use
\[ \boxed{x+y=Q} \]for total quantity and
\[ \boxed{c_1x+c_2y=c_fQ} \]for component quantity.
This creates a system of two equations in two unknowns.
Example: Solve a Two-Equation Mixture System
A \(40\)-liter mixture is to be \(32\%\) solute. It is made from \(20\%\) and \(50\%\) solutions.
Let
\[ x=\text{liters of }20\%\text{ solution} \]and
\[ y=\text{liters of }50\%\text{ solution}. \]Total quantity:
\[ x+y=40. \]Component quantity:
\[ 0.20x+0.50y=0.32(40). \]The right side is
\[ 12.8. \]Substitute
\[ x=40-y. \]Then
\[ 0.20(40-y)+0.50y=12.8. \] \[ 8-0.2y+0.5y=12.8. \] \[ 0.3y=4.8. \] \[ y=16. \]Therefore,
\[ x=24. \]Thus,
\[ \boxed{24\text{ L of }20\%\text{ solution}} \]and
\[ \boxed{16\text{ L of }50\%\text{ solution}}. \]Weighted-Average View of the Same Example
The final concentration \(32\%\) lies closer to \(20\%\) than to \(50\%\).
Therefore, we should expect more of the \(20\%\) solution than the \(50\%\) solution.
The solution
\[ 24\text{ L and }16\text{ L} \]satisfies that expectation.
Qualitative reasoning can often predict the direction of the answer before the algebra is completed.
Distance from the Target Concentration
In a two-source mixture, the relative amounts are connected to how far each source concentration lies from the target.
For concentrations \(c_1<c_f<c_2\),
\[ \boxed{ \frac{Q_1}{Q_2} = \frac{c_2-c_f}{c_f-c_1} }. \]This relationship can be derived from the ordinary mixture equation.
Deriving the Ratio Shortcut
Start with
\[ c_1Q_1+c_2Q_2=c_f(Q_1+Q_2). \]Expand:
\[ c_1Q_1+c_2Q_2=c_fQ_1+c_fQ_2. \]Rearrange:
\[ (c_f-c_1)Q_1=(c_2-c_f)Q_2. \]Therefore,
\[ \boxed{ \frac{Q_1}{Q_2} = \frac{c_2-c_f}{c_f-c_1} }. \]Example: Use the Ratio Shortcut
Mix \(20\%\) and \(50\%\) solutions to obtain \(35\%\).
The distance from \(20\%\) to \(35\%\) is
\[ 15 \]percentage points.
The distance from \(50\%\) to \(35\%\) is also
\[ 15. \]Therefore, the two source quantities must be equal:
\[ \boxed{1:1}. \]Another Ratio-Shortcut Example
Mix \(10\%\) and \(40\%\) solutions to obtain \(25\%\).
The target is \(15\) percentage points above \(10\%\) and \(15\) points below \(40\%\).
Therefore,
\[ \boxed{1:1}. \]Equal source amounts produce the midpoint concentration.
Unequal Ratio Example
Mix \(20\%\) and \(60\%\) solutions to obtain \(30\%\).
The distance from the high source to the target is
\[ 60-30=30. \]The distance from the target to the low source is
\[ 30-20=10. \]Therefore,
\[ \frac{Q_{\text{low}}}{Q_{\text{high}}} = \frac{30}{10} = 3. \]Thus the required ratio is
\[ \boxed{3:1}. \]Why the Lower Concentration May Need More Quantity
In the previous example, the target \(30\%\) is much closer to \(20\%\) than to \(60\%\).
Therefore, the mixture must contain more of the \(20\%\) source.
The ratio
\[ 3:1 \]reflects that weighted-average geometry.
Mixture Problems with Three Sources
Three-source problems follow the same conservation principle:
\[ \boxed{ c_1Q_1+c_2Q_2+c_3Q_3 = c_f(Q_1+Q_2+Q_3) }. \]Additional conditions are usually needed if several source quantities are unknown.
Example: Three Known Source Quantities
Combine:
- \(5\) liters at \(10\%\);
- \(10\) liters at \(30\%\);
- \(15\) liters at \(50\%\).
Component amount:
\[ 0.10(5)+0.30(10)+0.50(15). \] \[ 0.5+3+7.5=11. \]Total quantity:
\[ 5+10+15=30. \]Therefore,
\[ c_f=\frac{11}{30}\approx0.3667. \]Thus,
\[ \boxed{36.67\%\text{ approximately}}. \]Mixtures with Different Units
All quantities must use compatible units before they are added.
If one amount is given in liters and another in milliliters, convert first.
For example,
\[ 750\text{ mL}=0.75\text{ L}. \]Only then should the mixture equation be formed.
Mass-Based Mixtures
Mixture quantities do not need to be volumes.
Suppose a \(12\)-kg material is \(25\%\) component A.
The mass of component A is
\[ 0.25(12)=3. \]Therefore,
\[ \boxed{3\text{ kg of component A}}. \]Count-Based Mixtures
The same logic applies to counts.
If \(40\%\) of a \(250\)-item batch belongs to category A, then
\[ 0.40(250)=100. \]Thus,
\[ \boxed{100\text{ items}} \]belong to category A.
Rates, Prices, and Other Weighted Mixtures
The mixture framework extends beyond physical concentration.
A weighted-average equation has the same form:
\[ \boxed{ v_1Q_1+v_2Q_2 = v_f(Q_1+Q_2) }. \]Here \(v\) may represent concentration, price per unit, composition, quality score, or another per-unit property that combines linearly.
Later lessons apply this structure to alloy, composition, price, and value mixtures.
Checking by Total Quantity
If the final mixture is supposed to contain \(30\) liters and the solution gives
\[ 12\text{ L} \]and
\[ 18\text{ L}, \]then
\[ 12+18=30. \]The total-quantity condition is satisfied.
Checking by Component Quantity
Suppose \(12\) liters of \(20\%\) solution and \(18\) liters of \(40\%\) solution are claimed to produce a certain final mixture.
The component amount is
\[ 0.20(12)+0.40(18). \]Compute:
\[ 2.4+7.2=9.6. \]Since total quantity is \(30\),
\[ c_f=\frac{9.6}{30}=0.32. \]Therefore, the final concentration is
\[ \boxed{32\%}. \]Checking by Bounds
If a final concentration is formed only by mixing \(20\%\) and \(50\%\) sources, the result should lie between \(20\%\) and \(50\%\).
A calculated answer such as
\[ 72\% \]immediately signals an error in the model or arithmetic.
Checking by Direction
If more high-concentration material is added, the final concentration should move toward the higher source concentration.
If pure water is added to a solution, the solute concentration should decrease.
If pure solute is added, the concentration should increase.
These directional checks are often faster than recomputing the entire equation.
Common Error: Averaging Concentrations Directly
The expression
\[ \frac{c_1+c_2}{2} \]gives the correct final concentration only when equal source quantities are mixed.
In general, use the weighted average
\[ \boxed{ c_f = \frac{c_1Q_1+c_2Q_2}{Q_1+Q_2} }. \]Common Error: Using Percent Numbers Instead of Decimals
A \(30\%\) concentration should enter the algebra as
\[ 0.30, \]not
\[ 30. \]Thus \(20\) liters of \(30\%\) solution contain
\[ 0.30(20)=6 \]liters of component, not \(600\).
Common Error: Forgetting the Final Total Changes
If \(x\) liters are added to \(20\) existing liters, the final quantity is
\[ \boxed{20+x}, \]not \(20\).
The right side of the mixture equation must use the final total quantity.
Common Error: Forgetting Pure Material Has Concentration \(1\)
Pure solute contributes
\[ 1\cdot x=x \]units of solute when \(x\) units are added.
It should not be represented by \(100x\).
Common Error: Forgetting Pure Carrier Has Concentration \(0\)
Pure water in a salt-water problem contributes
\[ 0 \]salt.
It increases total solution quantity without increasing salt quantity.
Common Error: Mixing Incompatible Units
Quantities expressed in liters, milliliters, kilograms, or grams must be converted to compatible units before addition.
A correct component equation cannot be built from mismatched quantity units.
Common Error: Accepting a Negative Mixture Amount
A negative source quantity usually means the stated target cannot be produced under the assumed two-source mixture model.
Source quantities in ordinary physical mixture problems must satisfy
\[ \boxed{Q_i\ge0}. \]Common Error: Target Outside the Available Range
Mixing \(10\%\) and \(30\%\) solutions cannot produce a \(50\%\) solution without adding a stronger source or removing carrier material.
Check the target against the available source concentrations before solving.
A Reliable Procedure for Basic Mixture Problems
- Identify the conserved component.
- Convert all percentages to decimal fractions.
- Write each source quantity.
- Compute each component amount as concentration times quantity.
- Add the source component amounts.
- Set that total equal to final concentration times final quantity.
- Solve for the unknown.
- Check the total quantity and final concentration.
A Reliable Procedure for Fixed-Total Mixtures
- Let one source quantity be \(x\).
- Write the other source quantity as \(Q-x\).
- Use the target concentration to write the final component amount.
- Set source component total equal to final component total.
- Solve for \(x\).
- Compute the second source amount.
- Verify that both amounts are nonnegative and sum to \(Q\).
A Reliable Procedure for Weighted-Average Problems
- Identify the per-unit property being averaged.
- Multiply each source value by its quantity.
- Add the weighted contributions.
- Divide by the total quantity when the final average is requested.
- If a source quantity is unknown, write an equation before dividing.
- Check that the final weighted average lies within the feasible range.
Practice 1: Component Amount
How much solute is contained in \(18\) liters of a \(35\%\) solution?
Practice 2: Final Concentration
Mix \(12\) liters of a \(25\%\) solution with \(8\) liters of a \(50\%\) solution. Find the final concentration.
Practice 3: Unknown Source Quantity
How many liters of a \(60\%\) solution should be mixed with \(20\) liters of a \(30\%\) solution to obtain a \(40\%\) solution?
Practice 4: Unknown Source Concentration
Ten liters of an unknown solution are mixed with \(15\) liters of a \(20\%\) solution to obtain a \(36\%\) solution. Find the unknown concentration.
Practice 5: Fixed Final Total
How many liters of \(20\%\) and \(50\%\) solutions are needed to make \(24\) liters of a \(30\%\) solution?
Practice 6: Add Pure Solute
A \(16\)-liter solution is \(25\%\) solute. How much pure solute must be added to make the final mixture \(40\%\) solute?
Practice 7: Add Pure Water
A \(20\)-liter solution is \(50\%\) solute. How much pure water must be added to reduce the concentration to \(40\%\)?
Practice 8: Equal Quantities
Equal quantities of \(18\%\) and \(42\%\) solutions are mixed. Find the final concentration.
Practice 9: Ratio Shortcut
In what ratio should \(10\%\) and \(50\%\) solutions be mixed to obtain a \(20\%\) solution?
Practice 10: Three Sources
Mix \(5\) liters of \(20\%\), \(10\) liters of \(30\%\), and \(15\) liters of \(40\%\) solution. Find the final concentration.
Practice 11: Mass Mixture
A \(24\)-kg material is \(37.5\%\) component A. Find the mass of component A.
Practice 12: Final Total from Source Amounts
A mixture is made from \(14\) liters of a \(20\%\) solution and \(21\) liters of a \(40\%\) solution. Find the final amount and final concentration.
Practice 13: Feasibility Check
Can a \(55\%\) solution be produced using only \(20\%\) and \(45\%\) solutions with positive source amounts? Explain.
Practice 14: Two-Equation Mixture
A \(50\)-liter mixture is to be \(34\%\) solute using \(25\%\) and \(40\%\) solutions. Find the amount of each source.
Practice 15: Identify the Error
A student mixes \(10\) liters of \(20\%\) solution with \(30\) liters of \(40\%\) solution and says the final concentration is
\[ \frac{20\%+40\%}{2}=30\%. \]Explain the error and find the correct concentration.
Practice Answers
Practice 1
\[ 0.35(18)=6.3. \]Therefore,
\[ \boxed{6.3\text{ liters of solute}}. \]Practice 2
Component amount:
\[ 0.25(12)+0.50(8) = 3+4 = 7. \]Total quantity:
\[ 12+8=20. \]Therefore,
\[ c_f=\frac7{20}=0.35. \] \[ \boxed{35\%}. \]Practice 3
Let \(x\) be the amount of \(60\%\) solution.
\[ 0.60x+0.30(20)=0.40(x+20). \] \[ 0.60x+6=0.40x+8. \] \[ 0.20x=2. \] \[ x=10. \]Therefore,
\[ \boxed{10\text{ liters}}. \]Practice 4
Let \(c\) be the unknown concentration.
\[ 10c+0.20(15)=0.36(25). \] \[ 10c+3=9. \] \[ 10c=6. \] \[ c=0.60. \]Therefore,
\[ \boxed{60\%}. \]Practice 5
Let \(x\) be the amount of \(20\%\) solution.
\[ 0.20x+0.50(24-x)=0.30(24). \] \[ 0.20x+12-0.50x=7.2. \] \[ -0.30x=-4.8. \] \[ x=16. \]Therefore,
\[ \boxed{16\text{ L of }20\%\text{ solution}} \]and
\[ \boxed{8\text{ L of }50\%\text{ solution}}. \]Practice 6
Initial solute:
\[ 0.25(16)=4. \]Let \(x\) be pure solute added:
\[ 4+x=0.40(16+x). \] \[ 4+x=6.4+0.4x. \] \[ 0.6x=2.4. \] \[ x=4. \]Therefore,
\[ \boxed{4\text{ liters}}. \]Practice 7
Initial solute:
\[ 0.50(20)=10. \]Let \(x\) be water added:
\[ 10=0.40(20+x). \] \[ 10=8+0.4x. \] \[ 2=0.4x. \] \[ x=5. \]Therefore,
\[ \boxed{5\text{ liters}}. \]Practice 8
Equal quantities give the arithmetic mean:
\[ \frac{18+42}{2}=30. \]Therefore,
\[ \boxed{30\%}. \]Practice 9
Use the concentration-distance ratio:
\[ \frac{Q_{10}}{Q_{50}} = \frac{50-20}{20-10} = \frac{30}{10} = 3. \]Therefore,
\[ \boxed{3:1}. \]Practice 10
Component amount:
\[ 0.20(5)+0.30(10)+0.40(15) = 1+3+6 = 10. \]Total quantity:
\[ 30. \]Therefore,
\[ c_f=\frac{10}{30}=\frac13. \]Thus,
\[ \boxed{33\frac13\%}. \]Practice 11
\[ 0.375(24)=9. \]Therefore,
\[ \boxed{9\text{ kg}}. \]Practice 12
Final amount:
\[ 14+21=35. \]Component amount:
\[ 0.20(14)+0.40(21) = 2.8+8.4 = 11.2. \]Therefore,
\[ c_f=\frac{11.2}{35}=0.32. \]Thus,
\[ \boxed{35\text{ liters at }32\%}. \]Practice 13
No. A standard mixture of \(20\%\) and \(45\%\) sources with positive amounts must have final concentration between \(20\%\) and \(45\%\).
\[ \boxed{\text{Not possible under the stated model}}. \]Practice 14
Let \(x\) be the amount of \(25\%\) solution.
\[ 0.25x+0.40(50-x)=0.34(50). \] \[ 0.25x+20-0.40x=17. \] \[ -0.15x=-3. \] \[ x=20. \]Therefore,
\[ \boxed{20\text{ L of }25\%\text{ solution}} \]and
\[ \boxed{30\text{ L of }40\%\text{ solution}}. \]Practice 15
The source quantities are not equal, so the concentrations cannot be averaged directly.
Total component:
\[ 0.20(10)+0.40(30) = 2+12 = 14. \]Total quantity:
\[ 40. \]Therefore,
\[ c_f=\frac{14}{40}=0.35. \]Thus,
\[ \boxed{35\%}. \]What to Remember
A component amount is
\[ \boxed{A=cQ}. \]For two-source mixtures,
\[ \boxed{ c_1Q_1+c_2Q_2 = c_f(Q_1+Q_2) }. \]The final concentration is the weighted average
\[ \boxed{ c_f = \frac{c_1Q_1+c_2Q_2}{Q_1+Q_2} }. \]Mixture problems are fundamentally conservation problems: total quantity and component quantity must be tracked consistently.
The most important habit is to identify the conserved component before writing the equation.
Where We Go Next
The basic mixture model treats quantities as being combined. Many practical problems also add pure material, dilute a mixture, remove part of a mixture, or repeatedly replace one portion with another.
The next lesson, Concentration, Dilution, and Replacement, develops those operations in detail.