Overview
Interrupted work problems involve a job that begins under one set of conditions and is completed under another. A worker may leave before the job is finished, a machine may shut down, a crew may be replaced, or one process may complete part of a task before another takes over.
The central idea is to separate the job into two parts:
\[ \boxed{W_{\text{completed}}+W_{\text{remaining}}=1}. \]
Once the amount already completed is known, the remaining work can be found and assigned to the worker or process that finishes the job.
The Remaining-Work Principle
One complete job is represented by
\[ W=1. \]
If a worker completes some fraction \(W_c\) of the job, then the amount remaining is
\[ W_r=1-W_c. \]
Therefore,
\[ \boxed{W_{\text{remaining}}=1-W_{\text{completed}}}. \]
The remaining work can then be completed by another worker using the ordinary Work Principle:
\[ W=RT. \]
Example 1: One Worker Leaves
Worker A can complete a job in \(8\) hours. Worker A works alone for \(3\) hours and then leaves. Worker B finishes the remaining work in \(4\) hours. How long would Worker B take to complete the entire job alone?
Worker A's rate is
\[ R_A=\frac{1}{8}. \]
In \(3\) hours, Worker A completes
\[ W_A = \frac{1}{8}(3) = \frac{3}{8}. \]
The remaining work is
\[ 1-\frac{3}{8} = \frac{5}{8}. \]
Worker B completes \(\frac{5}{8}\) of the job in \(4\) hours. Therefore Worker B's rate is
\[ R_B = \frac{\frac{5}{8}}{4}. \]
Simplify:
\[ R_B = \frac{5}{8}\times\frac{1}{4} = \frac{5}{32}. \]
Worker B's full-job completion time is
\[ T_B = \frac{1}{\frac{5}{32}} = \frac{32}{5}. \]
Therefore,
\[ \boxed{T_B=6.4\text{ hours}}. \]
Think in Two Stages
Many interrupted-work problems can be understood as a two-stage process.
During Stage 1, the first worker completes
\[ W_1=R_1T_1. \]
The remaining work is
\[ W_r=1-W_1. \]
During Stage 2, the second worker completes that remaining work:
\[ W_r=R_2T_2. \]
Together,
\[ \boxed{R_1T_1+R_2T_2=1}. \]
This equation provides a compact algebraic model for many interrupted-work situations.
Example 2: A Machine Is Replaced
Machine A can complete a production run in \(10\) hours. It operates for \(4\) hours before shutting down. Machine B can complete the same production run in \(8\) hours. How long will Machine B need to finish the remaining work?
Machine A's rate is
\[ \frac{1}{10}. \]
In \(4\) hours, Machine A completes
\[ \frac{1}{10}(4) = \frac{2}{5}. \]
The remaining work is
\[ 1-\frac{2}{5} = \frac{3}{5}. \]
Machine B's rate is
\[ \frac{1}{8}. \]
Use
\[ T=\frac{W}{R}. \]
Then
\[ T = \frac{\frac{3}{5}}{\frac{1}{8}} = \frac{3}{5}\times 8 = \frac{24}{5}. \]
Therefore,
\[ \boxed{T=4.8\text{ hours}}. \]
Machine B needs \(4\) hours and \(48\) minutes to finish the production run.
Finding How Long the First Worker Worked
Sometimes the completion times of both workers are known, but the amount of time the first worker worked before leaving is unknown.
Suppose Worker A has rate \(R_A\), Worker B has rate \(R_B\), Worker A works for \(x\) hours, and Worker B then finishes the job in a known time \(t\).
The equation is
\[ R_Ax+R_Bt=1. \]
The unknown interruption time can then be found algebraically.
Example 3: Find When the First Worker Left
Worker A can complete a job in \(12\) hours. Worker B can complete the same job in \(8\) hours. Worker A begins the job alone and later leaves. Worker B then completes the remaining work in \(4\) hours. How long did Worker A work?
Worker A's rate is
\[ R_A=\frac{1}{12}. \]
Worker B's rate is
\[ R_B=\frac{1}{8}. \]
Worker B works for \(4\) hours, so Worker B completes
\[ W_B = \frac{1}{8}(4) = \frac{1}{2}. \]
Therefore Worker A must have completed the other half:
\[ W_A=\frac{1}{2}. \]
Let \(x\) represent the number of hours Worker A worked.
Then
\[ \frac{1}{12}x=\frac{1}{2}. \]
Multiply by \(12\):
\[ x=6. \]
Therefore,
\[ \boxed{6\text{ hours}}. \]
Writing the Entire Situation as One Equation
Instead of calculating the remaining work separately, an interrupted-work problem can often be represented by a single equation.
Suppose Worker A can complete a job in \(a\) hours and works for \(t_1\) hours. Worker B can complete the job in \(b\) hours and then works for \(t_2\) hours until the job is finished.
Their completed work is
\[ \frac{t_1}{a} + \frac{t_2}{b}. \]
Since the entire job is complete,
\[ \boxed{\frac{t_1}{a}+\frac{t_2}{b}=1}. \]
This is equivalent to finding the completed and remaining work in separate stages.
Example 4: Solve with One Equation
Worker A can complete a job in \(9\) hours. Worker A works for \(3\) hours and then leaves. Worker B can complete the job in \(12\) hours. How long must Worker B work to finish the job?
Let \(x\) represent Worker B's work time.
Write the equation:
\[ \frac{3}{9}+\frac{x}{12}=1. \]
Simplify:
\[ \frac{1}{3}+\frac{x}{12}=1. \]
Subtract \(\frac{1}{3}\):
\[ \frac{x}{12} = \frac{2}{3}. \]
Multiply by \(12\):
\[ x=8. \]
Therefore,
\[ \boxed{8\text{ hours}}. \]
Multiple Interruptions
Some problems involve more than one interruption. A worker may begin alone, another may take over, and a third may finish the remaining work.
The general model becomes
\[ R_1T_1+R_2T_2+R_3T_3=1. \]
For example, if three workers operate sequentially rather than simultaneously, their contributions still add because each completes part of the same job:
\[ \boxed{W_1+W_2+W_3=1}. \]
The important difference from a combined-rate problem is that the workers do not operate at the same time.
Example 5: Three Workers Work Sequentially
Worker A can complete a job in \(12\) hours, Worker B in \(8\) hours, and Worker C in \(6\) hours. Worker A works for \(2\) hours, Worker B then works for \(3\) hours, and Worker C finishes the job. How long must Worker C work?
Worker A completes
\[ W_A = \frac{1}{12}(2) = \frac{1}{6}. \]
Worker B completes
\[ W_B = \frac{1}{8}(3) = \frac{3}{8}. \]
The total completed work is
\[ \frac{1}{6}+\frac{3}{8}. \]
Using a common denominator of \(24\),
\[ \frac{4}{24}+\frac{9}{24} = \frac{13}{24}. \]
The remaining work is
\[ 1-\frac{13}{24} = \frac{11}{24}. \]
Worker C's rate is
\[ \frac{1}{6}. \]
Therefore,
\[ T_C = \frac{\frac{11}{24}}{\frac{1}{6}}. \]
Simplify:
\[ T_C = \frac{11}{24}\times 6 = \frac{11}{4}. \]
Thus,
\[ \boxed{T_C=\frac{11}{4}\text{ hours}} \]
or \(2\) hours and \(45\) minutes.
When the Completed Fraction Is Given Directly
Sometimes a problem directly states how much of the job has already been completed.
If \(65\%\) of the job is complete, then
\[ W_{\text{completed}}=0.65. \]
The remaining work is
\[ W_{\text{remaining}} = 1-0.65 = 0.35. \]
Equivalently,
\[ \boxed{35\%} \]
of the job remains.
The second worker's time can then be found from
\[ T=\frac{W_{\text{remaining}}}{R}. \]
Example 6: Finish a Partially Completed Job
A project is already \(60\%\) complete. A worker who can complete the entire project alone in \(10\) hours is assigned to finish it. How long will the remaining work take?
The remaining work is
\[ 1-0.60=0.40. \]
The worker's rate is
\[ R=\frac{1}{10}. \]
Use
\[ T=\frac{W}{R}. \]
Then
\[ T = \frac{0.40}{\frac{1}{10}} = 4. \]
Therefore,
\[ \boxed{4\text{ hours}}. \]
Interrupted Work vs. Staggered Work
Interrupted and staggered work problems are closely related, but they emphasize different changes in the timeline.
In a staggered work problem, a new worker or process usually joins after another has already started.
In an interrupted work problem, a worker or process stops before completion, often leaving another worker to finish the remainder.
Both can be modeled using
\[ R_1T_1+R_2T_2+\cdots=1, \]
but the story determines which rates are active during each stage.
Method for Solving Interrupted and Remaining-Work Problems
- Represent the complete job by \(1\).
- Identify the first worker's rate. If the worker completes the job in \(t\) hours, use \(R=\frac{1}{t}\).
- Find the work completed before the interruption. Use \(W=RT\).
- Find the remaining work. Use \(W_{\text{remaining}}=1-W_{\text{completed}}\).
- Identify the next worker's or process's rate.
- Use the Work Principle on the remaining work.
- Repeat if additional interruptions occur.
- Check that all completed portions add to \(1\).
Common Errors
- Using the full job instead of the remaining job. The second worker often completes only what remains, not an entire job.
- Forgetting work completed before the interruption. Earlier progress must be included in the total.
- Using the wrong work time for a worker. Each worker may operate for a different amount of time.
- Combining rates when workers are not working simultaneously. Sequential work should be represented as separate work contributions.
- Subtracting time instead of work. Remaining work is found by subtracting completed work from \(1\), not by subtracting completion times.
Practice Problems
- Worker A can complete a job in \(10\) hours. Worker A works for \(4\) hours and then leaves. What fraction of the job remains?
- Worker A can complete a job in \(8\) hours. After working for \(2\) hours, Worker A leaves. Worker B can complete the job alone in \(12\) hours. How long will Worker B need to finish the job?
- A machine can complete a production run in \(15\) hours. It operates for \(6\) hours before failing. What fraction of the production run remains?
- Worker A can complete a job in \(12\) hours. Worker B can complete it in \(9\) hours. Worker A begins alone and later leaves. Worker B then completes the job in \(6\) hours. How long did Worker A work?
- Machine A can complete a job in \(10\) hours. It works for \(3\) hours. Machine B then takes over and finishes the job in \(5\) hours. What is Machine B's full-job completion time?
- Worker A can complete a job in \(16\) hours, Worker B in \(12\) hours, and Worker C in \(8\) hours. Worker A works for \(4\) hours, Worker B then works for \(3\) hours, and Worker C finishes the job. How long does Worker C work?
- A project is \(70\%\) complete. A machine that can complete the entire project in \(6\) hours is used to finish it. How long will the remaining work take?
- Worker A can complete a job in \(9\) hours and Worker B in \(6\) hours. Worker A works for \(3\) hours and then Worker B finishes the job alone. How long does Worker B work?
Practice Answers
- Worker A completes \[ \frac{1}{10}(4) = \frac{2}{5}. \] Therefore, \[ 1-\frac{2}{5} = \boxed{\frac{3}{5}} \] remains.
- Worker A completes \[ \frac{1}{8}(2) = \frac{1}{4}. \] The remaining work is \[ \frac{3}{4}. \] Worker B's rate is \[ \frac{1}{12}. \] Therefore, \[ T = \frac{\frac{3}{4}}{\frac{1}{12}} = \boxed{9\text{ hours}}. \]
- The machine completes \[ \frac{1}{15}(6) = \frac{2}{5}. \] Therefore, \[ \boxed{\frac{3}{5}} \] of the production run remains.
- Worker B completes \[ \frac{1}{9}(6) = \frac{2}{3}. \] Therefore Worker A completed \[ 1-\frac{2}{3} = \frac{1}{3}. \] Worker A's rate is \[ \frac{1}{12}. \] Thus, \[ T_A = \frac{\frac{1}{3}}{\frac{1}{12}} = \boxed{4\text{ hours}}. \]
- Machine A completes \[ \frac{1}{10}(3) = \frac{3}{10}. \] The remaining work is \[ \frac{7}{10}. \] Machine B completes \(\frac{7}{10}\) in \(5\) hours, so \[ R_B = \frac{\frac{7}{10}}{5} = \frac{7}{50}. \] Therefore, \[ T_B = \frac{1}{\frac{7}{50}} = \boxed{\frac{50}{7}\text{ hours}} \] or approximately \(7.14\) hours.
- Worker A completes \[ \frac{1}{16}(4) = \frac{1}{4}. \] Worker B completes \[ \frac{1}{12}(3) = \frac{1}{4}. \] Together they complete \[ \frac{1}{2}. \] Therefore, \[ \frac{1}{2} \] remains. Worker C's rate is \[ \frac{1}{8}. \] Thus, \[ T_C = \frac{\frac{1}{2}}{\frac{1}{8}} = \boxed{4\text{ hours}}. \]
- The remaining work is \[ 1-0.70=0.30. \] The machine's rate is \[ \frac{1}{6}. \] Therefore, \[ T = \frac{0.30}{\frac{1}{6}} = 1.8. \] \[ \boxed{1.8\text{ hours}} \] or \(1\) hour and \(48\) minutes.
- Worker A completes \[ \frac{1}{9}(3) = \frac{1}{3}. \] The remaining work is \[ \frac{2}{3}. \] Worker B's rate is \[ \frac{1}{6}. \] Therefore, \[ T = \frac{\frac{2}{3}}{\frac{1}{6}} = \boxed{4\text{ hours}}. \]
Summary
Interrupted and remaining-work problems divide one complete job into portions performed during different stages.
The central relationship is
\[ \boxed{W_{\text{completed}}+W_{\text{remaining}}=1}. \]
If one worker completes part of the job,
\[ W_{\text{completed}}=RT, \]
then the remaining work is
\[ \boxed{W_{\text{remaining}}=1-RT}. \]
For two sequential workers, the entire situation can also be written as
\[ \boxed{R_1T_1+R_2T_2=1}. \]
The essential idea is to track how much work has been completed at each stage rather than treating each worker as though they complete an entire job.
The next problem type introduces unknown and algebraically related completion times, producing rational equations that must be solved for the individual rates.
Continue
Continue to 6. Unknown and Related-Rate Problems
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