Overview
Staggered work problems occur when two or more workers or processes do not operate during exactly the same time interval. One worker may begin alone and later be joined by another, or several machines may start and stop at different times.
The important idea is that each worker contributes an amount of work equal to that worker's rate multiplied by the amount of time that worker actually works.
For each worker or process,
\[ W_i = R_iT_i. \]
The total completed work is therefore
\[ \boxed{R_1T_1 + R_2T_2 + R_3T_3 + \cdots = W_{\text{total}}}. \]
For one complete job,
\[ \boxed{R_1T_1 + R_2T_2 + R_3T_3 + \cdots = 1}. \]
Different Workers May Have Different Times
In a basic combined-rate problem, all workers begin at the same time and continue until the job is finished. Their rates can therefore be combined immediately.
In a staggered problem, this is not always possible because different workers may operate for different lengths of time.
Suppose Worker A begins working alone and works for \(2\) hours. Worker B then joins Worker A, and both continue for another \(x\) hours.
Worker A works for a total of
\[ 2+x \]
hours, while Worker B works for only
\[ x \]
hours.
Those different time intervals must be represented separately in the work equation.
The General Staggered-Work Model
Suppose Worker A has rate \(R_A\) and works for \(T_A\) hours. Worker B has rate \(R_B\) and works for \(T_B\) hours.
Worker A completes
\[ W_A = R_AT_A, \]
and Worker B completes
\[ W_B = R_BT_B. \]
If together they complete one full job, then
\[ W_A+W_B=1. \]
Therefore,
\[ \boxed{R_AT_A+R_BT_B=1}. \]
This equation is the basic structure behind staggered work problems.
Example 1: One Worker Starts First
Worker A can complete a job in \(6\) hours. Worker B can complete the same job in \(4\) hours. Worker A works alone for \(2\) hours, after which Worker B joins. How much longer will they need to finish the job?
Worker A's rate is
\[ R_A=\frac{1}{6}. \]
Worker B's rate is
\[ R_B=\frac{1}{4}. \]
Let \(x\) represent the number of hours they work together after Worker B joins.
Worker A works for a total of
\[ 2+x \]
hours, while Worker B works for
\[ x \]
hours.
The total work equation is
\[ \frac{1}{6}(2+x)+\frac{1}{4}x=1. \]
Distribute:
\[ \frac{2}{6}+\frac{x}{6}+\frac{x}{4}=1. \]
Simplify:
\[ \frac{1}{3}+\frac{x}{6}+\frac{x}{4}=1. \]
Using a common denominator of \(12\),
\[ \frac{1}{3}+\frac{2x}{12}+\frac{3x}{12}=1. \]
Thus,
\[ \frac{1}{3}+\frac{5x}{12}=1. \]
Subtract \(\frac{1}{3}\):
\[ \frac{5x}{12}=\frac{2}{3}. \]
Multiply by \(12\):
\[ 5x=8. \]
Therefore,
\[ \boxed{x=\frac{8}{5}=1.6\text{ hours}}. \]
After Worker B joins, the two workers need \(1.6\) additional hours, or \(1\) hour and \(36\) minutes, to finish the job.
An Alternate Method: Find the Remaining Work First
The previous problem can also be solved in stages.
Worker A works alone for \(2\) hours at a rate of
\[ \frac{1}{6}. \]
So Worker A completes
\[ W = \frac{1}{6}(2) = \frac{1}{3} \]
of the job.
The remaining work is
\[ 1-\frac{1}{3} = \frac{2}{3}. \]
When Worker B joins, their combined rate is
\[ \frac{1}{6}+\frac{1}{4} = \frac{5}{12}. \]
Use
\[ T=\frac{W}{R}. \]
Then
\[ T = \frac{\frac{2}{3}}{\frac{5}{12}} = \frac{2}{3}\times\frac{12}{5} = \frac{8}{5}. \]
Again,
\[ \boxed{T=1.6\text{ hours}}. \]
Both methods are valid. One writes a single work equation, while the other divides the problem into stages.
Think in Time Intervals
Staggered problems become easier when the timeline is divided into intervals during which the active rates remain unchanged.
For example:
- Interval 1: Worker A works alone.
- Interval 2: Workers A and B work together.
Within each interval, calculate
\[ \text{Work}=\text{Rate}\times\text{Time}. \]
Then add the work completed during all intervals.
This interval-based approach becomes especially important in more advanced problems where several workers start and stop at different times.
Example 2: A Machine Joins Later
Machine A can complete a production run in \(10\) hours. Machine B can complete the same run in \(15\) hours. Machine A operates alone for \(3\) hours, after which Machine B is activated. How long after Machine B starts will the production run be completed?
Machine A's rate is
\[ R_A=\frac{1}{10}. \]
Machine B's rate is
\[ R_B=\frac{1}{15}. \]
In the first \(3\) hours, Machine A completes
\[ W_A = \frac{1}{10}(3) = \frac{3}{10}. \]
The remaining work is
\[ 1-\frac{3}{10} = \frac{7}{10}. \]
The combined rate after Machine B starts is
\[ \frac{1}{10}+\frac{1}{15}. \]
Using a common denominator,
\[ \frac{1}{10}+\frac{1}{15} = \frac{3}{30}+\frac{2}{30} = \frac{1}{6}. \]
Now solve for the time needed to complete the remaining \(\frac{7}{10}\) of the job:
\[ T = \frac{\frac{7}{10}}{\frac{1}{6}}. \]
Therefore,
\[ T = \frac{7}{10}\times 6 = \frac{42}{10} = 4.2. \]
Thus,
\[ \boxed{T=4.2\text{ hours}}. \]
The machines require \(4\) hours and \(12\) minutes after Machine B begins operating.
Example 3: Two Workers Start at Different Times
Worker A can complete a project in \(12\) hours. Worker B can complete it in \(8\) hours. Worker A begins at 8:00 a.m., and Worker B begins at 10:00 a.m. At what time will the project be completed?
Worker A works alone for the first \(2\) hours.
Worker A's rate is
\[ \frac{1}{12}. \]
The amount completed by 10:00 a.m. is
\[ \frac{1}{12}(2) = \frac{1}{6}. \]
The remaining work is
\[ 1-\frac{1}{6} = \frac{5}{6}. \]
Once Worker B joins, their combined rate is
\[ \frac{1}{12}+\frac{1}{8}. \]
Using a common denominator of \(24\),
\[ \frac{1}{12}+\frac{1}{8} = \frac{2}{24}+\frac{3}{24} = \frac{5}{24}. \]
The time required to complete the remaining work is
\[ T = \frac{\frac{5}{6}}{\frac{5}{24}}. \]
Simplify:
\[ T = \frac{5}{6}\times\frac{24}{5} = 4. \]
The workers need \(4\) hours after 10:00 a.m., so the project is completed at
\[ \boxed{2{:}00\text{ p.m.}}. \]
Problems with Three or More Stages
Some staggered problems involve several changes in the active workforce.
For example:
- Worker A works alone for \(2\) hours.
- Workers A and B work together for \(3\) hours.
- Worker C then joins both workers until the job is complete.
The work completed during each stage can be written separately:
\[ W_1=R_A(2), \]
\[ W_2=(R_A+R_B)(3), \]
and
\[ W_3=(R_A+R_B+R_C)x. \]
For one complete job,
\[ \boxed{W_1+W_2+W_3=1}. \]
The algebra may become longer, but the underlying Work Principle has not changed.
Example 4: Three Stages
Worker A can complete a job in \(12\) hours, Worker B in \(8\) hours, and Worker C in \(6\) hours. Worker A works alone for \(2\) hours. Worker B then joins for \(2\) more hours. After that, Worker C joins both workers. How long after Worker C joins will the job be completed?
Worker A's rate is
\[ \frac{1}{12}. \]
Worker B's rate is
\[ \frac{1}{8}. \]
Worker C's rate is
\[ \frac{1}{6}. \]
During the first stage, Worker A completes
\[ W_1 = \frac{1}{12}(2) = \frac{1}{6}. \]
During the second stage, Workers A and B have combined rate
\[ \frac{1}{12}+\frac{1}{8} = \frac{5}{24}. \]
In \(2\) hours they complete
\[ W_2 = \frac{5}{24}(2) = \frac{5}{12}. \]
The total work completed before Worker C joins is
\[ \frac{1}{6}+\frac{5}{12} = \frac{2}{12}+\frac{5}{12} = \frac{7}{12}. \]
The remaining work is
\[ 1-\frac{7}{12} = \frac{5}{12}. \]
When all three workers are active, their combined rate is
\[ \frac{1}{12}+\frac{1}{8}+\frac{1}{6}. \]
Using a common denominator of \(24\),
\[ \frac{2}{24}+\frac{3}{24}+\frac{4}{24} = \frac{9}{24} = \frac{3}{8}. \]
The time needed to complete the remaining work is
\[ T = \frac{\frac{5}{12}}{\frac{3}{8}}. \]
Thus,
\[ T = \frac{5}{12}\times\frac{8}{3} = \frac{10}{9}. \]
Therefore,
\[ \boxed{T=\frac{10}{9}\text{ hours}} \]
or approximately \(1\) hour and \(7\) minutes after Worker C joins.
Writing the Entire Problem as One Equation
Many staggered problems can be solved either by calculating each stage separately or by writing a single equation.
For example, if Worker A works \(2+x\) hours and Worker B works \(x\) hours, then
\[ \frac{1}{6}(2+x)+\frac{1}{4}x=1. \]
This equation already accounts for the different work times of the two workers.
The one-equation method is often efficient when the time expressions are simple. The stage-by-stage method is often easier to interpret when several changes occur.
Method for Solving Staggered Work Problems
- Represent the complete job by \(1\).
- Identify each worker's or process's rate. Convert completion times using \(R=\frac{1}{t}\).
- Divide the problem into time intervals. Create a new interval whenever someone begins or stops working.
- Determine which rates are active 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 work. For a complete job, set the sum equal to \(1\).
- Solve for the unknown time.
- Check that the time assignments are consistent with the problem.
Common Errors
- Assuming everyone works for the same amount of time. In staggered problems, each worker may have a different value of \(T\).
- Combining rates too early. Rates should be combined only during intervals when those workers are operating simultaneously.
- Forgetting work completed before another worker joins. Earlier work must be subtracted from the full job or included in the total equation.
- Confusing total elapsed time with an individual worker's time. A worker who joins later has worked fewer hours than someone who began earlier.
- Failing to update the active rate when the workforce changes. Each interval may have a different combined rate.
Practice Problems
- Worker A can complete a job in \(8\) hours. Worker B can complete it in \(12\) hours. Worker A works alone for \(2\) hours before Worker B joins. How much longer will they need to finish?
- Machine A can complete a job in \(9\) hours and Machine B in \(18\) hours. Machine A works alone for \(3\) hours before Machine B starts. How long after Machine B starts will the job be completed?
- Worker A completes a project in \(10\) hours and Worker B in \(15\) hours. Worker A begins \(2\) hours before Worker B. How long after Worker B begins will they finish the project?
- A pipe can fill a tank in \(6\) hours. A second pipe can fill it in \(12\) hours. The first pipe operates alone for \(2\) hours, after which the second pipe is opened. How much additional time is required to fill the tank?
- Worker A can complete a job in \(12\) hours. Worker B can complete it in \(6\) hours. Worker A works alone for \(3\) hours before Worker B joins. What fraction of the job remains when Worker B begins?
- Machine A can complete a production run in \(8\) hours and Machine B in \(16\) hours. Machine A operates alone for \(2\) hours, then both machines operate for \(3\) hours. What fraction of the job remains after the \(5\) total hours?
- Worker A completes a job in \(12\) hours, Worker B in \(8\) hours, and Worker C in \(6\) hours. Worker A works alone for \(2\) hours, then Worker B joins for \(2\) hours, and finally Worker C joins. How long after Worker C joins will they finish?
- A worker can complete a job in \(9\) hours. A second worker can complete it in \(6\) hours. The first worker begins at 7:00 a.m., and the second begins at 9:00 a.m. At what time will the job be completed?
Practice Answers
- Worker A completes \[ \frac{1}{8}(2)=\frac{1}{4}. \] The remaining work is \[ \frac{3}{4}. \] The combined rate is \[ \frac{1}{8}+\frac{1}{12} = \frac{5}{24}. \] Therefore, \[ T = \frac{\frac{3}{4}}{\frac{5}{24}} = \frac{18}{5}. \] \[ \boxed{3.6\text{ hours}} \]
- Machine A completes \[ \frac{1}{9}(3)=\frac{1}{3}. \] The remaining work is \[ \frac{2}{3}. \] The combined rate is \[ \frac{1}{9}+\frac{1}{18} = \frac{1}{6}. \] Thus, \[ T = \frac{\frac{2}{3}}{\frac{1}{6}} = \boxed{4\text{ hours}}. \]
- Worker A completes \[ \frac{1}{10}(2)=\frac{1}{5} \] before Worker B begins. The remaining work is \[ \frac{4}{5}. \] Their combined rate is \[ \frac{1}{10}+\frac{1}{15} = \frac{1}{6}. \] Therefore, \[ T = \frac{\frac{4}{5}}{\frac{1}{6}} = \boxed{\frac{24}{5}\text{ hours}} \] or \(4.8\) hours.
- The first pipe fills \[ \frac{1}{6}(2)=\frac{1}{3} \] of the tank. The remaining work is \[ \frac{2}{3}. \] The combined rate is \[ \frac{1}{6}+\frac{1}{12} = \frac{1}{4}. \] Therefore, \[ T = \frac{\frac{2}{3}}{\frac{1}{4}} = \boxed{\frac{8}{3}\text{ hours}}. \]
- Worker A completes \[ \frac{1}{12}(3) = \frac{1}{4}. \] Therefore the remaining work is \[ \boxed{\frac{3}{4}}. \]
- Machine A completes \[ \frac{1}{8}(2)=\frac{1}{4}. \] During the next \(3\) hours, the combined rate is \[ \frac{1}{8}+\frac{1}{16} = \frac{3}{16}. \] They complete \[ \frac{3}{16}(3) = \frac{9}{16}. \] Total completed work: \[ \frac{1}{4}+\frac{9}{16} = \frac{4}{16}+\frac{9}{16} = \frac{13}{16}. \] Remaining work: \[ \boxed{\frac{3}{16}}. \]
- Worker A first completes \[ \frac{1}{12}(2)=\frac{1}{6}. \] Workers A and B then complete \[ \left(\frac{1}{12}+\frac{1}{8}\right)(2) = \frac{5}{24}(2) = \frac{5}{12}. \] Total completed: \[ \frac{1}{6}+\frac{5}{12} = \frac{7}{12}. \] Remaining: \[ \frac{5}{12}. \] All three work at rate \[ \frac{1}{12}+\frac{1}{8}+\frac{1}{6} = \frac{3}{8}. \] Thus, \[ T = \frac{\frac{5}{12}}{\frac{3}{8}} = \boxed{\frac{10}{9}\text{ hours}}. \]
- The first worker completes \[ \frac{1}{9}(2)=\frac{2}{9} \] before 9:00 a.m. The remaining work is \[ \frac{7}{9}. \] Their combined rate is \[ \frac{1}{9}+\frac{1}{6} = \frac{5}{18}. \] The remaining time is \[ T = \frac{\frac{7}{9}}{\frac{5}{18}} = \frac{14}{5} = 2.8\text{ hours}. \] That is \(2\) hours and \(48\) minutes after 9:00 a.m. \[ \boxed{11{:}48\text{ a.m.}} \]
Summary
Staggered work problems involve workers or processes that operate for different lengths of time. Each contribution must therefore be represented separately.
The general model is
\[ \boxed{R_1T_1+R_2T_2+R_3T_3+\cdots=1} \]
for one complete job.
The most important strategy is to divide the problem into intervals during which the active rates remain unchanged. For each interval, use
\[ W=RT, \]
then add all completed work.
The next problem type develops this idea further by considering workers who stop before the job is finished or are replaced by someone else.
Continue
Continue to 5. Interrupted and Remaining-Work Problems
Return to 3. Opposing-Rate Problems