3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations
Key Takeaways
- 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.
3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations
Work and mixture problems are among the most structured word problems tested on the AFCT Arithmetic Reasoning section. While they may appear complex at first glance, both categories follow rigorous algebraic formulas and matrix setups.
1. Fundamentals of Work Problems ($W = R \cdot t$)
Work problems model the amount of work completed by individuals or machines operating at constant rates. The foundational work equation is:
- $W =$ Amount of work accomplished ($W = 1$ when completing one full job)
- $R =$ Work rate (fraction of the job completed per unit of time)
- $t =$ Time spent working
Individual Work Rate Representation
If Person A completes a full job in $A$ hours, Person A's work rate is $R_A = \frac{1}{A}$ of the job per hour. Similarly, if Person B completes the job in $B$ hours, $R_B = \frac{1}{B}$.
Combined Work Rate & The Two-Worker Shortcut
When Person A and Person B work together, their individual rates add together:
Since combined time $T = \frac{1}{R_{\text{combined}}}$, we derive the Product over Sum Shortcut for two workers:
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: $A = 3\text{ hours}$, $B = 6\text{ hours}$.
- Step 2: Apply product-over-sum formula: $T = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\text{ hours}$.
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:
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 $= \frac{1}{4}$ bladder/hr. Rate of drain valve $= \frac{1}{12}$ bladder/hr.
- Step 2: Calculate net filling rate: $R_{\text{net}} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$ bladder/hr.
- Step 3: Calculate time to fill 1 full bladder: $T = \frac{1}{R_{\text{net}}} = 6\text{ hours}$.
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:
The Mixture Table Setup
Organize information into a structured matrix:
| Component | Volume ($V$) | Concentration ($C$) | Pure Solute Amount ($C \cdot V$) |
|---|---|---|---|
| Solution 1 | $V_1$ | $C_1$ | $C_1 \cdot V_1$ |
| Solution 2 | $V_2$ | $C_2$ | $C_2 \cdot V_2$ |
| Final Mixture | $V_1 + V_2$ | $C_{\text{final}}$ | $C_{\text{final}} \cdot (V_1 + V_2)$ |
4. Dilution & Concentration Adjustments
When modifying existing solutions:
- Adding Pure Water (Dilution): Pure water contains $0%$ solute ($C = 0$).
- Adding Pure Chemical / Solute: Pure chemical contains $100%$ solute ($C = 1.00$).
- Evaporating Water: Evaporating pure water subtracts volume with $C = 0$.
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 $x =$ liters of pure coolant added ($C = 1.00$).
- Step 2: Set up solute equation: $(0.10 \times 10) + (1.00 \times x) = 0.40 \times (10 + x)$.
- Step 3: Expand and simplify: $1 + x = 4 + 0.40x$.
- Step 4: Rearrange terms: $x - 0.40x = 4 - 1 \implies 0.60x = 3$.
- Step 5: Solve for $x$: $x = \frac{3}{0.60} = 5\text{ liters}$.
5. Summary Formula Reference Matrix
| Problem Type | Standard Governing Formula | Key Operational Rule |
|---|---|---|
| Two Workers Together | $T = \frac{A \cdot B}{A + B}$ | Rates add ($R_{\text{total}} = R_A + R_B$) |
| Opposing Work / Leaks | $R_{\text{net}} = R_{\text{fill}} - R_{\text{drain}}$ | Rates subtract |
| Solution Mixture | $C_1 V_1 + C_2 V_2 = C_m V_m$ | Pure solute amount is conserved |
| Pure Water Dilution | $C_1 V_1 + 0 = C_m (V_1 + V_{\text{water}})$ | Added water solute contribution is 0 |
Specialist Miller can clean and service a vehicle in 3 hours, while Specialist Davis can clean and service the same vehicle in 6 hours. Working together at their respective rates, how long will it take them to service one vehicle?
An inlet pipe can fill a water storage tank in 4 hours. A secondary drainage valve can completely empty the full tank in 12 hours. If both the inlet pipe and drainage valve are left open simultaneously, how many hours will it take to fill the empty tank?
A mechanic has 10 liters of a 10% coolant solution. How many liters of pure (100%) coolant must be added to create a 40% coolant solution?
A chemist mixes 40 milliliters of a 15% saline solution with 60 milliliters of a 35% saline solution. What is the concentration of saline in the resulting mixture?