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3.1 Mixture, Concentration & Price-Blending Problems

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Armed Forces Qualification Test AFQT

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1 Chapter 1: AFQT Structure, Scoring & Enlistment Requirements

1.1 AFQT Overview, Subtest Breakdown & CAT-ASVAB vs. P&P Format 1.2 The AFQT Scoring Formula (2×VE + AR + MK) & Percentile Norming 1.3 Branch Minimum Scores, AFQT Categories & Retake Policies

2 Chapter 2: Arithmetic Reasoning (AR) — Ratios, Rates & Percentages

2.1 Ratios, Proportions & Scale Conversions 2.2 Percentages, Discounts, Interest & Tax Calculations 2.3 Rate, Distance, Time & Speed Word Problems 2.4 Work Rates & Combined Labor Problems

3 Chapter 3: Arithmetic Reasoning (AR) — Mixtures, Sequences & Applied Math

3.1 Mixture, Concentration & Price-Blending Problems 3.2 Sequences, Patterns & Series Calculations 3.3 Averages, Weighted Averages & Statistical Word Problems

4 Chapter 4: Word Knowledge (WK) — Vocabulary & Word Structure

4.1 High-Frequency AFQT Vocabulary & Synonym Recognition 4.2 Latin & Greek Roots, Prefixes & Suffixes for Word Decoding 4.3 Context Clues & Words with Multiple Meanings

5 Chapter 5: Paragraph Comprehension (PC) — Reading Analysis & Inference

5.1 Identifying Main Ideas & Primary Purpose 5.2 Locating Supporting Details & Direct Facts 5.3 Drawing Logical Inferences & Implied Conclusions 5.4 Analyzing Tone, Author Perspective & Contextual Vocabulary

6 Chapter 6: Mathematics Knowledge (MK) — Fundamentals & Linear Algebra

6.1 Arithmetic Operations, Order of Operations & Number Properties 6.2 Exponents, Scientific Notation & Square Roots / Radicals 6.3 Linear Equations, Inequalities & Absolute Value 6.4 Systems of Linear Equations & Word-to-Algebra Translation

7 Chapter 7: Mathematics Knowledge (MK) — Advanced Algebra & Functions

7.1 Polynomial Operations & Factoring Techniques 7.2 Quadratic Equations, Factoring & Quadratic Formula 7.3 Coordinate Geometry: Slope, Distance, Midpoint & Linear Graphs

8 Chapter 8: Mathematics Knowledge (MK) — Geometry & Spatial Math

8.1 Angles, Triangles & Pythagorean Theorem 8.2 Quadrilaterals, Polygons & Circle Geometry 8.3 3D Geometry: Surface Area & Volume of Solids 8.4 Basic Geometric Transformations & Trigonometric Ratios

3.1 Mixture, Concentration & Price-Blending Problems

Key Takeaways Explain This Topic AI

The fundamental mixture equation C1 * V1 + C2 * V2 = C3 * V3 governs all two-component liquid, chemical, and alloy blending problems on the AFQT AR subtest.

When diluting a solution with pure water, the added solvent has a solute concentration of 0% (C2 = 0), reducing the equation to C1 * V1 = C3 * (V1 + V_water).

When strengthening a solution with pure solute, the added substance has a concentration of 100% (C2 = 1.00), meaning every unit added increases both the solute amount and the total volume.

Price-blending and commodity mix problems use the weighted unit cost formula: (Price1 * Quantity1) + (Price2 * Quantity2) = Total Cost = (Target Price * Total Quantity).

The Alligation Alternate method provides a rapid visual ratio shortcut on the AFQT for finding component ratios without full algebraic substitution.

Last updated: July 2026

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 C_1 V_1 + C_2 V_2 = C_3 V_3 C 1 ​ V 1 ​ + C 2 ​ V 2 ​ = C 3 ​ V 3 ​ , where C C C represents concentration expressed as a decimal or percentage, and V V V represents volume. For pure water additions, set C 2 = 0 % C_2 = 0\% C 2 ​ = 0% . For pure solute additions, set C 2 = 100 % C_2 = 100\% 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.

The Universal Mixture Framework ( C 1 V 1 + C 2 V 2 = C 3 V 3 C_1 V_1 + C_2 V_2 = C_3 V_3 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

Symbol Meaning Mathematical Representation

C 1 C_1 C 1 ​ Concentration of Component 1 Expressed as decimal ( 0.25 0.25 0.25 ) or percentage ( 25 % 25\% 25% )

V 1 V_1 V 1 ​ Volume or weight of Component 1 Liters, gallons, quarts, ounces, or pounds

C 2 C_2 C 2 ​ Concentration of Component 2 Expressed as decimal ( 0.60 0.60 0.60 ) or percentage ( 60 % 60\% 60% )

  • V 2 V_2 V 2 ​ Volume or weight of Component 2 Added amount of second component
  • C 3 C_3 C 3 ​ Target concentration of final mixture Final desired percentage

V 3 V_3 V 3 ​ Total volume of final mixture Always equal to V 1 + V 2 V_1 + V_2 V 1 ​ + V 2 ​ Solute from Container 1 + Solute from Container 2 = Total Solute in Final Mixture \text{Solute from Container 1} + \text{Solute from Container 2} = \text{Total Solute in Final Mixture} Solute from Container 1 + Solute from Container 2 = Total Solute in Final Mixture ( C 1 × V 1 ) + ( C 2 × V 2 ) = C 3 × ( V 1 + V 2 ) (C_1 \times V_1) + (C_2 \times V_2) = C_3 \times (V_1 + V_2) ( C 1 ​ × V 1 ​ ) + ( C 2 ​ × V 2 ​ ) = C 3 ​ × ( V 1 ​ + V 2 ​ )

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 C_1 = 0.20 C 1 ​ = 0.20 (20% initial acid)
  • V 1 = 12 V_1 = 12 V 1 ​ = 12 liters
  • C 2 = 0.80 C_2 = 0.80 C 2 ​ = 0.80 (80% added acid)
  • V 2 = x V_2 = x V 2 ​ = x (unknown liters of 80% acid)
  • C 3 = 0.40 C_3 = 0.40 C 3 ​ = 0.40 (40% target acid)
  • V 3 = 12 + x V_3 = 12 + x 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 ) 0.20(12) + 0.80(x) = 0.40(12 + x) 0.20 ( 12 ) + 0.80 ( x ) = 0.40 ( 12 + x )

  • Step 3: Solve Algebraically 2.4 + 0.80 x = 4.8 + 0.40 x 2.4 + 0.80x = 4.8 + 0.40x 2.4 + 0.80 x = 4.8 + 0.40 x
  • Subtract 0.40 x 0.40x 0.40 x from both sides: 2.4 + 0.40 x = 4.8 2.4 + 0.40x = 4.8 2.4 + 0.40 x = 4.8
  • Subtract 2.4 2.4 2.4 from both sides: 0.40 x = 2.4 0.40x = 2.4 0.40 x = 2.4

Divide by 0.40 0.40 0.40 : x = 2.4 0.40 = 6 liters x = \frac{2.4}{0.40} = 6 \text{ liters} x = 0.40 2.4 ​ = 6 liters

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

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 % 0\% 0% ( C 2 = 0.00 C_2 = 0.00 C 2 ​ = 0.00 ). Therefore, the amount of pure solute added is 0 × V water = 0 0 \times V_{\text{water}} = 0 0 × V water ​ = 0 .

Dilution Equation Structure C 1 V 1 = C 3 ( V 1 + V water ) C_1 V_1 = C_3 (V_1 + V_{\text{water}}) C 1 ​ V 1 ​ = C 3 ​ ( V 1 ​ + V 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 C_1 = 0.30 C 1 ​ = 0.30 , V 1 = 40 V_1 = 40 V 1 ​ = 40 quarts
  • C 2 = 0.00 C_2 = 0.00 C 2 ​ = 0.00 (pure water has 0% antifreeze)
  • V 2 = x V_2 = x V 2 ​ = x quarts of pure water
  • C 3 = 0.12 C_3 = 0.12 C 3 ​ = 0.12 target concentration
  • V 3 = 40 + x V_3 = 40 + x V 3 ​ = 40 + x

Step 2: Set Up and Solve the Equation 0.30 ( 40 ) + 0 ( x ) = 0.12 ( 40 + x ) 0.30(40) + 0(x) = 0.12(40 + x) 0.30 ( 40 ) + 0 ( x ) = 0.12 ( 40 + x ) 12 = 4.8 + 0.12 x 12 = 4.8 + 0.12x 12 = 4.8 + 0.12 x 7.2 = 0.12 x 7.2 = 0.12x 7.2 = 0.12 x x = 7.2 0.12 = 60 quarts x = \frac{7.2}{0.12} = 60 \text{ quarts} x = 0.12 7.2 ​ = 60 quarts

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

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 % 100\% 100% ( C 2 = 1.00 C_2 = 1.00 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 solute ) = C 3 ( V 1 + V solute ) C_1 V_1 + 1.00(V_{\text{solute}}) = C_3 (V_1 + V_{\text{solute}}) C 1 ​ V 1 ​ + 1.00 ( V solute ​ ) = C 3 ​ ( V 1 ​ + V 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 C_1 = 0.40 C 1 ​ = 0.40 , V 1 = 15 V_1 = 15 V 1 ​ = 15 liters
  • C 2 = 1.00 C_2 = 1.00 C 2 ​ = 1.00 (pure alcohol)
  • V 2 = x V_2 = x V 2 ​ = x liters of pure alcohol
  • C 3 = 0.60 C_3 = 0.60 C 3 ​ = 0.60 target concentration
  • V 3 = 15 + x V_3 = 15 + x V 3 ​ = 15 + x

Step 2: Set Up and Solve 0.40 ( 15 ) + 1.00 ( x ) = 0.60 ( 15 + x ) 0.40(15) + 1.00(x) = 0.60(15 + x) 0.40 ( 15 ) + 1.00 ( x ) = 0.60 ( 15 + x ) 6 + x = 9 + 0.60 x 6 + x = 9 + 0.60x 6 + x = 9 + 0.60 x

Subtract 0.60 x 0.60x 0.60 x from both sides: 6 + 0.40 x = 9 6 + 0.40x = 9 6 + 0.40 x = 9

Subtract 6 6 6 from both sides: 0.40 x = 3 0.40x = 3 0.40 x = 3 x = 3 0.40 = 7.5 liters x = \frac{3}{0.40} = 7.5 \text{ liters} x = 0.40 3 ​ = 7.5 liters

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

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 . Cost of Component 1 + Cost of Component 2 = Total Cost of Blended Goods \text{Cost of Component 1} + \text{Cost of Component 2} = \text{Total Cost of Blended Goods} Cost of Component 1 + Cost of Component 2 = Total Cost of Blended Goods ( P 1 × Q 1 ) + ( P 2 × Q 2 ) = P blend × ( Q 1 + Q 2 ) (P_1 \times Q_1) + (P_2 \times Q_2) = P_{\text{blend}} \times (Q_1 + Q_2) ( P 1 ​ × Q 1 ​ ) + ( P 2 ​ × Q 2 ​ ) = P blend ​ × ( Q 1 ​ + Q 2 ​ )

  • Concept Chemical Mixture Price Blending
  • Component 1 Intensity Concentration % ( C 1 C_1 C 1 ​ ) Price per pound/unit ( P 1 P_1 P 1 ​ )
  • Component 1 Size Volume ( V 1 V_1 V 1 ​ ) Quantity/Weight ( Q 1 Q_1 Q 1 ​ )

Tracked Quantity Total Solute ( C 1 V 1 C_1 V_1 C 1 ​ V 1 ​ ) Total Dollar Value ( P 1 Q 1 P_1 Q_1 P 1 ​ Q 1 ​ )

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 = pounds of Arabica coffee A = \text{pounds of Arabica coffee} A = pounds of Arabica coffee ($14.00/lb)

The remaining weight is ( 30 − A ) = pounds of Robusta coffee (30 - A) = \text{pounds of Robusta coffee} ( 30 − A ) = pounds of Robusta coffee ($8.00/lb)

Target total weight = 30 lbs 30 \text{ lbs} 30 lbs at $10.00/lb = $300.00 total value

Step 2: Set Up the Total Value Equation 14 A + 8 ( 30 − A ) = 10 ( 30 ) 14A + 8(30 - A) = 10(30) 14 A + 8 ( 30 − A ) = 10 ( 30 ) 14 A + 240 − 8 A = 300 14A + 240 - 8A = 300 14 A + 240 − 8 A = 300 6 A + 240 = 300 6A + 240 = 300 6 A + 240 = 300 6 A = 60 6A = 60 6 A = 60 A = 10 lbs A = 10 \text{ lbs} A = 10 lbs

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

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

Place the higher concentration/price at the top-left corner.

Place the lower concentration/price at the bottom-left corner.

Place the desired target concentration/price in the center.

Subtract diagonally across to find the cross-differences (always positive). High ( 80 % ) Part High = ∣ 20 − 40 ∣ = 20 Target ( 40 % ) Low ( 20 % ) Part Low = ∣ 80 − 40 ∣ = 40 \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} High ( 80% ) Low ( 20% ) ​ Target ( 40% ) ​ Part High = ∣20 − 40∣ = 20 Part Low = ∣80 − 40∣ = 40 ​

Ratio of High to Low = 20 : 40 = 1 : 2 20 : 40 = 1 : 2 20 : 40 = 1 : 2 .

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

AFQT Exam Traps to Avoid

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.

Percentage vs. Decimal Errors: Multiplying 40 × 30 40 \times 30 40 × 30 instead of 40 × 0.30 40 \times 0.30 40 × 0.30 leads to absurd numbers. Keep units consistent across all terms.

Misidentifying Diluents: Water has 0 % 0\% 0% concentration of solute, but 100 % 100\% 100% concentration of water. Pay close attention to which chemical substance the question asks about.