The Central Idea
If one contributor completes work at rate \(r_1\) and another contributes at rate \(r_2\), their combined productive rate is
\[ \boxed{r_{\text{combined}}=r_1+r_2}. \]If one process undoes work at rate \(r_2\), the net rate is
\[ \boxed{r_{\text{net}}=r_1-r_2}. \]For one complete job,
\[ \boxed{1=r_{\text{net}}t}. \]Therefore,
\[ \boxed{t=\frac{1}{r_{\text{net}}}}. \]Review: Completion Time Becomes Rate
If worker A completes a job in \(a\) hours, then
\[ r_A=\frac1a. \]If worker B completes the same job in \(b\) hours, then
\[ r_B=\frac1b. \]These rates, not the completion times \(a\) and \(b\), are the quantities that combine.
Two Productive Workers
Suppose worker A completes one job in \(6\) hours and worker B completes it in \(4\) hours.
Their rates are
\[ \frac16 \]and
\[ \frac14 \]job per hour.
Together,
\[ r_{\text{combined}} = \frac16+\frac14 = \frac{2}{12}+\frac{3}{12} = \frac{5}{12}. \]Thus they complete
\[ \boxed{\frac{5}{12}\text{ of the job per hour}}. \]Find the Combined Completion Time
Continue with the combined rate
\[ \frac{5}{12}. \]For one complete job,
\[ 1=\frac{5}{12}t. \]Therefore,
\[ t=\frac{12}{5}=2.4. \]Thus the job takes
\[ \boxed{2.4\text{ hours}} \]when both workers operate together.
Why the Combined Time Is Shorter
Worker A alone takes \(6\) hours, and worker B alone takes \(4\) hours.
Together they take
\[ 2.4\text{ hours}. \]This makes sense because the combined productive rate
\[ \frac{5}{12} \]is greater than either individual rate.
For two positive contributors working independently,
\[ \boxed{ t_{\text{combined}} < \min(a,b) }. \]The General Two-Worker Formula
If one worker takes \(a\) hours and another takes \(b\) hours, then
\[ r_{\text{combined}} = \frac1a+\frac1b. \]Combine:
\[ \frac1a+\frac1b = \frac{a+b}{ab}. \]Therefore, the combined completion time is
\[ \boxed{ t=\frac{ab}{a+b} }. \]This formula assumes both contributors work productively and continuously on the same job at constant rates.
Example: Use the General Formula
One machine takes \(8\) hours and another takes \(12\) hours.
Then
\[ t= \frac{8(12)}{8+12} = \frac{96}{20} = 4.8. \]Therefore,
\[ \boxed{4.8\text{ hours}}. \]Three Productive Contributors
If three contributors have rates
\[ r_1,\qquad r_2,\qquad r_3, \]then
\[ \boxed{ r_{\text{combined}} = r_1+r_2+r_3 }. \]If their individual completion times are \(a\), \(b\), and \(c\), then
\[ \boxed{ r_{\text{combined}} = \frac1a+\frac1b+\frac1c }. \]Example: Three Workers
Three workers can individually complete a job in \(6\), \(8\), and \(12\) hours.
Their combined rate is
\[ \frac16+\frac18+\frac1{12}. \]Using denominator \(24\),
\[ \frac4{24}+\frac3{24}+\frac2{24} = \frac9{24} = \frac38. \]Therefore, the combined completion time is
\[ t=\frac{1}{3/8} = \frac83. \]Thus,
\[ \boxed{2\frac23\text{ hours}}. \]Work Tables for Combined Rates
A table keeps the contributors separate before their work is combined.
\[ \begin{array}{c|c|c|c} \text{contributor} & \text{rate} & \text{time} & \text{work}\\ \hline A & \frac1a & t & \frac{t}{a}\\ B & \frac1b & t & \frac{t}{b} \end{array} \]If both work for the same interval \(t\), then
\[ \frac{t}{a}+\frac{t}{b}=1. \]Factoring \(t\) gives
\[ t\left(\frac1a+\frac1b\right)=1. \]Combined Work as Fractions of the Job
Suppose worker A takes \(5\) hours and B takes \(10\) hours.
In \(2\) hours, A completes
\[ \frac25, \]while B completes
\[ \frac2{10}=\frac15. \]Together they complete
\[ \frac25+\frac15=\frac35. \]Therefore,
\[ \boxed{\frac35\text{ of the job}} \]is finished after \(2\) hours.
Unknown Combined Time
A worker takes \(9\) hours and another takes \(6\) hours. Let \(t\) be the time required together.
Their completed work is
\[ \frac{t}{9} \]and
\[ \frac{t}{6}. \]Since the whole job equals \(1\),
\[ \frac{t}{9}+\frac{t}{6}=1. \]Multiply by \(18\):
\[ 2t+3t=18. \] \[ 5t=18. \] \[ t=\frac{18}{5}=3.6. \]Therefore,
\[ \boxed{3.6\text{ hours}}. \]Unknown Individual Completion Time
Worker A completes a job in \(6\) hours. Working together, A and B complete the job in \(4\) hours. Find B's completion time.
Let B's time be \(x\) hours.
Then
\[ \frac16+\frac1x=\frac14. \]Subtract:
\[ \frac1x = \frac14-\frac16 = \frac{3}{12}-\frac{2}{12} = \frac1{12}. \]Therefore,
\[ \boxed{x=12\text{ hours}}. \]Unknown Rate from a Combined Rate
If the combined rate is known, an unknown individual rate can be found by subtraction.
In general,
\[ r_A+r_B=r_{\text{combined}}. \]Therefore,
\[ \boxed{ r_B=r_{\text{combined}}-r_A }. \]Once \(r_B\) is known, its completion time is
\[ \boxed{T_B=\frac1{r_B}}. \]Example: Find the Missing Machine Time
Machine A completes a job in \(10\) hours. Together, A and B complete it in \(6\) hours.
A's rate is
\[ \frac1{10}. \]The combined rate is
\[ \frac16. \]Therefore,
\[ r_B= \frac16-\frac1{10} = \frac5{30}-\frac3{30} = \frac1{15}. \]Thus,
\[ \boxed{B\text{ takes }15\text{ hours alone}}. \]Comparing Contributions
If two workers operate for the same amount of time, their completed work is proportional to their rates.
Suppose A's rate is
\[ \frac14 \]and B's rate is
\[ \frac16. \]Their rate ratio is
\[ \frac{1/4}{1/6} = \frac32. \]Therefore, over equal working times, A completes
\[ \boxed{\frac32} \]as much work as B.
Contribution Shares in a Completed Job
If A and B work together from start to finish, each contributor's fraction of the completed job equals its rate divided by the combined rate.
If
\[ r_A=\frac14, \qquad r_B=\frac16, \]then
\[ r_{\text{combined}} = \frac5{12}. \]A's share is
\[ \frac{1/4}{5/12} = \frac35, \]and B's share is
\[ \frac{1/6}{5/12} = \frac25. \]Their shares add to \(1\).
Opposing Work Rates
Some processes undo or remove work performed by another process.
Examples include:
- a drain emptying a tank while a pipe fills it;
- a leak removing fluid while a pump adds it;
- a process producing material while another removes material;
- a repair process operating while damage continues to accumulate.
In such cases, opposing rates subtract.
Net Work Rate
If a productive process has rate \(r_p\) and an opposing process has rate \(r_o\), then
\[ \boxed{ r_{\text{net}}=r_p-r_o }. \]The sign of the net rate determines the direction of change.
- If \(r_{\text{net}}>0\), net work is completed.
- If \(r_{\text{net}}=0\), there is no net progress.
- If \(r_{\text{net}}<0\), the opposing process dominates.
Example: Fill and Drain
A pipe can fill a tank in \(6\) hours. A drain can empty the same tank in \(9\) hours.
The filling rate is
\[ \frac16. \]The draining rate is
\[ \frac19. \]With both open,
\[ r_{\text{net}} = \frac16-\frac19 = \frac3{18}-\frac2{18} = \frac1{18}. \]Therefore, the tank fills in
\[ \boxed{18\text{ hours}}. \]Why the Net Completion Time Can Be Much Longer
The fill pipe alone needs \(6\) hours, but with the drain open the tank needs \(18\) hours.
The drain continuously removes part of the progress.
The net rate
\[ \frac1{18} \]is much smaller than the fill rate
\[ \frac16. \]A small positive net rate produces a large completion time.
When the Opposing Process Is Faster
Suppose a pipe fills a tank in \(10\) hours while a drain empties it in \(8\) hours.
The net rate is
\[ \frac1{10}-\frac18 = \frac4{40}-\frac5{40} = -\frac1{40}. \]The negative rate means the tank cannot fill from empty while both operate under this model.
Instead, the amount in the tank decreases at a net rate of
\[ \boxed{\frac1{40}\text{ tank per hour}}. \]No Net Progress
If a filling process and an opposing process have equal rates,
\[ r_p=r_o, \]then
\[ r_{\text{net}}=0. \]The system remains at the same work level while both rates continue unchanged.
A completion time based on
\[ \frac1{r_{\text{net}}} \]does not exist because division by zero is undefined.
General Opposing-Rate Formula
If one process completes a job in \(a\) hours and another reverses one full job in \(b\) hours, with \(a<b\) so the productive process is faster, then
\[ r_{\text{net}} = \frac1a-\frac1b. \]Combine:
\[ \frac1a-\frac1b = \frac{b-a}{ab}. \]Therefore, the net completion time is
\[ \boxed{ t=\frac{ab}{b-a} }. \]Example: Use the Opposing-Rate Formula
A fill pipe fills a tank in \(4\) hours, while a drain empties it in \(12\) hours.
Then
\[ t= \frac{4(12)}{12-4} = \frac{48}{8} = 6. \]Therefore,
\[ \boxed{6\text{ hours}}. \]Several Productive and Opposing Processes
With several processes, assign positive signs to productive rates and negative signs to opposing rates.
For example,
\[ \boxed{ r_{\text{net}} = r_1+r_2-r_3-r_4 }. \]The same principle extends to any number of constant independent rates.
Example: Two Fill Pipes and One Drain
Two pipes fill a tank in \(6\) hours and \(8\) hours. A drain empties it in \(24\) hours.
The net rate is
\[ \frac16+\frac18-\frac1{24}. \]Using denominator \(24\),
\[ \frac4{24}+\frac3{24}-\frac1{24} = \frac6{24} = \frac14. \]Therefore, the tank fills in
\[ \boxed{4\text{ hours}}. \]Unknown Opposing Rate
A pipe fills a tank in \(5\) hours. With a drain open, the tank fills in \(8\) hours. Find the drain's emptying time.
Let the drain rate be \(r_d\).
Then
\[ \frac15-r_d=\frac18. \]Therefore,
\[ r_d= \frac15-\frac18 = \frac8{40}-\frac5{40} = \frac3{40}. \]The drain's emptying time is
\[ \frac{1}{3/40} = \frac{40}{3}. \]Thus,
\[ \boxed{13\frac13\text{ hours}}. \]Unknown Productive Rate
A drain empties a tank in \(15\) hours. With a fill pipe open, the tank fills from empty in \(10\) hours. Find the fill pipe's rate.
Let \(r_f\) be the fill rate.
\[ r_f-\frac1{15}=\frac1{10}. \]Therefore,
\[ r_f = \frac1{10}+\frac1{15} = \frac3{30}+\frac2{30} = \frac16. \]Thus the fill pipe alone would fill the tank in
\[ \boxed{6\text{ hours}}. \]Combined Rate from Measured Output
Work problems do not always begin with completion times.
Suppose two machines together produce \(180\) units per hour, and one machine produces \(75\) units per hour.
The second machine's rate is
\[ 180-75=105. \]Therefore,
\[ \boxed{105\text{ units per hour}}. \]The same additive logic applies whether work is measured in whole jobs, components, gallons, or another consistent unit.
Production with Loss
A process produces \(120\) units per hour while defects or losses remove \(15\) units per hour.
The net production rate is
\[ 120-15=105. \]Over \(8\) hours, net production is
\[ 105(8)=840. \]Therefore,
\[ \boxed{840\text{ net units}}. \]Combined Rates Must Use the Same Work Unit
Rates may be added or subtracted only when they refer to the same kind of work.
For example,
\[ 30\text{ units/hour} \]cannot be added directly to
\[ \frac14\text{ job/hour} \]unless the size of one job is defined in units.
Normalize the work units before combining the rates.
Combined Rates Must Use the Same Time Unit
Suppose one machine works at
\[ 20\text{ units/min} \]and another at
\[ 900\text{ units/hr}. \]Convert the second rate:
\[ 900\text{ units/hr} = 15\text{ units/min}. \]Therefore, the combined rate is
\[ 20+15=35. \]Thus,
\[ \boxed{35\text{ units/min}}. \]Combined Work and Direct Variation
Once the combined rate is known and constant, total work varies directly with time:
\[ \boxed{ W=r_{\text{combined}}t }. \]If
\[ r_{\text{combined}}=\frac{7}{20}, \]then after \(2\) hours,
\[ W= \frac{7}{20}(2) = \frac7{10}. \]The combined-rate problem reduces to an ordinary work-rate problem after the rates have been consolidated.
Net Rate Is an Algebraic Sum
A useful general viewpoint is to assign a sign to every contributor.
Productive contributors receive positive rates, while opposing contributors receive negative rates.
Then
\[ \boxed{ r_{\text{net}}=\sum_i r_i }. \]This notation treats addition and subtraction as one unified algebraic operation.
Checking a Combined-Rate Answer
Suppose two workers take \(6\) and \(4\) hours alone, and the calculated combined time is \(2.4\) hours.
In \(2.4\) hours, A completes
\[ \frac{2.4}{6}=0.4, \]while B completes
\[ \frac{2.4}{4}=0.6. \]Together,
\[ 0.4+0.6=1. \]The whole job is recovered.
Checking an Opposing-Rate Answer
Suppose a fill pipe takes \(6\) hours and a drain takes \(9\) hours, with a calculated net fill time of \(18\) hours.
In \(18\) hours, the fill pipe contributes
\[ \frac{18}{6}=3 \]tank-equivalents of inflow.
The drain removes
\[ \frac{18}{9}=2. \]Net work:
\[ 3-2=1. \]The result is verified.
Common Error: Adding Completion Times
If two workers take \(6\) and \(4\) hours, adding
\[ 6+4=10 \]does not give their combined completion time.
Convert first:
\[ \frac16+\frac14=\frac5{12}. \]Then invert the combined rate:
\[ \boxed{\frac{12}{5}=2.4\text{ hours}}. \]Common Error: Averaging Completion Times
The average of \(6\) hours and \(4\) hours is
\[ 5\text{ hours}, \]but two workers acting together should complete the job faster than either one acting alone.
A \(5\)-hour result fails this basic reasonableness check.
Common Error: Adding Rates When One Process Opposes Another
A fill pipe and a drain do not both contribute positively to the same quantity.
If the fill rate is
\[ \frac16 \]and the drain rate is
\[ \frac19, \]the net rate is
\[ \boxed{ \frac16-\frac19 }. \]Common Error: Subtracting in the Wrong Order
If the tank is supposed to fill, use
\[ r_{\text{fill}}-r_{\text{drain}}. \]Reversing the order changes the sign and therefore changes the physical interpretation.
Define the positive direction of work before writing the net-rate equation.
Common Error: Ignoring an Impossible Net Rate
If the drain rate exceeds the fill rate, the tank cannot fill from empty while both processes remain active.
A negative net rate is not merely an arithmetic inconvenience. It describes a system moving in the opposite direction from the requested completion.
Common Error: Mixing Work Units
Rates such as
\[ \frac14\text{ job/hr} \]and
\[ 30\text{ parts/hr} \]cannot be added until “one job” is defined in parts.
The work quantities must refer to the same whole.
Common Error: Mixing Time Units
A rate in jobs per hour cannot be combined directly with a rate in jobs per minute.
Convert both to a common time unit before adding or subtracting.
Common Error: Forgetting to Invert the Final Rate
If a combined rate is
\[ \frac{3}{10}\text{ job/hr}, \]the completion time is not
\[ \frac{3}{10}\text{ hr}. \]Instead,
\[ t=\frac{1}{3/10} = \boxed{\frac{10}{3}\text{ hr}}. \]A Reliable Procedure for Combined Work Rates
- Define one complete job as \(1\).
- Convert each completion time \(T_i\) to rate \(1/T_i\).
- Make all work and time units consistent.
- Add the productive rates.
- Set \(1=r_{\text{combined}}t\) for one whole job.
- Solve for \(t\).
- Verify that each contributor's work fractions add to \(1\).
- Check that the combined time is shorter than every positive individual time.
A Reliable Procedure for Opposing Work Rates
- Choose which direction of work is positive.
- Convert all completion or reversal times into rates.
- Assign positive signs to productive rates.
- Assign negative signs to opposing rates.
- Add algebraically to obtain the net rate.
- Determine whether the net rate is positive, zero, or negative.
- If completion is possible, use \(1=r_{\text{net}}t\).
- Reconstruct the productive and opposing work to verify the result.
A Reliable Procedure for Unknown Individual Rates
- Write the known combined or net rate.
- Write all known individual rates.
- Represent the unknown contributor by a rate variable.
- Solve the rate equation.
- If the problem asks for completion time, take the reciprocal.
- Reject nonphysical rates or times.
- Substitute back into the combined-rate equation.
Practice 1: Two Workers
Worker A completes a job in \(8\) hours and worker B in \(12\) hours. Find their combined completion time.
Practice 2: Two Faster Workers
Two machines complete a job individually in \(5\) hours and \(7.5\) hours. Find the combined time.
Practice 3: Three Workers
Three workers complete a job individually in \(6\), \(9\), and \(18\) hours. Find the combined time.
Practice 4: Work Completed Together
Worker A takes \(10\) hours and worker B takes \(15\) hours. What fraction of the job do they complete together in \(3\) hours?
Practice 5: Unknown Worker Time
Worker A completes a job in \(8\) hours. Together, A and B complete it in \(5\) hours. How long does B take alone?
Practice 6: Another Unknown Worker
Worker A completes a job in \(12\) hours. Together, A and B complete it in \(7.5\) hours. Find B's completion time.
Practice 7: Fill and Drain
A pipe fills a tank in \(8\) hours while a drain empties it in \(12\) hours. How long does the tank take to fill with both open?
Practice 8: Two Fills and One Drain
Two pipes fill a tank in \(6\) hours and \(10\) hours. A drain empties it in \(15\) hours. Find the net fill time.
Practice 9: Impossible Fill
A pipe fills a tank in \(12\) hours and a drain empties it in \(8\) hours. Determine the net rate and explain what it means.
Practice 10: Unknown Drain
A pipe fills a tank in \(4\) hours. With a drain open, the tank fills in \(6\) hours. How long would the drain take to empty a full tank by itself?
Practice 11: Unknown Fill Pipe
A drain empties a tank in \(20\) hours. With a fill pipe open, the tank fills in \(12\) hours. How long would the fill pipe take alone?
Practice 12: Contribution Shares
Worker A takes \(4\) hours and worker B takes \(6\) hours. If they work together from start to finish, what fraction of the completed job is attributable to each worker?
Practice 13: Production Rates
Two machines produce \(72\) and \(48\) units per hour. How many units do they produce together in \(5.5\) hours?
Practice 14: Production with Loss
A process produces \(150\) units per hour while losses remove \(18\) units per hour. Find the net production over \(7\) hours.
Practice 15: Identify the Error
Two workers can complete a job in \(6\) hours and \(3\) hours. A student says their combined completion time is
\[ 6+3=9\text{ hours}. \]Explain the error and find the correct combined time.
Practice Answers
Practice 1
\[ r= \frac18+\frac1{12} = \frac3{24}+\frac2{24} = \frac5{24}. \] \[ t= \frac{1}{5/24} = \frac{24}{5}. \]Therefore,
\[ \boxed{4.8\text{ hours}}. \]Practice 2
\[ r= \frac15+\frac1{7.5}. \]Since
\[ \frac1{7.5}=\frac2{15}, \]we get
\[ r= \frac3{15}+\frac2{15} = \frac13. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 3
\[ r= \frac16+\frac19+\frac1{18} = \frac3{18}+\frac2{18}+\frac1{18} = \frac13. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 4
\[ r= \frac1{10}+\frac1{15} = \frac3{30}+\frac2{30} = \frac16. \]In \(3\) hours,
\[ W= \frac16(3)=\frac12. \]Therefore,
\[ \boxed{\frac12\text{ of the job}}. \]Practice 5
\[ \frac18+\frac1x=\frac15. \] \[ \frac1x = \frac15-\frac18 = \frac8{40}-\frac5{40} = \frac3{40}. \] \[ x=\frac{40}{3}. \]Therefore,
\[ \boxed{13\frac13\text{ hours}}. \]Practice 6
\[ \frac1{12}+\frac1x=\frac1{7.5}. \]Since
\[ \frac1{7.5}=\frac2{15}, \]then
\[ \frac1x = \frac2{15}-\frac1{12} = \frac8{60}-\frac5{60} = \frac1{20}. \]Therefore,
\[ \boxed{20\text{ hours}}. \]Practice 7
\[ r_{\text{net}} = \frac18-\frac1{12} = \frac3{24}-\frac2{24} = \frac1{24}. \]Therefore,
\[ \boxed{24\text{ hours}}. \]Practice 8
\[ r_{\text{net}} = \frac16+\frac1{10}-\frac1{15}. \]Using denominator \(30\),
\[ \frac5{30}+\frac3{30}-\frac2{30} = \frac6{30} = \frac15. \]Therefore,
\[ \boxed{5\text{ hours}}. \]Practice 9
\[ r_{\text{net}} = \frac1{12}-\frac18 = \frac2{24}-\frac3{24} = -\frac1{24}. \]Therefore,
\[ \boxed{ r_{\text{net}}=-\frac1{24}\text{ tank per hour} }. \]The drain is stronger than the fill pipe, so the tank cannot fill from empty while both remain open.
Practice 10
\[ \frac14-\frac1x=\frac16. \] \[ \frac1x = \frac14-\frac16 = \frac3{12}-\frac2{12} = \frac1{12}. \]Therefore,
\[ \boxed{12\text{ hours}}. \]Practice 11
\[ \frac1x-\frac1{20}=\frac1{12}. \] \[ \frac1x = \frac1{12}+\frac1{20} = \frac5{60}+\frac3{60} = \frac2{15}. \] \[ x=\frac{15}{2}. \]Therefore,
\[ \boxed{7.5\text{ hours}}. \]Practice 12
Combined rate:
\[ \frac14+\frac16=\frac5{12}. \]A's share:
\[ \frac{1/4}{5/12} = \frac35. \]B's share:
\[ \frac{1/6}{5/12} = \frac25. \]Therefore,
\[ \boxed{ A:\frac35,\qquad B:\frac25 }. \]Practice 13
\[ r=72+48=120\text{ units/hr}. \] \[ W=120(5.5)=660. \]Therefore,
\[ \boxed{660\text{ units}}. \]Practice 14
\[ r_{\text{net}}=150-18=132. \] \[ W=132(7)=924. \]Therefore,
\[ \boxed{924\text{ net units}}. \]Practice 15
Completion times do not add. Convert to rates:
\[ \frac16+\frac13 = \frac16+\frac26 = \frac12. \]Together they complete one-half of the job each hour.
Therefore,
\[ t=\frac1{1/2}=2. \]Thus,
\[ \boxed{2\text{ hours}}. \]What to Remember
Productive work rates add:
\[ \boxed{ r_{\text{combined}}=r_1+r_2+\cdots }. \]Opposing work rates subtract:
\[ \boxed{ r_{\text{net}} = r_{\text{productive}}-r_{\text{opposing}} }. \]If completion times are given, convert them first:
\[ \boxed{r=\frac1T}. \]For one complete job,
\[ \boxed{1=r_{\text{net}}t}. \]Therefore,
\[ \boxed{t=\frac1{r_{\text{net}}}} \]whenever the net rate is positive.
The most important habit is to combine rates rather than completion times, and to assign signs according to whether each process contributes to or opposes the desired work.
Where We Go Next
The problems in this lesson assumed that the relevant contributors operated over the same interval. Realistic work problems often involve delayed starts, early departures, interruptions, or unknown individual rates.
The next lesson, Staggered, Interrupted, and Unknown Work Rates, develops those piecewise work models.