3.1 Mixture, Concentration & Price-Blending Problems

3.1 Mixture, Concentration & Price-Blending Problems

Quick Answer: Mixture problems test your ability to track the total quantity of an active ingredient (solute, pure acid, alcohol, or pure metal) across different solutions. The universal master equation is $C_1 V_1 + C_2 V_2 = C_3 V_3$, where $C$ represents concentration expressed as a decimal or percentage, and $V$ represents volume. For pure water additions, set $C_2 = 0%$. For pure solute additions, set $C_2 = 100%$.

Mixture and price-blending problems are among the most frequently tested advanced quantitative reasoning questions on the Armed Services Vocational Aptitude Battery (ASVAB) AFQT. These items evaluate whether a candidate can isolate key variables, maintain ratio preservation under transformation, and model real-world physical systems algebraically under strict time constraints.

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The Universal Mixture Framework ($C_1 V_1 + C_2 V_2 = C_3 V_3$)

Every mixture word problem relies on a fundamental principle of conservation: the total amount of pure active ingredient (solute) before mixing must equal the total amount of pure active ingredient after mixing.

Core Equation Components

$\text{Solute from Container 1} + \text{Solute from Container 2} = \text{Total Solute in Final Mixture}$ $(C_1 \times V_1) + (C_2 \times V_2) = C_3 \times (V_1 + V_2)$

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Step-by-Step Worked Example: Standard Two-Solution Blend

Problem: A laboratory technician has 12 liters of a 20% acid solution and needs to mix it with an 80% acid solution to produce a 40% acid solution. How many liters of the 80% acid solution must be added?

Step 1: Identify Known and Unknown Variables

• $C_1 = 0.20$ (20% initial acid)
• $V_1 = 12$ liters
• $C_2 = 0.80$ (80% added acid)
• $V_2 = x$ (unknown liters of 80% acid)
• $C_3 = 0.40$ (40% target acid)
• $V_3 = 12 + x$ (total combined volume)

Step 2: Set Up the Conservation of Solute Equation

$0.20(12) + 0.80(x) = 0.40(12 + x)$

Step 3: Solve Algebraically

$2.4 + 0.80x = 4.8 + 0.40x$ Subtract $0.40x$ from both sides: $2.4 + 0.40x = 4.8$ Subtract $2.4$ from both sides: $0.40x = 2.4$ Divide by $0.40$: $x = \frac{2.4}{0.40} = 6 \text{ liters}$

Verification: 12 liters of 20% acid contains $2.4$ L pure acid. 6 liters of 80% acid contains $4.8$ L pure acid. Total acid = $2.4 + 4.8 = 7.2$ L. Total volume = $12 + 6 = 18$ L. Final strength = $7.2 / 18 = 0.40 = 40%$. The calculation is verified.

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Dilution Problems: Adding Pure Water or Pure Solvent

Dilution occurs when a liquid containing zero solute (such as pure distilled water) is added to a solution to lower its concentration.

Key Rule for Dilution: Pure solvent has a solute concentration of $0%$ ($C_2 = 0.00$). Therefore, the amount of pure solute added is $0 \times V_{\text{water}} = 0$.

Dilution Equation Structure

$C_1 V_1 = C_3 (V_1 + V_{\text{water}})$

Step-by-Step Worked Example: Radiator Dilution

Problem: An army motor pool mechanic has 40 quarts of an engine coolant mixture that is 30% antifreeze. How many quarts of pure water must be added to dilute the solution to a 12% antifreeze concentration?

Step 1: Identify Variables

• $C_1 = 0.30$, $V_1 = 40$ quarts
• $C_2 = 0.00$ (pure water has 0% antifreeze)
• $V_2 = x$ quarts of pure water
• $C_3 = 0.12$ target concentration
• $V_3 = 40 + x$

Step 2: Set Up and Solve the Equation

$0.30(40) + 0(x) = 0.12(40 + x)$ $12 = 4.8 + 0.12x$ $7.2 = 0.12x$ $x = \frac{7.2}{0.12} = 60 \text{ quarts}$

The mechanic must add 60 quarts of pure water to achieve the target 12% concentration.

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Concentration Fortification: Adding Pure Solute

Fortification involves adding pure, undiluted active ingredient (100% concentration) to raise the strength of a weak solution.

Key Rule for Fortification: Pure solute has a concentration of $100%$ ($C_2 = 1.00$). Every gallon or quart added contributes its full volume directly to the solute total.

Fortification Equation Structure

$C_1 V_1 + 1.00(V_{\text{solute}}) = C_3 (V_1 + V_{\text{solute}})$

Step-by-Step Worked Example: Disinfectant Fortification

Problem: A medical corps officer has 15 liters of a 40% alcohol disinfectant. How many liters of pure 100% isopropyl alcohol must be added to boost the solution to a 60% alcohol disinfectant?

Step 1: Identify Variables

• $C_1 = 0.40$, $V_1 = 15$ liters
• $C_2 = 1.00$ (pure alcohol)
• $V_2 = x$ liters of pure alcohol
• $C_3 = 0.60$ target concentration
• $V_3 = 15 + x$

Step 2: Set Up and Solve

$0.40(15) + 1.00(x) = 0.60(15 + x)$ $6 + x = 9 + 0.60x$ Subtract $0.60x$ from both sides: $6 + 0.40x = 9$ Subtract $6$ from both sides: $0.40x = 3$ $x = \frac{3}{0.40} = 7.5 \text{ liters}$

Adding 7.5 liters of pure alcohol raises the 15-liter batch from 40% to 60%.

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Price-Blending and Commodity Mixes

Price-blending problems apply the exact same mathematical model as chemical mixtures. Instead of tracking pure solute, you track total monetary value.

$\text{Cost of Component 1} + \text{Cost of Component 2} = \text{Total Cost of Blended Goods}$ $(P_1 \times Q_1) + (P_2 \times Q_2) = P_{\text{blend}} \times (Q_1 + Q_2)$

Step-by-Step Worked Example: Coffee Blend Unit Cost

Problem: A quartermaster needs to prepare a 30-pound coffee blend priced at $10.00 per pound by mixing premium Arabica coffee costing $14.00 per pound with commercial Robusta coffee costing $8.00 per pound. How many pounds of Arabica coffee are required?

Step 1: Set Up Variables

• Let $A = \text{pounds of Arabica coffee}$ ($14.00/lb)
• The remaining weight is $(30 - A) = \text{pounds of Robusta coffee}$ ($8.00/lb)
• Target total weight = $30 \text{ lbs}$ at $10.00/lb = $300.00 total value

Step 2: Set Up the Total Value Equation

$14A + 8(30 - A) = 10(30)$ $14A + 240 - 8A = 300$ $6A + 240 = 300$ $6A = 60$ $A = 10 \text{ lbs}$

The blend requires 10 pounds of Arabica coffee and 20 pounds of Robusta coffee ($30 - 10 = 20$).

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Shortcut Strategy: The Alligation Alternate Method

On timed AFQT exams, the Alligation Alternate grid allows you to determine component ratio proportions in under 30 seconds without writing multi-step algebraic equations.

How Alligation Works

1. Place the higher concentration/price at the top-left corner.
2. Place the lower concentration/price at the bottom-left corner.
3. Place the desired target concentration/price in the center.
4. Subtract diagonally across to find the cross-differences (always positive).

$\begin{array}{ccc} \text{High } (80\%) & & \text{Part High } = |20 - 40| = 20 \\ & \text{Target } (40\%) & \\ \text{Low } (20\%) & & \text{Part Low } = |80 - 40| = 40 \end{array}$

• Ratio of High to Low = $20 : 40 = 1 : 2$.
• For 12 liters of 20% (the Low component), you need half as much of the High component: $12 \times (1/2) = 6$ liters.

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AFQT Exam Traps to Avoid

1. Confusing Added Volume with Final Total Volume: Questions frequently ask for "how many gallons must be added" versus "what is the final volume of the mixture". Always check the exact wording before selecting an answer choice.
2. Percentage vs. Decimal Errors: Multiplying $40 \times 30$ instead of $40 \times 0.30$ leads to absurd numbers. Keep units consistent across all terms.
3. Misidentifying Diluents: Water has $0%$ concentration of solute, but $100%$ concentration of water. Pay close attention to which chemical substance the question asks about.