Overview
Replacement problems are more complex than ordinary dilution because material is removed before something new is added.
If a well-mixed solution is \(30\%\) salt and \(5\) liters are removed, the removed \(5\) liters are also \(30\%\) salt. Removing mixture therefore removes some of the tracked component along with the rest of the solution.
The general accounting idea is
\[ \boxed{ \text{final component} = \text{initial component} - \text{component removed} + \text{component added} }. \]The essential skill is keeping track of what leaves, what enters, and what remains.
Why Replacement Is Different from Dilution
In an ordinary dilution problem, water is simply added. The amount of the tracked component remains unchanged.
In a replacement problem, some of the original mixture is first removed.
Suppose a tank contains \(20\) liters of a \(40\%\) solution. The amount of tracked component is
\[ 0.40(20)=8. \]If \(5\) liters of the well-mixed solution are removed, then the removed portion contains
\[ 0.40(5)=2 \]liters of the component.
So the amount remaining is
\[ 8-2=6. \]This subtraction step is what distinguishes replacement problems from simple dilution.
The Well-Mixed Assumption
Replacement problems normally assume that the mixture is uniform before any portion is removed.
If the concentration throughout a container is \(C\), then any portion removed also has concentration \(C\).
Therefore, removing \(x\) units removes
\[ Cx \]units of the tracked component.
Without this assumption, we would not know the composition of the material being removed.
The Basic Replacement Model
Suppose a container initially holds \(T\) units of mixture at concentration \(C\).
The initial component amount is
\[ CT. \]Now suppose \(x\) units of the mixture are removed.
Because the mixture is uniform, the component amount removed is
\[ Cx. \]The component amount remaining is therefore
\[ CT-Cx. \]If \(x\) units of replacement material with concentration \(R\) are added, the replacement contributes
\[ Rx. \]If the container returns to its original total quantity \(T\), the final component amount is
\[ C_fT. \]Thus the standard one-step replacement equation is
\[ \boxed{ CT-Cx+Rx=C_fT }. \]A More Compact Form
The equation
\[ CT-Cx+Rx=C_fT \]can be rewritten as
\[ C(T-x)+Rx=C_fT. \]This form has an intuitive interpretation:
\[ \boxed{ \text{component remaining} + \text{component added} = \text{final component} }. \]Both forms describe exactly the same process.
Example 1: Remove Solution and Replace with Water
A tank contains \(20\) liters of a \(40\%\) salt solution. How many liters of the solution should be removed and replaced with water to reduce the concentration to \(30\%\)?
Let \(x\) represent the number of liters removed and replaced.
The initial amount of salt is
\[ 0.40(20)=8. \]Removing \(x\) liters of the \(40\%\) solution removes
\[ 0.40x \]liters of salt.
Water contributes no salt, so the amount added is
\[ 0(x)=0. \]Because the removed volume is replaced with the same volume of water, the final volume is still \(20\) liters.
At \(30\%\), the final salt amount must be
\[ 0.30(20)=6. \]Therefore,
\[ 8-0.40x=6. \]Subtract \(8\):
\[ -0.40x=-2. \]Thus,
\[ x=5. \]So
\[ \boxed{5\text{ liters}} \]of solution should be removed and replaced with water.
Checking the Result
The original mixture contains \(8\) liters of salt.
Removing \(5\) liters of a \(40\%\) solution removes
\[ 0.40(5)=2 \]liters of salt.
That leaves
\[ 8-2=6 \]liters of salt.
After replacing the removed \(5\) liters with water, the tank again contains \(20\) liters total.
Therefore,
\[ \frac{6}{20}=0.30. \]So the final concentration is
\[ \boxed{30\%}. \]Organizing the Replacement Process
| Stage | Concentration | Quantity | Component Amount |
|---|---|---|---|
| Initial mixture | \(0.40\) | \(20\) | \(8\) |
| Mixture removed | \(0.40\) | \(x\) | \(-0.40x\) |
| Water added | \(0\) | \(x\) | \(0\) |
| Final mixture | \(0.30\) | \(20\) | \(6\) |
The component column gives
\[ 8-0.40x+0=6. \]Replacement problems are often easiest when treated as a sequence of additions and subtractions in the component amount.
Example 2: Replacing with a Stronger Solution
A \(30\)-liter tank contains a \(20\%\) solution. How many liters should be removed and replaced with a \(50\%\) solution to increase the final concentration to \(30\%\)?
Let \(x\) be the amount removed and replaced.
The original component amount is
\[ 0.20(30)=6. \]Removing \(x\) liters removes
\[ 0.20x. \]The replacement solution adds
\[ 0.50x. \]The final component amount must be
\[ 0.30(30)=9. \]So
\[ 6-0.20x+0.50x=9. \]Combine like terms:
\[ 6+0.30x=9. \]Subtract \(6\):
\[ 0.30x=3. \]Therefore,
\[ x=10. \]Thus,
\[ \boxed{10\text{ liters}} \]must be removed and replaced with the \(50\%\) solution.
The Concentration Difference Drives the Change
In the previous example, every liter removed takes away component at a rate of
\[ 0.20 \]unit per liter.
Every liter added returns component at a rate of
\[ 0.50. \]So each liter replaced produces a net gain of
\[ 0.50-0.20=0.30 \]unit of component.
This explains why the equation simplified to
\[ 6+0.30x=9. \]In general, replacement changes the component amount at the net rate
\[ \boxed{R-C} \]per unit replaced, where \(C\) is the original concentration and \(R\) is the replacement concentration.
Solving the General Replacement Equation
Start with
\[ CT-Cx+Rx=C_fT. \]Combine the \(x\)-terms:
\[ CT+(R-C)x=C_fT. \]Subtract \(CT\):
\[ (R-C)x=T(C_f-C). \]Therefore,
\[ \boxed{ x= \frac{T(C_f-C)}{R-C} }. \]This formula gives the amount that must be removed and replaced when the original and final total quantities are the same.
It also exposes the logic of the process: the desired change in total component amount is divided by the net component change produced by each unit replaced.
Interpreting the Replacement Formula
Consider
\[ x= \frac{T(C_f-C)}{R-C}. \]The numerator
\[ T(C_f-C) \]represents the total change in component amount required.
The denominator
\[ R-C \]represents the change in component amount created by replacing one unit of the original mixture with one unit of the replacement material.
If the replacement concentration is only slightly different from the original concentration, a large amount must be replaced.
If the replacement concentration differs greatly from the original, less replacement is required.
Example 3: Replacing with Pure Component
A \(20\)-kilogram alloy is \(30\%\) copper. How many kilograms of alloy should be removed and replaced with pure copper to obtain a \(50\%\) copper alloy?
Here,
\[ C=0.30, \] \[ C_f=0.50, \]and pure copper has concentration
\[ R=1. \]Using the replacement formula,
\[ x= \frac{20(0.50-0.30)}{1-0.30}. \]So
\[ x= \frac{20(0.20)}{0.70}. \] \[ x=\frac{4}{0.70}. \] \[ x\approx5.71. \]Therefore, approximately
\[ \boxed{5.71\text{ kilograms}} \]of alloy should be removed and replaced with pure copper.
Example 4: Finding the Final Concentration
A \(25\)-liter tank contains a \(40\%\) solution. Five liters are removed and replaced with water. What is the final concentration?
The original component amount is
\[ 0.40(25)=10. \]Removing \(5\) liters removes
\[ 0.40(5)=2. \]So the component amount remaining is
\[ 10-2=8. \]Water adds no component.
The total volume returns to \(25\) liters.
Therefore,
\[ C_f=\frac{8}{25}. \] \[ C_f=0.32. \]The final concentration is
\[ \boxed{32\%}. \]A Fractional View of One Replacement
Suppose a container holds \(T\) units of mixture at concentration \(C\), and \(x\) units are removed.
The fraction of the original mixture that remains is
\[ \frac{T-x}{T}. \]Because the mixture is uniform, the same fraction of the original component remains.
If water is used as the replacement, the final concentration becomes
\[ \boxed{ C_f=C\left(\frac{T-x}{T}\right) }. \]For example, removing \(5\) liters from a \(25\)-liter tank leaves the fraction
\[ \frac{20}{25}=0.8. \]So a \(40\%\) solution becomes
\[ 0.40(0.8)=0.32, \]or
\[ \boxed{32\%}. \]Repeated Replacement
Replacement problems become especially interesting when the remove-and-replace process is repeated several times.
Suppose a container holds \(T\) units of mixture. Each time, \(x\) units are removed and replaced with water.
After one replacement, the fraction of the original component remaining is
\[ \frac{T-x}{T}. \]If the mixture is stirred thoroughly again and the same process is repeated, the same fraction of the new component amount remains.
After two replacements, the original component has been multiplied by
\[ \left(\frac{T-x}{T}\right)^2. \]After \(n\) replacements,
\[ \boxed{ A_n = A_0 \left(\frac{T-x}{T}\right)^n }. \]If the total quantity remains \(T\), then the concentration follows the same pattern:
\[ \boxed{ C_n = C_0 \left(\frac{T-x}{T}\right)^n }. \]Example 5: Two Successive Replacements
A \(20\)-liter tank contains a \(50\%\) solution. Five liters are removed and replaced with water. The mixture is stirred, and the process is repeated once more. What is the final concentration?
Each replacement leaves the fraction
\[ \frac{20-5}{20} = \frac{15}{20} = \frac34 \]of the existing component.
After two replacements,
\[ C_2 = 0.50\left(\frac34\right)^2. \]Therefore,
\[ C_2 = 0.50\left(\frac{9}{16}\right). \] \[ C_2=0.28125. \]So the final concentration is
\[ \boxed{28.125\%}. \]Checking the Repeated Replacement Step by Step
Initially, the tank contains
\[ 0.50(20)=10 \]liters of component.
During the first removal, \(5\) liters of \(50\%\) solution are removed:
\[ 0.50(5)=2.5. \]So
\[ 10-2.5=7.5 \]liters remain.
After water is added, the concentration is
\[ \frac{7.5}{20}=0.375. \]During the second removal, the \(5\) liters removed are now \(37.5\%\) component:
\[ 0.375(5)=1.875. \]The remaining component amount is
\[ 7.5-1.875=5.625. \]Therefore,
\[ \frac{5.625}{20}=0.28125, \]confirming the result
\[ \boxed{28.125\%}. \]The Concentration Changes After Each Replacement
A common error in repeated-replacement problems is to assume that every removed portion has the original concentration.
That is only true for the first removal.
After replacement and mixing, the concentration has changed. The next portion removed has the new concentration, not the original one.
For this reason, repeated replacements behave multiplicatively rather than linearly.
The factor
\[ \frac{T-x}{T} \]is applied again after each complete remove-mix-replace cycle.
Repeated Replacement with Another Solution
If the replacement material has concentration \(R\) instead of \(0\), the process moves the mixture gradually toward \(R\).
After one replacement of \(x\) units in a container of total quantity \(T\),
\[ C_1 = C_0\left(\frac{T-x}{T}\right) + R\left(\frac{x}{T}\right). \]This is a weighted average of the concentration that remains and the concentration of the replacement material.
Repeated application moves the mixture closer and closer to \(R\).
This provides another connection between mixture problems, weighted averages, functions, and repeated processes.
Example 6: Replacement with a Stronger Mixture
A \(40\)-liter tank contains a \(25\%\) solution. Eight liters are removed and replaced with a \(75\%\) solution. What is the final concentration?
The initial component amount is
\[ 0.25(40)=10. \]The amount removed is
\[ 0.25(8)=2. \]The amount remaining is
\[ 10-2=8. \]The replacement contributes
\[ 0.75(8)=6. \]So the final component amount is
\[ 8+6=14. \]The total volume remains \(40\) liters.
Therefore,
\[ C_f=\frac{14}{40}. \] \[ C_f=0.35. \]The final concentration is
\[ \boxed{35\%}. \]Replacement Is Not the Same as Addition
Suppose \(5\) liters of water are added to a \(20\)-liter mixture.
The final quantity becomes
\[ 25\text{ liters}. \]But if \(5\) liters are removed and then replaced with \(5\) liters of water, the final quantity remains
\[ 20\text{ liters}. \]These situations therefore require different equations.
For addition:
\[ CT=C_f(T+x). \]For remove-and-replace with water:
\[ CT-Cx=C_fT. \]Recognizing which physical process is occurring is essential before beginning the algebra.
Method for Replacement Problems
- Find the initial component amount. Use \[ A=CT. \]
- Determine the concentration of what is removed. For a well-mixed solution, the removed portion has the current concentration of the mixture.
- Calculate the component removed. Use \[ A_{\text{removed}}=Cx. \]
- Find the component amount remaining. Subtract the removed component from the initial component.
- Determine the concentration of the replacement material. Water may be \(0\%\); pure component may be \(100\%\); another mixture may have any concentration in between.
- Add the replacement component. Use \[ A_{\text{added}}=Rx. \]
- Determine the final total quantity. If equal amounts are removed and added, the original total remains unchanged.
- Apply conservation and solve. Use \[ CT-Cx+Rx=C_fT. \]
- For repeated replacement, update the concentration after every cycle. Do not continue using the original concentration.
Common Errors
- Removing volume but not component. Removing part of a mixture removes part of every uniformly distributed component.
- Using the replacement concentration for the removed portion. The removed portion has the concentration of the mixture at the moment it is removed.
- Changing the final total when equal quantities are removed and replaced. If \(x\) units leave and \(x\) units enter, the total remains \(T\).
- Treating replacement as simple dilution. Dilution adds material without first removing any of the original component.
- Using the original concentration in every repeated cycle. The concentration changes after each replacement.
- Adding concentrations directly. Track component amounts, not concentration values alone.
- Ignoring whether the target is possible. The final concentration must move toward the concentration of the replacement material.
Practice Problems
- A \(20\)-liter tank contains a \(50\%\) solution. How many liters should be removed and replaced with water to obtain a \(35\%\) solution?
- A \(30\)-liter tank contains a \(20\%\) solution. How many liters should be removed and replaced with a \(60\%\) solution to obtain a \(30\%\) solution?
- A \(25\)-liter solution is \(40\%\) component. Five liters are removed and replaced with water. Find the final concentration.
- A \(40\)-kilogram alloy is \(30\%\) copper. Eight kilograms are removed and replaced with pure copper. Find the new copper concentration.
- A \(50\)-liter tank contains a \(60\%\) solution. Ten liters are removed and replaced with a \(20\%\) solution. Find the final concentration.
- A \(24\)-liter tank contains a \(50\%\) solution. Six liters are removed and replaced with water. The mixture is stirred and the process is repeated once more. Find the final concentration.
- A \(20\)-liter tank contains a \(30\%\) solution. Five liters are removed and replaced with water three times, stirring after each replacement. Find the final concentration.
- A \(30\)-liter tank contains a \(25\%\) solution. How many liters should be removed and replaced with pure component to obtain a \(40\%\) solution?
Practice Answers
- \[ 0.50(20)-0.50x=0.35(20) \] \[ 10-0.50x=7 \] \[ 0.50x=3 \] \[ \boxed{x=6\text{ liters}}. \]
- \[ 0.20(30)-0.20x+0.60x=0.30(30) \] \[ 6+0.40x=9 \] \[ 0.40x=3 \] \[ \boxed{x=7.5\text{ liters}}. \]
- Initial component: \[ 0.40(25)=10. \] Removed: \[ 0.40(5)=2. \] Remaining: \[ 10-2=8. \] Thus, \[ C_f=\frac{8}{25}=0.32. \] \[ \boxed{32\%}. \]
- Initial copper: \[ 0.30(40)=12. \] Copper removed: \[ 0.30(8)=2.4. \] Copper remaining: \[ 12-2.4=9.6. \] Pure copper added: \[ 8. \] Final copper: \[ 9.6+8=17.6. \] Therefore, \[ C_f=\frac{17.6}{40}=0.44. \] \[ \boxed{44\%}. \]
- Initial component: \[ 0.60(50)=30. \] Removed: \[ 0.60(10)=6. \] Replacement contribution: \[ 0.20(10)=2. \] Final component: \[ 30-6+2=26. \] Thus, \[ C_f=\frac{26}{50}=0.52. \] \[ \boxed{52\%}. \]
- Each cycle leaves \[ \frac{24-6}{24} = \frac34 \] of the existing component. After two cycles, \[ C_2 = 0.50\left(\frac34\right)^2 \] \[ = 0.50\left(\frac{9}{16}\right) \] \[ =0.28125. \] \[ \boxed{28.125\%}. \]
- Each cycle leaves \[ \frac{20-5}{20} = \frac34 \] of the existing component. After three cycles, \[ C_3 = 0.30\left(\frac34\right)^3. \] \[ C_3 = 0.30\left(\frac{27}{64}\right) = 0.1265625. \] Therefore, \[ \boxed{12.65625\%}. \]
- \[ 0.25(30)-0.25x+x=0.40(30) \] \[ 7.5+0.75x=12. \] \[ 0.75x=4.5. \] \[ \boxed{x=6\text{ liters}}. \]
Summary
Replacement problems require tracking three stages:
\[ \boxed{ \text{initial component} - \text{component removed} + \text{component added} = \text{final component} }. \]For an initial quantity \(T\) with concentration \(C\), if \(x\) units are removed and replaced with a substance of concentration \(R\), then
\[ \boxed{ CT-Cx+Rx=C_fT }. \]Equivalently,
\[ \boxed{ C(T-x)+Rx=C_fT }. \]Solving for the amount replaced gives
\[ \boxed{ x= \frac{T(C_f-C)}{R-C} }. \]When the replacement substance is water,
\[ R=0. \]When replacement is repeated, the concentration must be updated after each cycle. With repeated replacement by water,
\[ \boxed{ C_n = C_0 \left(\frac{T-x}{T}\right)^n }. \]The central lesson is that replacement changes the component amount twice: once when material leaves and again when new material enters.
Continue the Exploration
The mixture problems studied so far have focused primarily on physical concentration. The next lesson extends the same conservation reasoning to quantities such as cost and value, where different products are blended to achieve a desired average price.
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Continue to: Price and Value Mixtures
Apply the mixture structure to commodities, products, and other situations in which quantity and unit value combine to determine total value.