2.4 Work Rates & Combined Labor Problems

2.4 Work Rates & Combined Labor Problems

Work rate word problems on the AFQT evaluate your ability to compute how fast individuals, teams, or machines complete tasks working individually or collaboratively. Understanding the relationship between work completed ($W$), rate of work ($R$), and time ($t$) is essential for high scores.

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The Core Work Rate Formula ($W = Rt$)

The fundamental work equation mirrors the distance formula ($d = rt$):

$$W = R \cdot t$$

Where:

• $W$ = Total work completed (often expressed as $1$ complete job).
• $R$ = Rate of work per unit of time (job/hour, job/day).
• $t$ = Time spent working.

Defining Individual Work Rates

If Worker A can complete $1$ entire job alone in $T$ hours, Worker A's hourly work rate ($R_A$) is:

$$R_A = \frac{1}{T} \text{ of the job per hour}$$

Example: If a mechanic takes $6$ hours to overhaul an engine, the mechanic overhauls $\frac{1}{6}$ of the engine per hour.

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Combined Labor: Two Workers Completing One Job

When two entities work simultaneously without interfering with each other, their individual rates add together to yield a combined rate ($R_{\text{combined}}$).

Deriving the Combined Work Equation

$$R_{\text{combined}} = R_A + R_B = \frac{1}{a} + \frac{1}{b}$$ $$\frac{1}{T_{\text{combined}}} = \frac{b + a}{a \cdot b}$$

The Product-Over-Sum Shortcut Formula

Inverting the fraction yields the famous Product-over-Sum formula for two workers:

$$T_{\text{combined}} = \frac{a \cdot b}{a + b}$$

Where:

• $a$ = Time required for Worker A alone.
• $b$ = Time required for Worker B alone.

Example: Tech A can configure a server in $4$ hours ($a = 4$), while Tech B can configure it in $6$ hours ($b = 6$).

$$T_{\text{combined}} = \frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4 \text{ hours (2 hours 24 minutes)}$$

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Multiple Workers and Variable Efficiency

When three or more workers collaborate, or when a crew consists of multiple identical workers, add all individual rates:

$$\text{Combined Rate } (3 \text{ workers}) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$

Identical Worker Crew Formula

If $n$ identical workers each take $T$ hours to finish a job alone, the total time required for the crew of $n$ workers is:

$$T_{\text{crew}} = \frac{T}{n}$$

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Partial Shifts and Staggered Start Times

AFQT word problems often involve scenarios where one worker starts earlier, and a second worker joins later.

4-Step Algorithm for Staggered Work Problems

1. Find individual rates: $R_A = \frac{1}{a}$ and $R_B = \frac{1}{b}$.
2. Calculate initial work completed: Multiply Worker A's rate by the solo time ($t_{\text{solo}}$). $$W_{\text{initial}} = R_A \cdot t_{\text{solo}}$$
3. Determine remaining work: Subtract initial work from $1$ full job. $$W_{\text{remaining}} = 1 - W_{\text{initial}}$$
4. Divide remaining work by combined rate: Find remaining joint time ($t_{\text{joint}}$). $$t_{\text{joint}} = \frac{W_{\text{remaining}}}{R_A + R_B}$$

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Tank Filling and Pipe Flow Problems

Pipe flow problems operate on the exact same principles as labor rates. However, when a drain or outlet pipe is open, its rate subtracts from the total filling rate.

$$\text{Net Rate} = R_{\text{inlet 1}} + R_{\text{inlet 2}} - R_{\text{outlet}}$$ $$\frac{1}{T_{\text{net}}} = \frac{1}{A} + \frac{1}{B} - \frac{1}{C}$$

Example: Pipe A fills a water tank in $3$ hours ($+1/3$), Pipe B fills it in $6$ hours ($+1/6$), and Drain C empties it in $4$ hours ($-1/4$).

$$\text{Net Rate} = \frac{1}{3} + \frac{1}{6} - \frac{1}{4} = \frac{4}{12} + \frac{2}{12} - \frac{3}{12} = \frac{3}{12} = \frac{1}{4}$$ $$\text{Time to fill tank} = 4 \text{ hours}$$

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Step-by-Step Worked AFQT Examples

Worked Example 1: Combined Labor Shortcut

Problem: Specialist Miller can dig a tactical trench in $5$ hours alone. Specialist Davis can dig the same trench in $7.5$ hours alone. If they work together at their respective constant rates, how many hours will it take them to dig the trench?

• Step 1: Identify individual times $a = 5$ and $b = 7.5$.
• Step 2: Apply Product-over-Sum shortcut. $$T = \frac{5 \times 7.5}{5 + 7.5} = \frac{37.5}{12.5} = 3 \text{ hours}$$

Worked Example 2: Staggered Work Shifts

Problem: Carpenter A can build a command ramp in $10$ hours, and Carpenter B can build it in $15$ hours. Carpenter A begins working alone for $4$ hours. Carpenter B then joins Carpenter A, and they work together until the ramp is completed. What is the total time spent building the ramp?

• Step 1: Find individual rates: $R_A = \frac{1}{10}$ and $R_B = \frac{1}{15}$.
• Step 2: Calculate work done by Carpenter A in first $4$ hours. $$W_{\text{A solo}} = 4 \times \frac{1}{10} = \frac{4}{10} = \frac{2}{5} \text{ of the job}$$
• Step 3: Calculate remaining work. $$W_{\text{remaining}} = 1 - \frac{2}{5} = \frac{3}{5} \text{ of the job}$$
• Step 4: Find combined rate $R_A + R_B$. $$R_{\text{combined}} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \text{ job/hour}$$
• Step 5: Calculate joint time for remaining work. $$t_{\text{joint}} = \frac{3/5}{1/6} = \frac{3}{5} \times 6 = \frac{18}{5} = 3.6 \text{ hours}$$
• Step 6: Add initial solo time to joint time for total duration. $$\text{Total Time} = 4 \text{ hours} + 3.6 \text{ hours} = 7.6 \text{ hours (7 hours 36 minutes)}$$

Worked Example 3: Pipe Filling with Drain

Problem: An inlet pipe can fill a fuel bladder in $4$ hours. A drainage valve can completely empty a full bladder in $12$ hours. If the valve is accidentally left open while the inlet pipe is filling an empty bladder, how many hours will it take to fill the bladder?

• Step 1: Identify inlet rate ($+1/4$) and outlet rate ($-1/12$).
• Step 2: Subtract drain rate from inlet rate to find net rate. $$\text{Net Rate} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \text{ bladder/hour}$$
• Step 3: Invert net rate to find total filling time. $$\text{Total Time} = 6 \text{ hours}$$

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Common AFQT Traps & Shortcut Strategies

1. The Direct Addition Fallacy: Never add completion times together! If Worker A takes $4$ hours and Worker B takes $6$ hours, working together takes LESS than $4$ hours, never $4 + 6 = 10$ hours!
2. The Product-over-Sum Restriction: Remember that $T = \frac{ab}{a+b}$ ONLY works for TWO workers. For three workers, you must find a common denominator and sum their fractional rates: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.