Overview
Mixed work-rate problems combine two or more of the structures developed throughout this lesson set. A problem may involve several workers, changing start times, leaks or drains, unknown completion times, changing workforce size, or multiple stages of work.
The mathematics is still based on the same Work Principle:
\[ \boxed{W=RT}. \]
The main challenge is no longer simply solving an equation. It is deciding how the situation should be divided into stages, which rates are active during each stage, and whether those rates contribute positively or negatively toward the goal.
The General Work-Rate Model
For several workers or processes, the total work can be written as
\[ W_1+W_2+W_3+\cdots=W_{\text{total}}. \]
Since each amount of work is rate multiplied by time,
\[ R_1T_1+R_2T_2+R_3T_3+\cdots=W_{\text{total}}. \]
For one complete job,
\[ \boxed{R_1T_1+R_2T_2+R_3T_3+\cdots=1}. \]
Rates that oppose the goal can be represented as negative quantities, while changes in workforce or operating conditions can be handled by dividing the problem into separate time intervals.
Model Selection Comes First
In a mixed problem, the first task is to identify which work-rate cases are present.
- Single rate: one process operates alone.
- Combined rates: two or more positive rates operate simultaneously.
- Opposing rates: one or more rates work against the desired result.
- Staggered work: workers or processes begin at different times.
- Interrupted work: a worker or process stops before the job is completed.
- Unknown or related rates: completion times must be represented with variables.
- Workforce productivity: the number of equally productive workers or machines changes.
A mixed problem may contain several of these features at once.
Divide the Problem into Intervals
The most reliable strategy for a complicated work problem is to divide the timeline whenever the active rates change.
Suppose:
- Worker A works alone for \(2\) hours;
- Worker B then joins;
- after \(3\) more hours, Worker A leaves;
- Worker B finishes the job alone.
The problem contains three distinct intervals:
\[ \text{Interval 1: } R_A(2) \]
\[ \text{Interval 2: } (R_A+R_B)(3) \]
\[ \text{Interval 3: } R_Bx. \]
For one complete job,
\[ \boxed{ R_A(2)+(R_A+R_B)(3)+R_Bx=1 }. \]
This interval structure is often more important than the algebra that follows.
Example 1: One Worker Joins and Another Leaves
Worker A can complete a job in \(12\) hours, and Worker B can complete the same job in \(8\) hours. Worker A works alone for \(2\) hours. Worker B then joins, and they work together for \(3\) hours. Worker A then leaves. How much longer will Worker B need to finish the job?
The rates are
\[ R_A=\frac{1}{12} \]
and
\[ R_B=\frac{1}{8}. \]
During the first \(2\) hours, Worker A completes
\[ W_1 = \frac{1}{12}(2) = \frac{1}{6}. \]
During the next \(3\) hours, their combined rate is
\[ \frac{1}{12}+\frac{1}{8} = \frac{2}{24}+\frac{3}{24} = \frac{5}{24}. \]
They complete
\[ W_2 = \frac{5}{24}(3) = \frac{15}{24} = \frac{5}{8}. \]
The total completed work is
\[ \frac{1}{6}+\frac{5}{8}. \]
Using a common denominator of \(24\),
\[ \frac{4}{24}+\frac{15}{24} = \frac{19}{24}. \]
The remaining work is
\[ 1-\frac{19}{24} = \frac{5}{24}. \]
Worker B finishes alone at rate
\[ \frac{1}{8}. \]
Therefore,
\[ T = \frac{\frac{5}{24}}{\frac{1}{8}} = \frac{5}{24}\times 8 = \frac{5}{3}. \]
Thus Worker B needs
\[ \boxed{\frac{5}{3}\text{ hours}} \]
or \(1\) hour and \(40\) minutes to finish the job.
Example 2: Combined Rates with a Leak
Pipe A can fill a tank in \(6\) hours, and Pipe B can fill it in \(8\) hours. A drain can empty the full tank in \(12\) hours. Pipe A operates alone for \(2\) hours. Then Pipe B and the drain are opened while Pipe A remains open. How much longer will the tank take to fill?
Pipe A's rate is
\[ \frac{1}{6}. \]
During the first \(2\) hours, Pipe A fills
\[ \frac{1}{6}(2) = \frac{1}{3} \]
of the tank.
The remaining amount is
\[ 1-\frac{1}{3} = \frac{2}{3}. \]
After Pipe B and the drain are opened, the net rate becomes
\[ R_{\text{net}} = \frac{1}{6} + \frac{1}{8} - \frac{1}{12}. \]
Using a common denominator of \(24\),
\[ R_{\text{net}} = \frac{4}{24} + \frac{3}{24} - \frac{2}{24} = \frac{5}{24}. \]
The time required to fill the remaining \(\frac{2}{3}\) is
\[ T = \frac{\frac{2}{3}}{\frac{5}{24}}. \]
Thus,
\[ T = \frac{2}{3}\times\frac{24}{5} = \frac{16}{5}. \]
Therefore,
\[ \boxed{T=3.2\text{ hours}}. \]
The tank requires an additional \(3\) hours and \(12\) minutes after Pipe B and the drain are opened.
Example 3: Unknown Rate with a Delayed Start
Worker A can complete a job in \(x\) hours. Worker B requires \(4\) hours longer than Worker A. Worker A works alone for \(2\) hours, and then Worker B joins. Together they finish the remaining work in \(3\) hours. Find their individual completion times.
Worker A's completion time is
\[ x, \]
so Worker A's rate is
\[ \frac{1}{x}. \]
Worker B's completion time is
\[ x+4, \]
so Worker B's rate is
\[ \frac{1}{x+4}. \]
Worker A works for a total of
\[ 2+3=5 \]
hours, while Worker B works for \(3\) hours.
The work equation is
\[ \frac{5}{x}+\frac{3}{x+4}=1. \]
Multiply by \(x(x+4)\):
\[ 5(x+4)+3x=x(x+4). \]
Expand:
\[ 5x+20+3x=x^2+4x. \]
Combine like terms:
\[ 8x+20=x^2+4x. \]
Move everything to one side:
\[ x^2-4x-20=0. \]
Factor:
\[ (x-2)(x+10)=0. \]
The algebra gives
\[ x=2 \qquad\text{or}\qquad x=-10. \]
The negative solution is impossible. However, \(x=2\) would mean Worker A completes the entire job in \(2\) hours, yet the problem states that Worker A works alone for \(2\) hours and then continues working for another \(3\) hours. That is inconsistent with the physical situation because the job would already be complete before Worker B joined.
Therefore, the stated conditions produce no physically valid solution.
This example demonstrates an important feature of mixed problems: an algebraic solution must still be checked against the original situation.
Check Whether the Story Is Possible
Mixed work-rate problems require more than symbolic manipulation. The final solution must be consistent with the timeline and physical meaning of the problem.
Useful checks include:
- Did the job already finish before a later stage supposedly began?
- Is a completion time negative or zero?
- Does a denominator become zero?
- Does a leak make filling faster rather than slower?
- Does adding workers increase the completion time under an equal-productivity assumption?
- Does a worker operate longer than the total job would take that worker alone?
A mathematically valid root can still be invalid in context.
Example 4: Changing Workforce Size
Eight workers can complete a project in \(15\) days. They work for \(5\) days, after which \(4\) additional equally productive workers join. How many more days will the enlarged crew need to finish the project?
The total amount of labor required is
\[ 8(15)=120 \]
worker-days.
During the first \(5\) days, the original crew completes
\[ 8(5)=40 \]
worker-days of work.
The remaining work requires
\[ 120-40=80 \]
worker-days.
The new workforce contains
\[ 8+4=12 \]
workers.
Let \(t\) represent the remaining time:
\[ 12t=80. \]
Therefore,
\[ t=\frac{80}{12} = \frac{20}{3}. \]
Thus,
\[ \boxed{\frac{20}{3}\text{ days}} \]
or approximately \(6.67\) additional days are required.
Example 5: A Worker Leaves and a Faster Worker Joins
Worker A can complete a job in \(10\) hours. Worker B can complete it in \(15\) hours. They work together for \(3\) hours, after which Worker B leaves and Worker C joins. Worker C works twice as fast as Worker A. How much longer will Workers A and C need to finish the job?
Worker A's rate is
\[ R_A=\frac{1}{10}. \]
Worker B's rate is
\[ R_B=\frac{1}{15}. \]
During the first \(3\) hours, the combined rate is
\[ \frac{1}{10}+\frac{1}{15} = \frac{3}{30}+\frac{2}{30} = \frac{1}{6}. \]
They complete
\[ W_1 = \frac{1}{6}(3) = \frac{1}{2}. \]
Therefore,
\[ \frac{1}{2} \]
of the job remains.
Worker C works twice as fast as Worker A, so
\[ R_C = 2\left(\frac{1}{10}\right) = \frac{1}{5}. \]
Workers A and C now have combined rate
\[ \frac{1}{10}+\frac{1}{5} = \frac{1}{10}+\frac{2}{10} = \frac{3}{10}. \]
The remaining time is
\[ T = \frac{\frac{1}{2}}{\frac{3}{10}} = \frac{1}{2}\times\frac{10}{3} = \frac{5}{3}. \]
Therefore,
\[ \boxed{\frac{5}{3}\text{ hours}} \]
or \(1\) hour and \(40\) minutes are required after Worker C joins.
One Equation or Several Stages?
Some mixed problems can be represented by one equation, while others are easier to solve stage by stage.
A single-equation approach may look like
\[ R_1T_1+R_2T_2+R_3T_3=1. \]
A stage-based approach instead calculates each contribution separately:
\[ W_1=R_1T_1, \]
\[ W_2=R_2T_2, \]
\[ W_3=R_3T_3, \]
and then uses
\[ W_1+W_2+W_3=1. \]
Neither method is inherently better. The preferred method is the one that makes the changing rates and time intervals easiest to track accurately.
A General Strategy for Mixed Problems
- Define the complete job. Usually set \(W=1\).
- Identify all workers or processes.
- Convert completion times into rates. Use \(R=\frac{1}{t}\).
- Assign positive or negative signs. Determine which rates contribute toward or oppose the goal.
- Build a timeline. Divide the problem whenever someone starts, stops, joins, leaves, or changes productivity.
- Determine the active rate during each interval.
- Calculate the work completed during each interval. Use \(W=RT\).
- Add the work from all intervals.
- Set the total equal to the required amount of work.
- Solve the resulting equation.
- Check all restrictions and physical conditions.
- Interpret the answer in the language of the original problem.
Practice Problems
- Worker A can complete a job in \(8\) hours and Worker B in \(12\) hours. They work together for \(2\) hours, after which Worker A leaves. How much longer will Worker B need to finish the job?
- A pipe fills a tank in \(5\) hours. It operates alone for \(1\) hour, after which a second pipe that fills the tank in \(10\) hours and a drain that empties it in \(20\) hours are opened. How much longer will the tank take to fill?
- Worker A can complete a job in \(10\) hours. Worker B works twice as fast as Worker A. Worker A works alone for \(2\) hours, then Worker B joins. How much additional time is required?
- Twelve workers can complete a project in \(18\) days. They work for \(6\) days, after which \(3\) workers leave. How many additional days are needed?
- Worker A can complete a job in \(x\) hours, while Worker B takes \(2\) hours longer. Worker A works alone for \(1\) hour, and then both workers work together for \(3\) hours to finish the job. Write and solve the work equation.
- A pump fills a tank in \(8\) hours, while a leak empties it in \(24\) hours. The pump operates with the leak for \(3\) hours. The leak is then repaired. How much additional time will the pump need to fill the tank?
- Machine A completes a job in \(12\) hours and Machine B in \(18\) hours. They work together for \(4\) hours. Machine B then stops, and Machine C, whose rate is twice Machine A's rate, begins operating with Machine A. How much additional time is required?
- Six workers can complete a job in \(20\) days. After \(5\) days, the workforce doubles. After another \(4\) days, half of the enlarged workforce leaves. How much additional time is required to finish the job?
Practice Answers
- The initial combined rate is \[ \frac{1}{8}+\frac{1}{12} = \frac{5}{24}. \] In \(2\) hours they complete \[ \frac{5}{24}(2) = \frac{5}{12}. \] The remaining work is \[ \frac{7}{12}. \] Worker B's rate is \[ \frac{1}{12}. \] Therefore, \[ T = \frac{\frac{7}{12}}{\frac{1}{12}} = \boxed{7\text{ hours}}. \]
- The first pipe completes \[ \frac{1}{5}(1) = \frac{1}{5}. \] The remaining work is \[ \frac{4}{5}. \] The new net rate is \[ \frac{1}{5} + \frac{1}{10} - \frac{1}{20}. \] \[ R_{\text{net}} = \frac{4}{20} + \frac{2}{20} - \frac{1}{20} = \frac{1}{4}. \] Therefore, \[ T = \frac{\frac{4}{5}}{\frac{1}{4}} = \boxed{\frac{16}{5}\text{ hours}} \] or \(3.2\) hours.
- Worker A's rate is \[ \frac{1}{10}. \] Worker B works twice as fast, so \[ R_B=\frac{1}{5}. \] Worker A completes \[ \frac{1}{10}(2)=\frac{1}{5} \] before Worker B joins. The remaining work is \[ \frac{4}{5}. \] Their combined rate is \[ \frac{1}{10}+\frac{1}{5} = \frac{3}{10}. \] Thus, \[ T = \frac{\frac{4}{5}}{\frac{3}{10}} = \frac{8}{3}. \] \[ \boxed{\frac{8}{3}\text{ hours}} \] or \(2\) hours and \(40\) minutes.
- Total labor required: \[ 12(18)=216 \] worker-days. Completed during the first \(6\) days: \[ 12(6)=72. \] Remaining labor: \[ 216-72=144. \] The remaining workforce is \[ 12-3=9. \] Therefore, \[ 9t=144. \] \[ \boxed{t=16\text{ days}}. \]
- Worker A's rate is \[ \frac{1}{x}, \] and Worker B's rate is \[ \frac{1}{x+2}. \] Worker A works for a total of \(4\) hours, while Worker B works for \(3\) hours. The equation is \[ \frac{4}{x}+\frac{3}{x+2}=1. \] Multiply by \(x(x+2)\): \[ 4(x+2)+3x=x(x+2). \] \[ 4x+8+3x=x^2+2x. \] \[ 7x+8=x^2+2x. \] \[ x^2-5x-8=0. \] Using the quadratic formula, \[ x = \frac{5\pm\sqrt{25+32}}{2} = \frac{5\pm\sqrt{57}}{2}. \] Reject the negative solution. \[ \boxed{x=\frac{5+\sqrt{57}}{2}\approx 6.27\text{ hours}}. \] Worker B requires approximately \[ \boxed{8.27\text{ hours}}. \]
- The net rate during the first \(3\) hours is \[ \frac{1}{8}-\frac{1}{24} = \frac{2}{24} = \frac{1}{12}. \] The amount filled is \[ \frac{1}{12}(3) = \frac{1}{4}. \] The remaining work is \[ \frac{3}{4}. \] After the leak is repaired, the pump works at rate \[ \frac{1}{8}. \] Thus, \[ T = \frac{\frac{3}{4}}{\frac{1}{8}} = \boxed{6\text{ hours}}. \]
- The initial combined rate is \[ \frac{1}{12}+\frac{1}{18} = \frac{5}{36}. \] In \(4\) hours, \[ W = \frac{5}{36}(4) = \frac{5}{9}. \] The remaining work is \[ \frac{4}{9}. \] Machine C works twice as fast as Machine A: \[ R_C = 2\left(\frac{1}{12}\right) = \frac{1}{6}. \] Machines A and C have combined rate \[ \frac{1}{12}+\frac{1}{6} = \frac{1}{4}. \] Therefore, \[ T = \frac{\frac{4}{9}}{\frac{1}{4}} = \boxed{\frac{16}{9}\text{ hours}}. \]
- The job requires \[ 6(20)=120 \] worker-days. During the first \(5\) days, \[ 6(5)=30 \] worker-days are completed. The workforce then doubles to \(12\). During the next \(4\) days, \[ 12(4)=48 \] worker-days are completed. Total completed labor: \[ 30+48=78. \] Remaining: \[ 120-78=42. \] Half of the \(12\)-worker crew leaves, leaving \[ 6 \] workers. Therefore, \[ 6t=42. \] \[ \boxed{t=7\text{ additional days}}. \]
Final Synthesis
Mixed work-rate problems show that the different cases studied throughout this lesson set are not separate mathematical systems. They are variations of one general framework.
The Work Principle remains
\[ \boxed{W=RT}. \]
For several contributions,
\[ \boxed{ R_1T_1+R_2T_2+R_3T_3+\cdots=W_{\text{total}} }. \]
The problem determines how the rates and times should be interpreted. Rates may be added or subtracted, workers may operate during different intervals, workforce size may change, and unknown completion times may introduce rational equations.
The strongest problem-solving strategy is therefore not to memorize a separate formula for every case. Instead:
\[ \boxed{ \text{Identify the work} \rightarrow \text{identify the rates} \rightarrow \text{identify the times} \rightarrow \text{build the model} \rightarrow \text{solve and check} } \]
Once this structure is understood, even complicated work-rate problems can be reduced to combinations of familiar ideas.
Lesson Complete
You have now worked through the major classes of work-rate problems and applied the Work Principle to single rates, combined rates, opposing rates, staggered work, interrupted work, unknown rates, workforce productivity, and mixed situations.
Return to 7. Workforce and Productivity Problems