The Central Idea
Work, rate, and time are related by the same general structure used in distance-rate-time problems:
\[ \boxed{\text{work}=\text{rate}\cdot\text{time}}. \]Using \(W\) for the amount of work completed, \(r\) for the work rate, and \(t\) for time,
\[ \boxed{W=rt}. \]The main difference is that a complete job is usually represented by
\[ \boxed{W=1}. \]Representing One Complete Job
In most work problems, the entire task is treated as one unit of work:
\[ 1\text{ job}. \]Half the job is
\[ \frac12, \]one-quarter of the job is
\[ \frac14, \]and \(80\%\) of the job is
\[ 0.80. \]Fractions and decimals therefore describe how much of the whole task has been completed.
Completion Time and Work Rate
Suppose one worker can complete one job in \(T\) hours.
Its work rate is
\[ \boxed{r=\frac{1}{T}\text{ job per hour}}. \]Completion time and work rate are reciprocals.
A smaller completion time means a larger work rate, while a larger completion time means a smaller work rate.
Example: Convert Completion Time to Rate
A machine can complete one job in \(8\) hours.
Its rate is
\[ r=\frac{1}{8}. \]Therefore,
\[ \boxed{ r=\frac18\text{ job per hour} }. \]This means the machine completes one-eighth of the job during each hour of constant operation.
Example: Find Work Completed after a Given Time
A worker completes one job in \(6\) hours. How much of the job is completed in \(2\) hours?
The rate is
\[ \frac16\text{ job per hour}. \]Therefore,
\[ W=rt = \frac16(2) = \frac13. \]Thus,
\[ \boxed{\frac13\text{ of the job}} \]is completed.
Example: Find the Time from a Work Rate
A machine works at a rate of
\[ \frac{1}{12}\text{ job per hour}. \]How long does one complete job take?
Use
\[ 1=\frac{1}{12}t. \]Therefore,
\[ t=12. \]Thus,
\[ \boxed{12\text{ hours}}. \]The Three Equivalent Forms
From
\[ W=rt, \]we can solve for any variable:
\[ \boxed{W=rt}, \] \[ \boxed{r=\frac{W}{t}}, \]and
\[ \boxed{t=\frac{W}{r}}. \]When the whole job is \(W=1\),
\[ \boxed{t=\frac{1}{r}}. \]Why Work Rates Are Often Fractions
A worker usually completes only part of a full job during one unit of time.
If a job takes \(10\) hours, then in one hour the worker completes
\[ \frac{1}{10} \]of the job.
In \(3\) hours, the completed fraction is
\[ 3\left(\frac{1}{10}\right) = \frac{3}{10}. \]Fractions are therefore the natural language of work-rate problems.
Work Tables
A work table organizes the same three quantities used in the equation \(W=rt\).
\[ \begin{array}{c|c|c|c} \text{worker or machine} & \text{rate} & \text{time} & \text{work}\\ \hline A & r_A & t_A & r_At_A\\ B & r_B & t_B & r_Bt_B \end{array} \]Each row represents the amount of work completed by one contributor.
Later lessons will use these tables to combine several work rates.
Rate from a Known Fraction of a Job
A worker completes
\[ \frac{3}{5} \]of a job in \(4\) hours.
The rate is
\[ r=\frac{W}{t} = \frac{3/5}{4}. \]Therefore,
\[ r=\frac{3}{20}. \]Thus,
\[ \boxed{ r=\frac{3}{20}\text{ job per hour} }. \]Completion Time from a Partial-Work Observation
If a worker completes
\[ \frac{3}{20} \]of a job each hour, then the time for one whole job is
\[ t=\frac{1}{3/20}. \]Dividing by a fraction gives
\[ t=\frac{20}{3}. \]Therefore,
\[ \boxed{ t=6\frac23\text{ hours} }. \]Percent of a Job
Work completed can also be expressed as a percent.
If a worker completes \(35\%\) of a job in \(2\) hours, then
\[ W=0.35. \]The rate is
\[ r=\frac{0.35}{2}=0.175. \]Therefore,
\[ \boxed{ r=0.175\text{ job per hour} }. \]From Percent Rate to Completion Time
Suppose a machine completes \(12.5\%\) of a job each hour.
Convert to decimal:
\[ 12.5\%=0.125=\frac18. \]Therefore, the machine completes
\[ \frac18 \]of the job per hour.
The full job takes
\[ \boxed{8\text{ hours}}. \]Work Rate Is a Unit Rate
A work rate is another example of a unit rate.
The expression
\[ \frac{1}{5}\frac{\text{job}}{\text{hour}} \]means one-fifth of a job per hour.
Multiplying by time gives work:
\[ \frac{1}{5}\frac{\text{job}}{\text{hour}} (3\text{ hr}) = \frac35\text{ job}. \]The hour units cancel exactly as they do in distance-rate-time problems.
Units Must Match
If a rate is measured in jobs per hour, time must be expressed in hours.
Suppose a machine works at
\[ \frac14\text{ job per hour} \]for \(90\) minutes.
Convert:
\[ 90\text{ min}=1.5\text{ hr}. \]Then
\[ W=\frac14(1.5)=\frac38. \]Therefore,
\[ \boxed{\frac38\text{ of the job}} \]is completed.
Work Rate and Direct Variation
At a constant work rate,
\[ W=rt. \]Thus work completed varies directly with time.
For fixed \(r\),
\[ \frac{W}{t}=r. \]Doubling the working time doubles the amount of work completed, provided the rate remains constant.
Completion Time and Rate Vary Inversely
For one complete job,
\[ 1=rt. \]Therefore,
\[ \boxed{t=\frac{1}{r}}. \]Completion time varies inversely with work rate.
Doubling a constant work rate cuts the completion time in half.
Example: Doubling the Rate
A machine normally completes one job in \(12\) hours.
Its original rate is
\[ \frac{1}{12}. \]If an improved process doubles the rate, the new rate is
\[ \frac{2}{12}=\frac16. \]Therefore, the new completion time is
\[ \boxed{6\text{ hours}}. \]Example: Increasing the Rate by \(25\%\)
A process completes a job in \(10\) hours.
Its original rate is
\[ r=\frac{1}{10}. \]Increasing the rate by \(25\%\) gives
\[ r_{\text{new}} = 1.25\left(\frac{1}{10}\right) = \frac18. \]Therefore, the new completion time is
\[ \boxed{8\text{ hours}}. \]A \(25\%\) rate increase does not produce a \(25\%\) reduction in time.
Example: Decreasing the Rate
A machine normally completes a job in \(6\) hours.
Its rate is
\[ \frac16. \]If its rate decreases by \(20\%\), the new rate is
\[ 0.80\left(\frac16\right) = \frac{2}{15}. \]The new completion time is
\[ \frac{1}{2/15} = \frac{15}{2}. \]Therefore,
\[ \boxed{7.5\text{ hours}}. \]Work Completed over Several Intervals
If the rate changes over time, calculate the work completed during each interval separately.
Suppose a machine works for \(2\) hours at
\[ \frac16\text{ job per hour} \]and then \(3\) hours at
\[ \frac18\text{ job per hour}. \]The total work completed is
\[ \frac16(2)+\frac18(3). \]Therefore,
\[ W=\frac13+\frac38 = \frac{17}{24}. \]Thus,
\[ \boxed{\frac{17}{24}\text{ of the job}} \]is completed.
Remaining Work
If a fraction \(W\) of a job has already been completed, the remaining fraction is
\[ \boxed{1-W}. \]If
\[ W=\frac{17}{24}, \]then the remaining work is
\[ 1-\frac{17}{24} = \frac{7}{24}. \]Therefore,
\[ \boxed{\frac{7}{24}\text{ of the job remains}}. \]Time Needed to Finish the Remaining Work
Suppose
\[ \frac{7}{24} \]of a job remains, and a machine works at
\[ \frac18\text{ job per hour}. \]Use
\[ t=\frac{W}{r}. \]Then
\[ t= \frac{7/24}{1/8} = \frac{7}{24}\cdot8 = \frac73. \]Therefore,
\[ \boxed{2\frac13\text{ hours}}. \]Jobs Need Not Be Identical in Size
The convention \(W=1\) works when one full task is the chosen unit.
If a problem compares different-sized jobs, define the work unit clearly.
For example, a production process might be measured in
\[ \text{units produced per hour} \]rather than fractions of one job.
The underlying equation remains
\[ W=rt. \]Production Problems
A machine produces \(45\) components per hour.
In \(6\) hours, it produces
\[ 45(6)=270. \]Therefore,
\[ \boxed{270\text{ components}}. \]This is structurally the same work-rate relationship, but the work unit is an individual component rather than one complete job.
Example: Find Production Time
A machine produces \(32\) units per hour. How long does it take to produce \(240\) units?
Use
\[ t=\frac{W}{r}. \]Then
\[ t=\frac{240}{32}=7.5. \]Therefore,
\[ \boxed{7.5\text{ hours}}. \]Example: Find Required Production Rate
A plant must produce \(900\) units in \(12\) hours.
The required average production rate is
\[ r=\frac{900}{12}=75. \]Therefore,
\[ \boxed{75\text{ units per hour}}. \]Efficiency as a Rate Multiplier
Some work problems describe one worker or machine as a multiple of another worker's rate.
If machine B works \(1.5\) times as fast as machine A, and A has rate \(r\), then
\[ \boxed{r_B=1.5r}. \]If A requires time \(T\) for one job, then
\[ r=\frac1T. \]Therefore,
\[ r_B=\frac{1.5}{T}. \]Example: Faster Worker and Completion Time
Worker A completes a job in \(12\) hours. Worker B works twice as fast as A.
A's rate is
\[ \frac1{12}. \]B's rate is
\[ 2\left(\frac1{12}\right) = \frac16. \]Therefore, B completes the job in
\[ \boxed{6\text{ hours}}. \]Example: Slower Worker and Completion Time
Worker B works at \(80\%\) of worker A's rate. A completes a job in \(10\) hours.
A's rate is
\[ \frac1{10}. \]B's rate is
\[ 0.8\left(\frac1{10}\right) = \frac{2}{25}. \]Therefore, B's completion time is
\[ \frac{1}{2/25} = \frac{25}{2} = 12.5. \]Thus,
\[ \boxed{12.5\text{ hours}}. \]Rate Comparisons and Time Comparisons
For the same fixed job, completion times and work rates are reciprocal.
If worker A is twice as fast as worker B, then
\[ r_A=2r_B. \]Therefore,
\[ T_A=\frac{1}{r_A} = \frac{1}{2r_B} = \frac12T_B. \]Twice the rate means half the completion time.
Example: Relating Two Completion Times
Worker A works \(25\%\) faster than worker B.
Then
\[ r_A=1.25r_B. \]Their completion times satisfy
\[ T_A = \frac{1}{1.25r_B} = 0.8T_B. \]Therefore, A requires
\[ \boxed{80\%} \]of B's completion time.
Work Models with Variables
Suppose a worker requires \(x\) hours to complete a job.
The worker's rate is
\[ \boxed{\frac1x}. \]If the worker operates for \(t\) hours, the completed work is
\[ \boxed{\frac{t}{x}}. \]This expression appears repeatedly in more advanced work problems.
Example: Variable Completion Time
A machine can complete one job in \(x\) hours.
How much of the job does it complete in \(3\) hours?
Its rate is
\[ \frac1x. \]Therefore,
\[ W=\frac1x(3). \]Thus,
\[ \boxed{W=\frac3x}. \]Example: Completion Time Given a Fraction Completed
A machine completes
\[ \frac{5}{12} \]of a job in \(2.5\) hours. How long would one full job take at the same constant rate?
First find the rate:
\[ r= \frac{5/12}{2.5}. \]Since
\[ 2.5=\frac52, \]we get
\[ r= \frac{5}{12}\cdot\frac25 = \frac16. \]Therefore, one full job takes
\[ \boxed{6\text{ hours}}. \]Example: How Much Work Remains?
A worker who can complete a job in \(9\) hours works for \(5\) hours.
Work completed:
\[ \frac19(5)=\frac59. \]Work remaining:
\[ 1-\frac59=\frac49. \]Therefore,
\[ \boxed{\frac49\text{ of the job remains}}. \]Example: Finish Time at a New Rate
A worker completes
\[ \frac59 \]of a job. The remaining
\[ \frac49 \]is completed at a new rate of
\[ \frac{2}{9}\text{ job per hour}. \]The additional time is
\[ t= \frac{4/9}{2/9}=2. \]Therefore,
\[ \boxed{2\text{ additional hours}}. \]Work Rates Can Be Added
If two independent workers contribute positively to the same job at the same time, their work completed during an interval can be added.
For example, in one hour, rates
\[ \frac16 \]and
\[ \frac14 \]would contribute a total of
\[ \frac16+\frac14 = \frac{5}{12} \]of a job.
The next lesson develops combined and opposing work rates systematically.
Why Completion Times Are Not Added
If one worker takes \(6\) hours and another takes \(4\) hours, their completion times are not the quantities that combine directly.
Their rates are
\[ \frac16 \]and
\[ \frac14. \]Work contributions are measured in fractions of a job per unit time, so the rates are the natural quantities to combine.
Constant Rate Is an Assumption
The basic equation
\[ W=rt \]assumes that \(r\) remains constant over the relevant interval.
Real workers and machines may speed up, slow down, stop, or change efficiency.
Applied algebra problems usually idealize these effects into constant-rate segments that can be modeled separately.
Efficiency and Capacity Limits
A calculated work rate is meaningful only within the assumptions of the model.
A machine rated at \(40\) units per hour may not sustain that output indefinitely, and two workers together may not always work independently without interference.
Algebraic work models should therefore be interpreted as structured approximations of the stated conditions.
Checking with the Whole Job
If a worker's rate is claimed to be
\[ \frac17\text{ job per hour}, \]then after \(7\) hours,
\[ W= \frac17(7)=1. \]The model reproduces one whole job, confirming the reciprocal relationship.
Checking with Fractions
Suppose a worker takes \(8\) hours per job and works \(3\) hours.
The completed fraction is
\[ \frac38. \]Since \(3<8\), the completed amount should be less than one whole job.
This quick comparison provides a useful reasonableness check.
Checking with Units
A rate of
\[ \frac16\frac{\text{job}}{\text{hr}} \]multiplied by
\[ 4\text{ hr} \]gives
\[ \frac46\text{ job} = \frac23\text{ job}. \]The time unit cancels, leaving work.
If the units do not simplify to the requested quantity, recheck the setup.
Common Error: Using Completion Time as the Rate
If a job takes \(5\) hours, the work rate is not
\[ 5\text{ jobs per hour}. \]The correct rate is the reciprocal:
\[ \boxed{ \frac15\text{ job per hour} }. \]Common Error: Forgetting that the Whole Job Is \(1\)
If a worker completes
\[ \frac38 \]of a job, the remaining work is not
\[ 8-3. \]The complete job is \(1\), so
\[ 1-\frac38 = \frac58. \]Therefore,
\[ \boxed{\frac58\text{ remains}}. \]Common Error: Adding Completion Times
Completion times are reciprocals of rates.
If one worker takes \(6\) hours and another takes \(3\) hours, the useful quantities are
\[ \frac16 \]and
\[ \frac13 \]job per hour.
Adding \(6+3\) does not describe their combined productivity.
Common Error: Confusing Rate Increase with Time Decrease
If a rate increases by \(20\%\), the completion time does not automatically decrease by \(20\%\).
Since
\[ t=\frac1r, \]the new time is found by taking the reciprocal of the new rate.
A \(20\%\) rate increase multiplies rate by \(1.2\), so time is multiplied by
\[ \frac1{1.2} = \frac56. \]Common Error: Mixing Hours and Minutes
If a work rate is measured per hour and the work interval is \(30\) minutes, convert:
\[ 30\text{ min}=0.5\text{ hr}. \]A rate of
\[ \frac15\text{ job per hour} \]for \(30\) minutes completes
\[ \frac15(0.5) = \frac1{10} \]of the job.
Common Error: Treating \(50\%\) Faster as Half the Time
If one worker is \(50\%\) faster, the new rate is
\[ 1.5r. \]The new time is
\[ \frac{1}{1.5r} = \frac23\left(\frac1r\right). \]Therefore, a \(50\%\) increase in rate reduces completion time to
\[ \boxed{\frac23} \]of the original time, not one-half.
Common Error: Allowing Work to Exceed the Task without Interpretation
If a model for one finite job gives
\[ W>1, \]then the worker has enough time to complete more than one equivalent job, or the model has been applied beyond the completion point.
For a single-task problem, stop the interpretation when
\[ W=1. \]A Reliable Procedure for Basic Work Problems
- Define one complete job as \(W=1\).
- Identify the completion time or work rate.
- If completion time is given, convert it to rate using \(r=1/T\).
- Make all time units compatible.
- Use \(W=rt\).
- Solve for the requested quantity.
- Interpret the result as work, rate, or time.
- Check that the fraction of work and units make sense.
A Reliable Procedure for Partial-Work Problems
- Write the completed portion as a fraction or decimal of one job.
- Write the remaining portion as \(1-W\) when needed.
- Use \(r=W/t\) to find a rate from observed work.
- Use \(t=W/r\) to find the time needed for a known amount of work.
- Separate intervals when the work rate changes.
- Add completed work across intervals.
- Verify that the total completed plus remaining work equals \(1\).
A Reliable Procedure for Rate-Comparison Problems
- Choose a base rate \(r\).
- Translate statements such as “twice as fast” into rate multipliers.
- Translate percent rate changes using factors such as \(1.25r\) or \(0.8r\).
- Use reciprocal relationships to convert rates back to completion times.
- Do not apply the same percentage change directly to time.
- Check that faster rates correspond to shorter times.
Practice 1: Convert Time to Work Rate
A worker can complete one job in \(9\) hours. Find the work rate.
Practice 2: Work Completed
A machine completes one job in \(8\) hours. What fraction of the job does it complete in \(3\) hours?
Practice 3: Find Completion Time
A machine works at a rate of
\[ \frac{1}{14}\text{ job per hour}. \]How long does one full job take?
Practice 4: Partial Work to Rate
A worker completes
\[ \frac25 \]of a job in \(3\) hours. Find the worker's rate.
Practice 5: Partial Work to Full Completion Time
Using Practice 4, find the time required for one full job at the same rate.
Practice 6: Percent of a Job
A machine completes \(30\%\) of a job in \(2\) hours. Find its rate as a decimal number of jobs per hour.
Practice 7: Remaining Work
A worker who completes one job in \(10\) hours works for \(6\) hours. What fraction of the job remains?
Practice 8: Finish the Remaining Work
A project has
\[ \frac38 \]of its work remaining. A machine completes work at
\[ \frac18 \]of a job per hour. How long will the remaining work take?
Practice 9: Production Rate
A machine produces \(54\) units per hour for \(7.5\) hours. How many units are produced?
Practice 10: Required Production Rate
A facility must produce \(1{,}080\) units in \(12\) hours. Find the required average production rate.
Practice 11: Faster Worker
Worker A completes a job in \(15\) hours. Worker B works \(50\%\) faster than A. Find B's completion time.
Practice 12: Slower Worker
Worker A completes a job in \(8\) hours. Worker B works at \(80\%\) of A's rate. Find B's completion time.
Practice 13: Rate Increase
A machine completes a job in \(20\) hours. Its rate increases by \(25\%\). Find the new completion time.
Practice 14: Changing Rates across Intervals
A machine works \(2\) hours at
\[ \frac15 \]job per hour and then \(3\) hours at
\[ \frac1{10} \]job per hour. What fraction of the job is completed?
Practice 15: Identify the Error
A worker can complete a job in \(4\) hours. A student says the worker's rate is \(4\) jobs per hour.
Explain the error and give the correct rate.
Practice Answers
Practice 1
The rate is the reciprocal of the completion time:
\[ \boxed{ r=\frac19\text{ job per hour} }. \]Practice 2
\[ W=\frac18(3)=\frac38. \]Therefore,
\[ \boxed{\frac38\text{ of the job}}. \]Practice 3
\[ 1=\frac1{14}t. \] \[ t=14. \]Therefore,
\[ \boxed{14\text{ hours}}. \]Practice 4
\[ r=\frac{2/5}{3} = \frac{2}{15}. \]Therefore,
\[ \boxed{ r=\frac{2}{15}\text{ job per hour} }. \]Practice 5
\[ t=\frac{1}{2/15} = \frac{15}{2}. \]Therefore,
\[ \boxed{7.5\text{ hours}}. \]Practice 6
\[ r=\frac{0.30}{2}=0.15. \]Therefore,
\[ \boxed{0.15\text{ job per hour}}. \]Practice 7
The worker completes
\[ \frac1{10}(6)=\frac35. \]Therefore, the remaining work is
\[ 1-\frac35=\frac25. \]Thus,
\[ \boxed{\frac25}. \]Practice 8
\[ t= \frac{3/8}{1/8}=3. \]Therefore,
\[ \boxed{3\text{ hours}}. \]Practice 9
\[ W=54(7.5)=405. \]Therefore,
\[ \boxed{405\text{ units}}. \]Practice 10
\[ r=\frac{1080}{12}=90. \]Therefore,
\[ \boxed{90\text{ units per hour}}. \]Practice 11
A's rate is
\[ \frac1{15}. \]B's rate is
\[ 1.5\left(\frac1{15}\right) = \frac1{10}. \]Therefore,
\[ \boxed{10\text{ hours}}. \]Practice 12
A's rate is
\[ \frac18. \]B's rate is
\[ 0.8\left(\frac18\right) = \frac1{10}. \]Therefore,
\[ \boxed{10\text{ hours}}. \]Practice 13
Original rate:
\[ \frac1{20}. \]New rate:
\[ 1.25\left(\frac1{20}\right) = \frac1{16}. \]Therefore,
\[ \boxed{16\text{ hours}}. \]Practice 14
First interval:
\[ \frac15(2)=\frac25. \]Second interval:
\[ \frac1{10}(3)=\frac3{10}. \]Total:
\[ \frac25+\frac3{10} = \frac4{10}+\frac3{10} = \frac7{10}. \]Therefore,
\[ \boxed{\frac7{10}\text{ of the job}}. \]Practice 15
Four hours is the time for one job, not the number of jobs completed each hour.
The correct rate is the reciprocal:
\[ \boxed{ \frac14\text{ job per hour} }. \]What to Remember
Work, rate, and time satisfy
\[ \boxed{W=rt}. \]For one complete job,
\[ \boxed{W=1}. \]If a worker completes one job in \(T\) units of time, the work rate is
\[ \boxed{r=\frac1T}. \]Work completed in time \(t\) is
\[ \boxed{W=\frac{t}{T}}. \]Completion time and work rate are reciprocal quantities:
\[ \boxed{t=\frac1r} \]for one whole job.
The most important habit is to convert completion times into rates before combining or comparing productive activity.
Where We Go Next
A single work rate describes one worker or machine. Many practical problems involve several contributors operating together, or one process undoing work performed by another.
- Single, Combined, and Opposing Work Rates, develops the algebra for adding productive rates, subtracting opposing rates, and solving for unknown completion times.
- Staggered, Interrupted, and Unknown Work Rates, develops those piecewise work models.