The Central Idea
The basic work relationship remains
\[ \boxed{W=rt}. \]When rates or active contributors change over time, divide the problem into intervals:
\[ \boxed{ W_{\text{total}} = W_1+W_2+\cdots+W_n }. \]Since each interval satisfies \(W_i=r_it_i\),
\[ \boxed{ W_{\text{total}} = r_1t_1+r_2t_2+\cdots+r_nt_n }. \]For one complete job,
\[ \boxed{W_{\text{total}}=1}. \]Why Piecewise Work Models Are Necessary
Suppose worker A works alone for \(2\) hours and then worker B joins.
During the first interval, only A contributes.
During the second interval, both A and B contribute.
Because the active rate changes, one single constant-rate equation does not describe the entire job.
Instead, write one work expression for each interval.
Staggered Starts
A staggered start occurs when contributors begin working at different times.
If worker A begins \(h\) hours before worker B, and B works for \(t\) hours, then A works for
\[ \boxed{t+h} \]hours in total.
If their rates are \(r_A\) and \(r_B\), then the completed work is
\[ r_A(t+h)+r_Bt. \]Example: One Worker Starts Early
Worker A completes a job in \(8\) hours. Worker B completes it in \(12\) hours. A works alone for \(2\) hours, then B joins. How long after B joins does the job finish?
Let \(t\) be the number of hours both work together.
A's total work is
\[ \frac{t+2}{8}. \]B's work is
\[ \frac{t}{12}. \]Therefore,
\[ \frac{t+2}{8}+\frac{t}{12}=1. \]Multiply by \(24\):
\[ 3(t+2)+2t=24. \] \[ 3t+6+2t=24. \] \[ 5t=18. \] \[ t=3.6. \]Therefore,
\[ \boxed{3.6\text{ hours after B joins}}. \]The Same Problem by Intervals
During A's first \(2\) hours alone, A completes
\[ \frac28=\frac14. \]The remaining work is
\[ 1-\frac14=\frac34. \]Once B joins, their combined rate is
\[ \frac18+\frac1{12} = \frac5{24}. \]The remaining completion time is
\[ t= \frac{3/4}{5/24} = \frac{18}{5} = 3.6. \]The interval method and the single-equation method agree.
Delayed Starts as Head Starts
A delayed-start work problem is structurally similar to a pursuit problem.
The early worker builds a work “lead” before the second contributor begins.
If worker A works alone for \(h\) hours at rate \(r_A\), the work already completed is
\[ \boxed{r_Ah}. \]The remaining work is
\[ \boxed{1-r_Ah}. \]Early Departures
Sometimes two workers begin together, but one leaves before the job is finished.
If A and B work together for \(h\) hours and then A continues alone for \(t\) additional hours, total work is
\[ \boxed{ (r_A+r_B)h+r_At=1 }. \]Example: One Worker Leaves Early
Worker A completes a job in \(10\) hours, and worker B in \(15\) hours. They work together for \(3\) hours, then B leaves. How much longer does A need?
Their combined rate is
\[ \frac1{10}+\frac1{15} = \frac16. \]Work completed in \(3\) hours is
\[ \frac16(3)=\frac12. \]Therefore,
\[ \frac12 \]of the job remains.
A's rate is
\[ \frac1{10}. \]So the remaining time is
\[ t= \frac{1/2}{1/10}=5. \]Therefore,
\[ \boxed{5\text{ additional hours}}. \]Interrupted Work
An interruption is an interval during which a contributor temporarily stops working.
A stopped worker contributes
\[ \boxed{0} \]work during the interruption.
The job is therefore modeled with separate active and inactive intervals.
Example: Machine Stops Temporarily
A machine completes one job in \(8\) hours. It works for \(3\) hours, stops for \(1\) hour, and then resumes at the same rate.
Before the stop, it completes
\[ \frac38 \]of the job.
During the stop, it completes
\[ 0. \]The remaining work is
\[ 1-\frac38=\frac58. \]At rate
\[ \frac18, \]the remaining work takes
\[ \frac{5/8}{1/8}=5 \]hours.
Total elapsed time is
\[ 3+1+5=9. \]Therefore,
\[ \boxed{9\text{ hours elapsed}}. \]Working Time versus Elapsed Time
A work interruption creates a distinction between active working time and elapsed clock time.
In the previous example, the machine works for
\[ 3+5=8 \]hours, but the total elapsed time is
\[ 9 \]hours because of the one-hour interruption.
Read carefully which quantity the problem asks for.
Interrupted Combined Work
Suppose two workers operate together, then one pauses while the other continues.
If both work for \(t_1\) hours and only A works for \(t_2\) hours, then
\[ \boxed{ (r_A+r_B)t_1+r_At_2=1 }. \]Example: One Worker Pauses
Worker A completes a job in \(6\) hours and worker B in \(9\) hours. They work together for \(2\) hours. Then B stops, and A finishes the job. How long does A work alone?
Combined rate:
\[ \frac16+\frac19 = \frac5{18}. \]Work completed together:
\[ \frac5{18}(2) = \frac59. \]Remaining work:
\[ 1-\frac59=\frac49. \]A's rate is
\[ \frac16. \]Therefore,
\[ t= \frac{4/9}{1/6} = \frac{8}{3}. \]Thus,
\[ \boxed{2\frac23\text{ hours}}. \]Rate Changes during the Job
A contributor may remain active but change productivity.
If a worker operates at rate \(r_1\) for time \(t_1\) and then at rate \(r_2\) for time \(t_2\), then
\[ \boxed{ W=r_1t_1+r_2t_2 }. \]The same piecewise structure handles speedups, slowdowns, efficiency changes, and reduced-capacity operation.
Example: Productivity Changes Midway
A machine works at
\[ \frac18 \]job per hour for \(4\) hours, then improves to
\[ \frac16 \]job per hour.
Work completed in the first interval:
\[ \frac18(4)=\frac12. \]Remaining work:
\[ \frac12. \]At the new rate, the remaining time is
\[ \frac{1/2}{1/6}=3. \]Therefore, total working time is
\[ \boxed{7\text{ hours}}. \]Unknown Individual Work Rate
Suppose worker A has known rate \(r_A\), worker B has unknown rate \(r\), and both work for the same time \(t\).
For one job,
\[ \boxed{ r_At+rt=1 }. \]Solving for \(r\),
\[ \boxed{ r=\frac{1-r_At}{t} }. \]Example: Find an Unknown Rate
Worker A completes a job in \(8\) hours. A and B together complete the job in \(5\) hours. Find B's rate.
A's rate is
\[ \frac18. \]The combined rate is
\[ \frac15. \]Therefore,
\[ r_B= \frac15-\frac18 = \frac8{40}-\frac5{40} = \frac3{40}. \]Thus,
\[ \boxed{ r_B=\frac3{40}\text{ job per hour} }. \]Convert an Unknown Rate to Completion Time
Continue with
\[ r_B=\frac3{40}. \]B's completion time is
\[ T_B= \frac{1}{3/40} = \frac{40}{3}. \]Therefore,
\[ \boxed{ T_B=13\frac13\text{ hours} }. \]Unknown Completion Time inside a Rate Equation
If worker B takes \(x\) hours alone, then B's rate is
\[ \frac1x. \]This reciprocal expression can be inserted directly into a combined work equation.
For example,
\[ \frac18+\frac1x=\frac15. \]Solving this rational equation determines B's completion time.
Unknown Start Delay
Sometimes the work rates are known but the delay before another contributor joins is unknown.
If A starts first and works alone for \(h\) hours, then both work together for \(t\) hours, the equation is
\[ \boxed{ r_Ah+(r_A+r_B)t=1 }. \]If all other quantities are known, solve for \(h\).
Example: Find the Delay
Worker A completes a job in \(10\) hours and worker B in \(15\) hours. A begins first. After B joins, they work together for \(4\) hours and finish the job. How long did A work alone?
Let \(h\) be A's solo time.
A's solo work is
\[ \frac{h}{10}. \]Their combined rate is
\[ \frac1{10}+\frac1{15} = \frac16. \]Combined work for \(4\) hours is
\[ \frac16(4)=\frac23. \]Therefore,
\[ \frac{h}{10}+\frac23=1. \] \[ \frac{h}{10}=\frac13. \] \[ h=\frac{10}{3}. \]Thus,
\[ \boxed{3\frac13\text{ hours}}. \]Unknown Early Departure Time
Suppose A and B begin together, B leaves after \(h\) hours, and A finishes alone in \(t\) additional hours.
The work equation is
\[ \boxed{ (r_A+r_B)h+r_At=1 }. \]If \(t\) and the rates are known, solve for the departure time \(h\).
Example: Find When a Worker Left
Worker A can complete a job in \(12\) hours, and B in \(8\) hours. They begin together. B leaves, and A then works alone for \(4\) more hours. The job is completed. How long did they work together?
Let \(h\) be their shared working time.
\[ \left(\frac1{12}+\frac18\right)h+\frac1{12}(4)=1. \]The combined rate is
\[ \frac2{24}+\frac3{24} = \frac5{24}. \]Also,
\[ \frac4{12}=\frac13. \]Therefore,
\[ \frac5{24}h+\frac13=1. \] \[ \frac5{24}h=\frac23. \] \[ h=\frac{16}{5}=3.2. \]Therefore,
\[ \boxed{3.2\text{ hours together}}. \]Unknown Rate after an Interruption
A machine may resume at a different unknown rate after being interrupted.
If it completes \(W_1\) before the interruption and has \(1-W_1\) remaining, then
\[ \boxed{ 1-W_1=rt }. \]Therefore,
\[ \boxed{ r=\frac{1-W_1}{t} }. \]Example: Find the Post-Interruption Rate
A machine completes
\[ \frac25 \]of a job before stopping. After restarting, it finishes the remaining work in \(3\) hours. Find its new rate.
Remaining work:
\[ 1-\frac25=\frac35. \]Therefore,
\[ r= \frac{3/5}{3} = \frac15. \]Thus,
\[ \boxed{ r=\frac15\text{ job per hour} }. \]Unknown Number of Identical Workers
If \(n\) identical workers each work at rate \(r\), their combined rate is
\[ \boxed{nr}. \]For one complete job,
\[ nr t=1. \]Therefore,
\[ \boxed{ n=\frac{1}{rt} }. \]Example: How Many Workers Are Needed?
One worker completes a job in \(24\) hours. How many identical workers are needed to complete the job in \(4\) hours, assuming perfectly additive rates?
One worker's rate is
\[ \frac1{24}. \]Let \(n\) be the number of workers:
\[ n\left(\frac1{24}\right)(4)=1. \] \[ \frac{n}{6}=1. \] \[ n=6. \]Therefore,
\[ \boxed{6\text{ workers}}. \]Unknown Time with Several Intervals
If the job has several known intervals plus one unknown final interval, compute the known work first.
If known work totals \(W_k\), then the remaining work is
\[ 1-W_k. \]If the final active rate is \(r_f\), then
\[ \boxed{ t_f=\frac{1-W_k}{r_f} }. \]Example: Three-Stage Work Problem
Worker A takes \(12\) hours alone. Worker B takes \(18\) hours alone. A works alone for \(2\) hours, then A and B work together for \(3\) hours, and finally B works alone until completion. How long does B work alone at the end?
Stage 1:
\[ W_1= \frac{2}{12} = \frac16. \]Stage 2:
\[ W_2= 3\left(\frac1{12}+\frac1{18}\right). \]Since
\[ \frac1{12}+\frac1{18} = \frac3{36}+\frac2{36} = \frac5{36}, \]we get
\[ W_2= \frac{15}{36} = \frac5{12}. \]Total known work:
\[ \frac16+\frac5{12} = \frac2{12}+\frac5{12} = \frac7{12}. \]Remaining work:
\[ \frac5{12}. \]B's rate is
\[ \frac1{18}. \]Therefore,
\[ t= \frac{5/12}{1/18} = \frac{15}{2} = 7.5. \]Thus,
\[ \boxed{7.5\text{ hours}}. \]Piecewise Work Tables
For complicated timing, an interval table is often more useful than a worker table.
\[ \begin{array}{c|c|c|c} \text{interval} & \text{active rate} & \text{time} & \text{work}\\ \hline 1 & r_1 & t_1 & r_1t_1\\ 2 & r_2 & t_2 & r_2t_2\\ 3 & r_3 & t_3 & r_3t_3 \end{array} \]The final equation is
\[ r_1t_1+r_2t_2+r_3t_3=1. \]Timeline Diagrams
A simple timeline can prevent errors in staggered problems.
Mark:
- when each contributor starts;
- when anyone stops;
- when rates change;
- the completion point.
Each span between changes becomes one interval in the work equation.
Piecewise Rates and Algebraic Modeling
The general structure is
\[ \boxed{ \sum_{i=1}^{n} r_i t_i=1 }. \]Some \(r_i\) may be combined rates, some may be single-worker rates, and some may be zero during interruptions.
The notation is compact, but the interpretation of each interval must remain clear.
Unknown Rates Can Produce Rational Equations
If an unknown completion time \(x\) appears as a rate
\[ \frac1x, \]the work equation becomes rational.
For example,
\[ \frac{t}{8}+\frac{t}{x}=1. \]Use the rational-equation methods developed earlier in Algebraic Relationships to solve for \(x\).
Example: Unknown Time in a Staggered Problem
Worker A takes \(6\) hours alone. Worker B takes \(x\) hours alone. A works alone for \(1\) hour, then both work together for \(2\) hours and finish the job. Find \(x\).
A's total time is \(3\) hours, so A completes
\[ \frac36=\frac12. \]B works \(2\) hours, so B completes
\[ \frac2x. \]Therefore,
\[ \frac12+\frac2x=1. \] \[ \frac2x=\frac12. \] \[ x=4. \]Thus,
\[ \boxed{4\text{ hours}}. \]Unknown Rate with an Early Departure
Worker A takes \(10\) hours alone. Worker B has unknown rate \(r\). They work together for \(2\) hours, then B leaves. A works alone for \(5\) more hours and the job finishes.
A works a total of \(7\) hours:
\[ \frac7{10}. \]B's contribution is
\[ 2r. \]Therefore,
\[ \frac7{10}+2r=1. \] \[ 2r=\frac3{10}. \] \[ r=\frac3{20}. \]Therefore,
\[ \boxed{ r=\frac3{20}\text{ job per hour} }. \]Find the Unknown Completion Time from That Rate
If
\[ r=\frac3{20}, \]then the corresponding completion time is
\[ T= \frac1{3/20} = \frac{20}{3}. \]Therefore,
\[ \boxed{6\frac23\text{ hours}}. \]Opposing Rates with Staggered Timing
A productive and opposing process may not begin simultaneously.
Suppose a fill pipe works alone for \(h\) hours, then a drain opens and both operate for \(t\) hours.
The work equation is
\[ \boxed{ r_fh+(r_f-r_d)t=1 }. \]Example: Drain Opens Later
A pipe fills a tank in \(6\) hours. After \(2\) hours, a drain that empties the full tank in \(12\) hours is opened. How much longer does filling take?
Initial work:
\[ \frac26=\frac13. \]Remaining work:
\[ \frac23. \]Net rate after the drain opens:
\[ \frac16-\frac1{12} = \frac1{12}. \]Therefore,
\[ t= \frac{2/3}{1/12} = 8. \]Thus,
\[ \boxed{8\text{ additional hours}}. \]Opposing Process Stops Early
If a drain or other opposing process operates only during part of the job, separate the interval in which it is active from the interval in which it is not.
For example,
\[ (r_f-r_d)t_1+r_ft_2=1. \]The signs and active contributors may change from interval to interval.
Example: Drain Closes before Completion
A fill pipe fills a tank in \(8\) hours, and a drain empties it in \(24\) hours. Both operate for \(3\) hours, then the drain is closed. How long does the fill pipe need afterward?
Net rate during the first interval:
\[ \frac18-\frac1{24} = \frac2{24} = \frac1{12}. \]Work completed in \(3\) hours:
\[ \frac1{12}(3)=\frac14. \]Remaining work:
\[ \frac34. \]Fill rate after the drain closes:
\[ \frac18. \]Therefore,
\[ t= \frac{3/4}{1/8}=6. \]Thus,
\[ \boxed{6\text{ additional hours}}. \]Negative Net Work Can Occur Temporarily
During an interval, opposing work may exceed productive work.
Then
\[ r_{\text{net}}<0. \]The completed amount decreases during that interval.
A later interval with positive net rate may still finish the job, provided the model permits the work quantity to remain meaningful.
Checking Staggered Work Problems
Reconstruct every contributor's total active time.
Then compute each total contribution:
\[ W_A=r_At_A, \qquad W_B=r_Bt_B. \]For a completed job,
\[ \boxed{ W_A+W_B+\cdots=1 }. \]Checking Interrupted Problems
Verify that stopped intervals contribute zero work.
Also distinguish total active time from elapsed time.
If a machine works \(7\) hours and is stopped for \(2\) hours, elapsed time is
\[ 9\text{ hours}, \]but its work contribution is based only on the \(7\) active hours.
Checking Unknown Rates
After solving for an unknown rate, convert it back to a completion time if appropriate and ask whether the result is plausible.
A worker with a larger rate should require a shorter time.
Substitute the recovered rate into the original work equation to verify that the total work equals \(1\).
Common Error: Giving Everyone the Same Working Time
In staggered problems, contributors may have different active durations.
If A starts \(2\) hours early and B later works for \(t\) hours, then A's time is
\[ \boxed{t+2}, \]not \(t\).
Common Error: Counting an Interruption as Productive Time
If a machine is stopped, its rate during that interval is
\[ 0. \]Including the stopped time in
\[ W=rt \]as though the machine were active overstates the completed work.
Common Error: Forgetting Work Already Completed
When a second worker joins late, the first worker has already completed part of the job.
The later combined interval applies only to the remaining work.
Do not restart the job mathematically from zero.
Common Error: Forgetting Remaining Work
If
\[ W_{\text{done}}=\frac58, \]then the final interval must complete
\[ 1-\frac58=\frac38, \]not a whole job.
Common Error: Mixing Total Time and Segment Time
If the first interval lasts \(2\) hours and the second lasts \(t\) hours, the contributor active throughout works for
\[ t+2. \]A contributor active only during the second interval works for
\[ t. \]Label each duration explicitly.
Common Error: Treating a Stop as Negative Work
An interruption usually contributes
\[ 0 \]work, not negative work.
Negative work occurs only when an opposing process actively reverses or removes progress.
Common Error: Accepting a Negative Completion Time
A negative algebraic time does not represent a valid future work interval.
If the model produces
\[ t<0, \]either the wrong algebraic branch was selected or the stated conditions are incompatible with the requested completion.
Common Error: Solving for a Rate but Reporting It as a Time
A result such as
\[ r=\frac3{20} \]means
\[ \frac3{20}\text{ job per hour}. \]If the question asks for completion time, invert:
\[ T=\frac{20}{3}. \]A Reliable Procedure for Staggered Starts
- Identify when each contributor begins.
- Convert each completion time to a work rate.
- Determine each contributor's total active time.
- Write one work expression for each contributor or interval.
- Add the work expressions and set the total equal to \(1\).
- Solve for the unknown delay or shared working time.
- Reconstruct the total contribution of each worker.
- Verify that all contributions sum to one complete job.
A Reliable Procedure for Interrupted Work
- Mark every point where a worker or machine starts, stops, or changes rate.
- Divide the timeline into intervals.
- Determine the active net rate during each interval.
- Assign zero rate to inactive contributors.
- Compute \(W_i=r_it_i\) for each interval.
- Add the interval work.
- Distinguish working time from total elapsed time.
- Verify the total completed work.
A Reliable Procedure for Unknown Work Rates
- Represent an unknown completion time \(x\) by rate \(1/x\).
- Or represent an unknown rate directly by \(r\).
- Write all known contributions using \(W=rt\).
- Add productive work and subtract opposing work as required.
- Set the final total equal to the required work amount.
- Solve the resulting linear or rational equation.
- Convert between rate and completion time when needed.
- Substitute back to verify the complete job.
Practice 1: Delayed Start
Worker A completes a job in \(6\) hours and worker B in \(12\) hours. A works alone for \(1\) hour, then B joins. How long after B joins does the job finish?
Practice 2: Longer Head Start
Worker A takes \(10\) hours and worker B takes \(15\) hours. A works alone for \(3\) hours before B joins. Find the additional time needed.
Practice 3: Early Departure
Worker A takes \(8\) hours and worker B takes \(12\) hours. They work together for \(2\) hours, then B leaves. How much longer does A need?
Practice 4: Temporary Interruption
A machine completes a job in \(10\) hours. It works for \(4\) hours, stops for \(2\) hours, then resumes. Find the total elapsed time to completion.
Practice 5: One Worker Pauses
Worker A takes \(6\) hours and worker B takes \(8\) hours. They work together for \(2\) hours, then B stops and A finishes alone. Find A's additional time.
Practice 6: Rate Change
A machine works at
\[ \frac1{10} \]job per hour for \(3\) hours, then at
\[ \frac15 \]job per hour. How much longer does it need after the rate change?
Practice 7: Unknown Worker Rate
Worker A takes \(10\) hours alone. A and B together complete the job in \(6\) hours. Find B's rate and B's completion time.
Practice 8: Unknown Delay
Worker A takes \(8\) hours and worker B takes \(12\) hours. A starts first. After B joins, they work together for \(3\) hours and finish. How long did A work alone?
Practice 9: Unknown Departure Time
Worker A takes \(10\) hours and worker B takes \(5\) hours. They begin together. B leaves, and A then works alone for \(4\) additional hours. How long did they work together?
Practice 10: Three-Stage Job
Worker A takes \(12\) hours and worker B takes \(24\) hours. A works alone for \(2\) hours, then both work together for \(4\) hours. B then finishes alone. How long does B work alone at the end?
Practice 11: Drain Opens Later
A pipe fills a tank in \(8\) hours. After \(2\) hours, a drain that empties the tank in \(16\) hours is opened. How much longer does filling take?
Practice 12: Drain Closes Early
A fill pipe fills a tank in \(6\) hours and a drain empties it in \(18\) hours. Both operate for \(2\) hours, then the drain closes. How much longer does filling take?
Practice 13: Unknown Completion Time
Worker A takes \(8\) hours. Worker B takes \(x\) hours. A works alone for \(2\) hours, then both work together for \(3\) hours and finish the job. Find \(x\).
Practice 14: Number of Identical Workers
One worker completes a job in \(30\) hours. How many identical workers are needed to complete the job in \(5\) hours, assuming perfectly additive rates?
Practice 15: Identify the Error
Worker A begins \(2\) hours before worker B. After B starts, they work together for \(4\) hours. A student assigns both workers \(4\) hours of work time.
Explain the error and state the correct work times for A and B.
Practice Answers
Practice 1
A completes
\[ \frac16 \]during the first hour.
Remaining work:
\[ \frac56. \]Combined rate:
\[ \frac16+\frac1{12} = \frac14. \]Therefore,
\[ t= \frac{5/6}{1/4} = \frac{10}{3}. \]Thus,
\[ \boxed{3\frac13\text{ hours}}. \]Practice 2
A completes
\[ \frac3{10} \]before B joins.
Remaining work:
\[ \frac7{10}. \]Combined rate:
\[ \frac1{10}+\frac1{15} = \frac16. \]Therefore,
\[ t= \frac{7/10}{1/6} = \frac{21}{5} = 4.2. \]Thus,
\[ \boxed{4.2\text{ hours}}. \]Practice 3
Combined rate:
\[ \frac18+\frac1{12} = \frac5{24}. \]Work completed in \(2\) hours:
\[ \frac5{24}(2)=\frac5{12}. \]Remaining work:
\[ \frac7{12}. \]A alone works at
\[ \frac18. \]Therefore,
\[ t= \frac{7/12}{1/8} = \frac{14}{3}. \]Thus,
\[ \boxed{4\frac23\text{ hours}}. \]Practice 4
The machine requires \(10\) active hours total.
A \(2\)-hour interruption adds directly to elapsed time:
\[ 10+2=12. \]Therefore,
\[ \boxed{12\text{ hours elapsed}}. \]Practice 5
Combined rate:
\[ \frac16+\frac18 = \frac7{24}. \]Work completed in \(2\) hours:
\[ \frac7{24}(2)=\frac7{12}. \]Remaining:
\[ \frac5{12}. \]A's rate is
\[ \frac16. \]Therefore,
\[ t= \frac{5/12}{1/6} = \frac52. \]Thus,
\[ \boxed{2.5\text{ hours}}. \]Practice 6
First-stage work:
\[ \frac1{10}(3)=\frac3{10}. \]Remaining work:
\[ \frac7{10}. \]New rate:
\[ \frac15. \]Therefore,
\[ t= \frac{7/10}{1/5} = \frac72. \]Thus,
\[ \boxed{3.5\text{ hours}}. \]Practice 7
\[ r_B= \frac16-\frac1{10} = \frac5{30}-\frac3{30} = \frac1{15}. \]Therefore,
\[ \boxed{ r_B=\frac1{15}\text{ job per hour} } \]and
\[ \boxed{T_B=15\text{ hours}}. \]Practice 8
Combined rate:
\[ \frac18+\frac1{12} = \frac5{24}. \]Work completed together in \(3\) hours:
\[ \frac5{24}(3)=\frac58. \]Therefore, A must complete
\[ 1-\frac58=\frac38 \]alone before B joins.
Since A's rate is
\[ \frac18, \]the solo time is
\[ \frac{3/8}{1/8}=3. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 9
Let \(h\) be their shared time.
\[ \left(\frac1{10}+\frac15\right)h+\frac1{10}(4)=1. \] \[ \frac3{10}h+\frac25=1. \] \[ \frac3{10}h=\frac35. \] \[ h=2. \]Therefore,
\[ \boxed{2\text{ hours together}}. \]Practice 10
A alone for \(2\) hours:
\[ \frac2{12}=\frac16. \]Combined rate:
\[ \frac1{12}+\frac1{24} = \frac18. \]Four hours together:
\[ \frac18(4)=\frac12. \]Total completed:
\[ \frac16+\frac12 = \frac23. \]Remaining:
\[ \frac13. \]B alone works at
\[ \frac1{24}. \]Therefore,
\[ t= \frac{1/3}{1/24}=8. \]Thus,
\[ \boxed{8\text{ hours}}. \]Practice 11
Initial work:
\[ \frac28=\frac14. \]Remaining:
\[ \frac34. \]Net rate after the drain opens:
\[ \frac18-\frac1{16} = \frac1{16}. \]Therefore,
\[ t= \frac{3/4}{1/16}=12. \]Thus,
\[ \boxed{12\text{ additional hours}}. \]Practice 12
Net rate during the first \(2\) hours:
\[ \frac16-\frac1{18} = \frac2{18} = \frac19. \]Work completed:
\[ \frac19(2)=\frac29. \]Remaining:
\[ \frac79. \]Fill rate after the drain closes:
\[ \frac16. \]Therefore,
\[ t= \frac{7/9}{1/6} = \frac{14}{3}. \]Thus,
\[ \boxed{4\frac23\text{ additional hours}}. \]Practice 13
A works a total of \(5\) hours:
\[ \frac58. \]B works \(3\) hours:
\[ \frac3x. \]Therefore,
\[ \frac58+\frac3x=1. \] \[ \frac3x=\frac38. \] \[ x=8. \]Therefore,
\[ \boxed{8\text{ hours}}. \]Practice 14
One worker's rate is
\[ \frac1{30}. \]Let \(n\) workers operate for \(5\) hours:
\[ n\left(\frac1{30}\right)(5)=1. \] \[ \frac{n}{6}=1. \] \[ n=6. \]Therefore,
\[ \boxed{6\text{ workers}}. \]Practice 15
Worker A begins \(2\) hours earlier and then also works during the \(4\) shared hours.
Therefore,
\[ \boxed{t_A=6\text{ hours}} \]while
\[ \boxed{t_B=4\text{ hours}}. \]Assigning both workers \(4\) hours ignores A's two-hour head start.
What to Remember
Staggered and interrupted work problems are piecewise work problems:
\[ \boxed{ W_{\text{total}} = \sum_i r_i t_i }. \]For one complete job,
\[ \boxed{ \sum_i r_i t_i=1 }. \]Delayed starts create different active times. Interruptions create zero-rate intervals. Early departures change the active combined rate. Unknown completion times appear through reciprocal rates such as
\[ \boxed{\frac1x}. \]The most important habit is to build the timeline first, identify which rates are active during each interval, and only then write the work equation.
Where We Go Next
Work-rate problems combine contributions over time. Another major family of applied algebra problems combines quantities by composition: amounts of ingredients are mixed while total quantity and total substance are conserved.
The next lesson, Mixture Problems, develops that conservation framework.