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3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations
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3.1 Word Problems: Rates, Ratios, Proportions & Percentages 3.2 Financial & Distance Math: Interest, Profit/Loss, Speed & Time 3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations 3.4 Multi-Step Problem Solving & Algebraic Word Problems
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3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations
Key Takeaways Explain This Topic AI
The Work formula W = R * t establishes that Work equals Rate times Time; individual rates add when working together (R_total = R1 + R2).
The combined time T for two entities working together to complete 1 job is given by the shortcut T = (A * B) / (A + B).
When entities work against each other (e.g., inlet pipe filling while a drain leaks), subtract the rates: R_net = R_fill - R_drain.
Mixture concentration problems rely on solute balance: Concentration1 * Volume1 + Concentration2 * Volume2 = Concentration_final * Volume_final.
Pure water added to a solution has a concentration of 0%; adding pure solute (100% concentration) increases chemical density rapidly.
Last updated: July 2026
3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations
Work and mixture problems are among the most structured word problems tested on the Arithmetic Reasoning practice. While they may appear complex at first glance, both categories follow rigorous algebraic formulas and matrix setups.
1. Fundamentals of Work Problems ( W = R ⋅ t W = R \cdot t W = R ⋅ t )
Work problems model the amount of work completed by individuals or machines operating at constant rates. The foundational work equation is: W = R ⋅ t W = R \cdot t W = R ⋅ t
- W = W = W = Amount of work accomplished ( W = 1 W = 1 W = 1 when completing one full job)
- R = R = R = Work rate (fraction of the job completed per unit of time)
- t = t = t = Time spent working
- Individual Work Rate Representation
If Person A completes a full job in A A A hours, Person A's work rate is R A = 1 A R_A = \frac{1}{A} R A = A 1 of the job per hour. Similarly, if Person B completes the job in B B B hours, R B = 1 B R_B = \frac{1}{B} R B = B 1 .
Combined Work Rate & The Two-Worker Shortcut
When Person A and Person B work together, their individual rates add together: R combined = R A + R B = 1 A + 1 B = A + B A ⋅ B R_{\text{combined}} = R_A + R_B = \frac{1}{A} + \frac{1}{B} = \frac{A + B}{A \cdot B} R combined = R A + R B = A 1 + B 1 = A ⋅ B A + B
Since combined time T = 1 R combined T = \frac{1}{R_{\text{combined}}} T = R combined 1 , we derive the Product over Sum Shortcut for two workers: T = A ⋅ B A + B T = \frac{A \cdot B}{A + B} T = A + B A ⋅ B
Step-by-Step Worked Example 1: Specialist Miller can clean and service a tactical vehicle in 3 hours, while Specialist Davis can service the same vehicle in 6 hours. Working together, how long will it take them to service one vehicle?
Step 1: Identify individual times: A = 3 hours A = 3\text{ hours} A = 3 hours , B = 6 hours B = 6\text{ hours} B = 6 hours .
Step 2: Apply product-over-sum formula: T = 3 × 6 3 + 6 = 18 9 = 2 hours T = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\text{ hours} T = 3 + 6 3 × 6 = 9 18 = 2 hours .
- 2. Advanced Work Scenarios: Opposing Rates & Staggered Starts
- Opposing Work Rates (Inlet Pipes and Drains)
When one process performs work while another undoes work (such as a pipe filling a reservoir while a leak drains it), subtract the rates : R net = R fill − R drain = 1 Fill Time − 1 Drain Time R_{\text{net}} = R_{\text{fill}} - R_{\text{drain}} = \frac{1}{\text{Fill Time}} - \frac{1}{\text{Drain Time}} R net = R fill − R drain = Fill Time 1 − Drain Time 1
Step-by-Step Worked Example 2: An inlet pipe can fill a fuel storage bladder in 4 hours. A secondary drain valve can completely empty the full bladder in 12 hours. If both the inlet pipe and drain valve are open, how long will it take to fill the empty bladder?
Step 1: Rate of inlet pipe = 1 4 = \frac{1}{4} = 4 1 bladder/hr. Rate of drain valve = 1 12 = \frac{1}{12} = 12 1 bladder/hr.
Step 2: Calculate net filling rate: R net = 1 4 − 1 12 = 3 12 − 1 12 = 2 12 = 1 6 R_{\text{net}} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6} R net = 4 1 − 12 1 = 12 3 − 12 1 = 12 2 = 6 1 bladder/hr.
Step 3: Calculate time to fill 1 full bladder: T = 1 R net = 6 hours T = \frac{1}{R_{\text{net}}} = 6\text{ hours} T = R net 1 = 6 hours .
3. Mixture & Solution Concentration Problems
Mixture problems involve combining two or more solutions of different concentration percentages to produce a final mixture with a target concentration.
The Pure Solute Balance Equation
The core principle is that the total amount of pure substance (solute) before mixing equals the total amount of pure substance in the final mixture: Solute Amount = Concentration (decimal) × Total Liquid Volume \text{Solute Amount} = \text{Concentration (decimal)} \times \text{Total Liquid Volume} Solute Amount = Concentration (decimal) × Total Liquid Volume C 1 ⋅ V 1 + C 2 ⋅ V 2 = C final ⋅ ( V 1 + V 2 ) C_1 \cdot V_1 + C_2 \cdot V_2 = C_{\text{final}} \cdot (V_1 + V_2) C 1 ⋅ V 1 + C 2 ⋅ V 2 = C final ⋅ ( V 1 + V 2 )
The Mixture Table Setup
Organize information into a structured matrix:
- Component Volume ( V V V ) Concentration ( C C C ) Pure Solute Amount ( C ⋅ V C \cdot V C ⋅ V )
- Solution 1 V 1 V_1 V 1 C 1 C_1 C 1 C 1 ⋅ V 1 C_1 \cdot V_1 C 1 ⋅ V 1
- Solution 2 V 2 V_2 V 2 C 2 C_2 C 2 C 2 ⋅ V 2 C_2 \cdot V_2 C 2 ⋅ V 2
Final Mixture V 1 + V 2 V_1 + V_2 V 1 + V 2 C final C_{\text{final}} C final C final ⋅ ( V 1 + V 2 ) C_{\text{final}} \cdot (V_1 + V_2) C final ⋅ ( V 1 + V 2 )
4. Dilution & Concentration Adjustments
When modifying existing solutions:
Adding Pure Water (Dilution): Pure water contains 0 % 0\% 0% solute ( C = 0 C = 0 C = 0 ).
Adding Pure Chemical / Solute: Pure chemical contains 100 % 100\% 100% solute ( C = 1.00 C = 1.00 C = 1.00 ).
Evaporating Water: Evaporating pure water subtracts volume with C = 0 C = 0 C = 0 .
Step-by-Step Worked Example 3 (Adding Pure Solute): A mechanic has 10 liters of a 10% radiator coolant solution. How many liters of pure (100%) coolant must be added to create a 40% coolant solution?
Step 1: Let x = x = x = liters of pure coolant added ( C = 1.00 C = 1.00 C = 1.00 ).
Step 2: Set up solute equation: ( 0.10 × 10 ) + ( 1.00 × x ) = 0.40 × ( 10 + x ) (0.10 \times 10) + (1.00 \times x) = 0.40 \times (10 + x) ( 0.10 × 10 ) + ( 1.00 × x ) = 0.40 × ( 10 + x ) .
Step 3: Expand and simplify: 1 + x = 4 + 0.40 x 1 + x = 4 + 0.40x 1 + x = 4 + 0.40 x .
Step 4: Rearrange terms: x − 0.40 x = 4 − 1 ⟹ 0.60 x = 3 x - 0.40x = 4 - 1 \implies 0.60x = 3 x − 0.40 x = 4 − 1 ⟹ 0.60 x = 3 .
Step 5: Solve for x x x : x = 3 0.60 = 5 liters x = \frac{3}{0.60} = 5\text{ liters} x = 0.60 3 = 5 liters .
5. Summary Formula Reference Matrix
Problem Type Standard Governing Formula Key Operational Rule
Two Workers Together T = A ⋅ B A + B T = \frac{A \cdot B}{A + B} T = A + B A ⋅ B Rates add ( R total = R A + R B R_{\text{total}} = R_A + R_B R total = R A + R B )
Opposing Work / Leaks R net = R fill − R drain R_{\text{net}} = R_{\text{fill}} - R_{\text{drain}} R net = R fill − R drain Rates subtract
Solution Mixture C 1 V 1 + C 2 V 2 = C m V m C_1 V_1 + C_2 V_2 = C_m V_m C 1 V 1 + C 2 V 2 = C m V m Pure solute amount is conserved
Pure Water Dilution C 1 V 1 + 0 = C m ( V 1 + V water ) C_1 V_1 + 0 = C_m (V_1 + V_{\text{water}}) C 1 V 1 + 0 = C m ( V 1 + V water ) Added water solute contribution is 0