5.2 Work Rates: Combined Rates, Sequential Tasks, and Opposing Rates

5.2 Work Rates: Combined Rates, Sequential Tasks, and Opposing Rates

Quick Summary: Work-rate problems on the GMAT Focus Quantitative Reasoning measure are mathematically identical to uniform motion problems, governed by the master equation $\text{Work} = \text{Rate} \times \text{Time}$ ($W = R \times T$). Rather than struggling with reciprocal fractions (such as $\frac{1}{A} + \frac{1}{B} = \frac{1}{T}$), high-scoring test-takers assign total work units using the least common multiple (LCM) of the given times. This 'Smart Number' strategy converts reciprocal fractions into clean integer operations, allowing you to solve combined labor, sequential hand-offs, opposing inlet/outlet flows, and worker-pool scaling in under two minutes.

Work-rate problems frequently appear on the GMAT because they rigorously test two core business analytics skills: calculating throughput capacity and managing operational workflows under resource constraints. Without calculator access, executing operations with fractions such as $\frac{1}{12} + \frac{1}{15} + \frac{1}{20}$ invites arithmetic errors. By understanding rate structures and applying smart number substitutions, you can bypass complex fraction arithmetic entirely.

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The Fundamental Work Equation: W = R × T

The governing framework for all work problems connects three foundational variables:

$\text{Work Done} = \text{Rate of Work} \times \text{Time Spent} \quad (W = R \times T)$

From this relationship, the formulas for rate and time are:

$\text{Rate} = \frac{\text{Work}}{\text{Time}} \quad (R = \frac{W}{T}) \qquad \text{and} \qquad \text{Time} = \frac{\text{Work}}{\text{Rate}} \quad (T = \frac{W}{R})$

The Standard Fractional Formulation (1 Job)

When a problem defines the work as completing one single job ($W = 1$):

• If Worker A completes a job alone in $t_A$ hours, Worker A's rate is $R_A = \frac{1}{t_A}$ jobs per hour.
• If Worker B completes the same job alone in $t_B$ hours, Worker B's rate is $R_B = \frac{1}{t_B}$ jobs per hour.
• Working together simultaneously, their rates add directly: $R_{\text{combined}} = R_A + R_B = \frac{1}{t_A} + \frac{1}{t_B} = \frac{t_A + t_B}{t_A t_B}$
• The combined time $T$ required to finish the 1 job is the reciprocal of the combined rate: $T_{\text{combined}} = \frac{1}{R_{\text{combined}}} = \frac{t_A t_B}{t_A + t_B}$

The Two-Worker Product-Over-Sum Shortcut: For two workers with completion times $a$ and $b$, their combined time is always:
$T = \frac{a \times b}{a + b}$ Example: If Machine 1 takes 6 hours and Machine 2 takes 12 hours, combined time is $\frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \text{ hours}$.

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The Smart Number Technique: Bypassing Fractions with the LCM

While the product-over-sum shortcut works well for two workers, problems involving three workers, partial shifts, or changing worker pools become mathematically cumbersome when using fractions. The fastest, most robust approach on the GMAT is assigning a concrete Smart Number to the total work.

The 4-Step Smart Number LCM Method

1. Find the LCM: Determine the Least Common Multiple (LCM) of all given completion times. Let this LCM represent the total quantity of work (e.g., "widgets," "pages," "gallons," or "units").
2. Derive Integer Rates: Compute each worker's individual rate in units per hour: $R_i = \frac{\text{Total Units (LCM)}}{t_i}$
3. Combine the Integer Rates: Sum the integer rates according to who is working during each phase of the task.
4. Calculate Time: Divide the target work units by the combined rate: $\text{Time} = \frac{\text{Work Units}}{\text{Combined Rate}}$

Direct Comparison: Fractional Setup vs. Smart Number Method

Problem: Worker A completes a task in 12 hours, Worker B in 15 hours, and Worker C in 20 hours. How long do they take together?

• Fractional Method:
  $R_{\text{total}} = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{5}{60} + \frac{4}{60} + \frac{3}{60} = \frac{12}{60} = \frac{1}{5} \implies T = 5 \text{ hours}$
• Smart Number LCM Method:
  1. $\text{LCM}(12, 15, 20) = 60 \text{ units of work}$.
  2. Rates: $R_A = \frac{60}{12} = 5$, $R_B = \frac{60}{15} = 4$, $R_C = \frac{60}{20} = 3$ units/hr.
  3. Combined Rate: $5 + 4 + 3 = 12$ units/hr.
  4. Time: $\frac{60}{12} = 5 \text{ hours}$.

The LCM method transforms fraction reduction into basic integer arithmetic, drastically reducing scratchpad errors under time constraints.

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Sequential and Staggered Work Tasks

In many GMAT word problems, individuals do not work simultaneously from start to finish. A worker may begin alone, be joined later by a colleague, or depart before the job is finished. To solve these problems without confusion, partition the timeline into distinct, non-overlapping phases.

The Two-Phase Staggered Framework

1. Phase 1 (Solo Work):
   • Identify who is working and for what duration ($T_1$).
   • Calculate the units completed during Phase 1:
     $W_1 = R_1 \times T_1$
2. Determine Remaining Work:
   • Subtract Phase 1 output from total required work:
     $W_{\text{rem}} = W_{\text{total}} - W_1$
3. Phase 2 (Joint / Handoff Work):
   • Identify the new active rate $R_{\text{phase 2}}$ (e.g., $R_1 + R_2$ if Worker 2 joins; $R_2$ alone if Worker 1 leaves).
   • Calculate the time needed to finish the remaining units:
     $T_2 = \frac{W_{\text{rem}}}{R_{\text{phase 2}}}$
4. Answer the Exact Question Asked:
   • If the prompt asks for time required to finish the remaining work, the answer is $T_2$.
   • If the prompt asks for total elapsed time from start to completion, the answer is $T_1 + T_2$.

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Opposing Rates: Inlets, Outlets, and Net Flow Dynamics

Tank and cistern problems are standard work problems where entities work in direct opposition to one another:

• Inlet Pipes (Positive Rates): Add liquid to the tank ($+R_{\text{in}}$).
• Outlet Pipes / Leaks (Negative Rates): Remove liquid from the tank ($-R_{\text{out}}$).

$\text{Net Rate} = R_{\text{net}} = R_{\text{inlet}} - R_{\text{outlet}} = \sum R_{\text{fill}} - \sum R_{\text{drain}}$

The Three Regimes of Net Flow

1. Filling Condition ($R_{\text{inlet}} > R_{\text{outlet}}$): The net rate is positive. The tank will eventually fill from empty in:
   $T_{\text{fill}} = \frac{\text{Capacity}}{R_{\text{net}}}$
2. Emptying Condition ($R_{\text{outlet}} > R_{\text{inlet}}$): The net rate is negative. If the tank has an initial volume of liquid, it will empty in:
   $T_{\text{empty}} = \frac{\text{Initial Volume}}{R_{\text{outlet}} - R_{\text{inlet}}}$
3. Static Condition ($R_{\text{inlet}} = R_{\text{outlet}}$): The liquid level remains permanently unchanged.

Smart Number Tank Application: Always assign the tank capacity as the LCM of all fill times and drain times. If Pipe A fills a tank in 4 hours, Pipe B fills it in 6 hours, and Drain C empties it in 8 hours, let tank capacity be $\text{LCM}(4, 6, 8) = 24$ gallons. Rates: $R_A = +6$, $R_B = +4$, $R_C = -3$ gal/hr. Net Rate $= 6 + 4 - 3 = +7$ gal/hr. Time to fill from empty $= \frac{24}{7}$ hours.

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Worker-Pool Proportionality: The Man-Hours Equation

When a problem features groups of identical laborers, machines, or teams, use the Proportional Scaling Model:

$\text{Total Work Done} = (\text{Number of Workers}) \times (\text{Time}) \times (\text{Individual Worker Rate})$

Assuming all workers produce at the identical constant rate, the ratio between two production scenarios is constant:

$\frac{W_1}{M_1 \times T_1} = \frac{W_2}{M_2 \times T_2} \quad \implies \quad M_1 \times T_1 \times W_2 = M_2 \times T_2 \times W_1$

where $M$ is the number of workers (or machines), $T$ is the time elapsed (days, hours), and $W$ is the total work output (items produced, ditches dug, reports prepared).

Inverse and Direct Variation within Worker Pools

• Workers vs. Time (Inverse Variation): For a constant amount of work, doubling the number of workers halves the required time ($M \times T = \text{constant}$).
• Workers vs. Output (Direct Variation): For a fixed time period, doubling workers doubles output ($\frac{W}{M} = \text{constant}$).

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Worked Problem Solving Examples

Example 1: Three Workers with Distinct Rates (LCM Method)

Problem: Working alone at their constant respective rates, Alice can complete a software audit in 6 hours, Bob can complete it in 8 hours, and Charlie can complete it in 12 hours. If all three work together simultaneously at their respective rates, how many hours and minutes will it take them to complete the audit?

Step-by-Step Solution:

1. Assign Total Work Units Using LCM:
   • Times are 6, 8, and 12 hours.
   • $\text{LCM}(6, 8, 12) = 24 \text{ units of audit work}$.
2. Calculate Individual Hourly Rates:
   • Alice: $R_A = \frac{24}{6} = 4 \text{ units/hour}$.
   • Bob: $R_B = \frac{24}{8} = 3 \text{ units/hour}$.
   • Charlie: $R_C = \frac{24}{12} = 2 \text{ units/hour}$.
3. Calculate Combined Hourly Rate:
   • $R_{\text{combined}} = 4 + 3 + 2 = 9 \text{ units/hour}$.
4. Calculate Time to Complete:
   • $T = \frac{\text{Total Work}}{R_{\text{combined}}} = \frac{24}{9} = \frac{8}{3} \text{ hours} = 2\frac{2}{3} \text{ hours}$.
5. Convert Fractional Hours to Minutes:
   • $\frac{2}{3} \text{ hour} = \frac{2}{3} \times 60 = 40 \text{ minutes}$.
   • Total time: 2 hours and 40 minutes.

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Example 2: Staggered Production Schedule

Problem: Machine P can manufacture a batch of 1,200 electronic components in 3 hours, while Machine Q requires 6 hours to manufacture the identical batch. Machine P is turned on alone at 8:00 AM. At 9:30 AM, Machine Q is also turned on, and both machines run simultaneously until the batch of 1,200 components is completed. At what time is the batch finished?

Step-by-Step Solution:

1. Determine Hourly Production Rates:
   • Rate of P: $R_P = \frac{1200}{3} = 400 \text{ components/hour}$.
   • Rate of Q: $R_Q = \frac{1200}{6} = 200 \text{ components/hour}$.
2. Calculate Production During Solo Phase (8:00 AM – 9:30 AM):
   • Solo duration: $1.5 \text{ hours} = \frac{3}{2} \text{ hours}$.
   • Output of Machine P:
     $W_{\text{solo}} = 400 \times 1.5 = 600 \text{ components}$
3. Calculate Remaining Components to Manufacture:
   • $W_{\text{rem}} = 1200 - 600 = 600 \text{ components}$.
4. Calculate Combined Production Rate from 9:30 AM Onward:
   • $R_{\text{joint}} = R_P + R_Q = 400 + 200 = 600 \text{ components/hour}$.
5. Determine Joint Operating Time:
   • $T_{\text{joint}} = \frac{W_{\text{rem}}}{R_{\text{joint}}} = \frac{600}{600} = 1 \text{ hour}$.
6. Establish the Final Completion Time:
   • Adding 1 hour of joint operation to 9:30 AM yields 10:30 AM.

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Example 3: Opposing Flow Rates in a Cistern with a Drain

Problem: An empty water cistern has two inlet pipes, Pipe X and Pipe Y, and one drain pipe, Pipe Z. Pipe X can fill the cistern alone in 4 hours, and Pipe Y can fill it alone in 6 hours. Drain Pipe Z can empty a completely full cistern in 8 hours. If the cistern is empty and all three pipes are opened simultaneously, how many hours will it take to fill the cistern completely?

Step-by-Step Solution:

1. Assign Smart Number Capacity via LCM:
   • Given times: 4, 6, and 8 hours.
   • $\text{LCM}(4, 6, 8) = 24 \text{ gallons (or volume units)}$.
2. Determine Integer Hourly Flow Rates:
   • Pipe X (fill): $R_X = +\frac{24}{4} = +6 \text{ gal/hr}$.
   • Pipe Y (fill): $R_Y = +\frac{24}{6} = +4 \text{ gal/hr}$.
   • Pipe Z (drain): $R_Z = -\frac{24}{8} = -3 \text{ gal/hr}$.
3. Compute the Net Inflow Rate:
   • $R_{\text{net}} = R_X + R_Y + R_Z = (+6) + (+4) + (-3) = +7 \text{ gal/hr}$.
4. Calculate Time to Fill:
   • $T = \frac{\text{Capacity}}{R_{\text{net}}} = \frac{24}{7} \text{ hours} = 3\frac{3}{7} \text{ hours}$.
5. Convert Mixed Fraction:
   • $3 \text{ hours}$ and $\frac{3}{7} \times 60 \approx 25.7 \text{ minutes}$, so the cistern fills in exactly $3\frac{3}{7}$ hours (or $\frac{24}{7}$ hours).

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High-Frequency GMAT Traps & Pacing Strategies

• Trap 1: Adding Times Instead of Rates: Never add completion times together! If Worker 1 takes 4 hours and Worker 2 takes 6 hours, working together will never take $4 + 6 = 10$ hours, nor will it take the average of 5 hours. Two workers together must always finish in less time than the fastest individual worker alone ($T < 4$ hours).
• Trap 2: Answering "Remaining Time" When Asked for "Total Time": In staggered work problems, questions often ask for the total elapsed project time. Calculating the joint phase time ($T_2$) and selecting it before adding the initial solo phase ($T_1$) is a classic distractor trap.
• Trap 3: Sign Inversion on Drains and Leaks: When setting up net rate equations, ensure that drains, leaks, and waste rates carry negative signs. Adding a leak rate instead of subtracting it leads directly to attractive trap options.
• Trap 4: Assuming Equal Individual Contributions in Teams: If a prompt states that "a crew of 5 workers completed a task in 8 hours," do not assume each worker did $\frac{1}{5}$ unless the prompt confirms they work at identical rates. If individual rates differ, you must solve for each worker's distinct rate.