Mixture Problems
Model mixture problems by tracking the conserved substance or total value, then distinguish dilution, evaporation, blending, and replacement situations.
Mixture Problems
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Track the amount that is being conserved
This is a Mathematics modeling topic. The liquid, alloy, food, or price context supplies quantities for a weighted equation; no Chemistry knowledge is required.
For concentration problems, track the amount of pure substance:
pure amount = concentration × total amountFor price blends, track total value:
total value = unit price × quantityThen write:
amount before + amount added − amount removed = amount afterConcentration
Write a percent as a decimal before multiplying.
Dilution
Adding water changes total volume but not pure solute.
Evaporation
Only water leaves; the pure solute remains.
Price blend
Equate the sum of ingredient values to the blend’s total value.
Replacement
Removing mixture removes every component in the same current proportion.
Build a quantity–rate–pure amount table
For every ingredient, record:
quantity × concentration = pure amountAdd the pure-amount column, not the concentration column. If the total mixture is fixed, express one quantity as total−x.
For repeated replacement, multiply by the fraction retained at every step.
Why it works
Concentrations cannot usually be averaged directly. The conserved pure amount or total value determines the correct weighted result.
Five forms you should recognize
Problem: Two liters of a 20% salt solution are mixed with 3 liters of a 40% solution. Find the final concentration.
Setup: Add the pure salt, then divide by the combined volume.
0.20(2) + 0.40(3) = 1.6 L salt1.6÷(2 + 3) = 0.32 = 32%Why: Concentrations are weighted by their quantities.
Problem: How much water must be added to 10 L of a 30% solution to make it 20%?
Setup: The solution initially contains 0.30(10) = 3 L of pure substance. Water changes the total volume, not that 3 L.
final total = 3÷0.20 = 15 Lwater added = 15−10 = 5 LProblem: A merchant has 20 kg of rice worth ₱80 per kilogram. How many kilograms worth ₱140 per kilogram must be added to produce a blend worth ₱100 per kilogram?
Setup: Total ingredient value equals total blend value.
80(20) + 140x = 100(20 + x)1600 + 140x = 2000 + 100x → x = 10 kgProblem: A 25 L syrup mixture is 12% sugar. How much water must evaporate for it to become 15% sugar?
Setup: The 3 L of sugar remains; only water leaves.
0.12(25) = 3 L sugarfinal total = 3÷0.15 = 20 Lwater evaporated = 25−20 = 5 LProblem: A 20 L container holds a 30% salt solution. Five liters are removed and replaced with water. Find the new concentration.
Setup: Removing 5 of 20 liters removes one-fourth of the salt.
salt remaining = 0.30(20)(1520) = 4.5 Lnew concentration = 4.5÷20 = 22.5%Why: Replacement restores the total volume but adds no salt.
Check before you commit
- Averaging concentrations without considering quantities
- Adding water to the pure-solute amount
- Treating evaporation as removal of the whole mixture
- Forgetting that removed mixture contains solute
- Using the original concentration during a second replacement step
- Mixing pesos per kilogram with total pesos
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Foundations
Build the core procedure with immediate explanations.
Core Practice
Use mixed forms with less scaffolding.
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Apply the competency in unfamiliar representations.
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Mixture Problems FAQ
When can I simply average two concentrations?
Only when equal quantities are mixed. Otherwise use a weighted average based on quantity.
What remains constant during dilution?
The amount of pure solute remains constant; total volume changes.
Why does repeated replacement use multiplication?
Each step retains a fraction of the amount present immediately before that step.
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