Overview
Single-rate problems are the simplest class of work-rate problems. They involve one worker, machine, pipe, crew, or other process operating at a constant rate. The goal may be to determine how much work is completed, how fast the work is being performed, or how long it will take to complete the job.
These problems are governed directly by the Work Principle:
\[ \boxed{W = RT} \]
where \(W\) is the amount of work completed, \(R\) is the work rate, and \(T\) is the time spent working.
Three Forms of the Work Principle
The Work Principle can be rearranged depending on which quantity is unknown.
To find work:
\[ \boxed{W = RT} \]
To find rate:
\[ \boxed{R = \frac{W}{T}} \]
To find time:
\[ \boxed{T = \frac{W}{R}} \]
Before solving a problem, identify which of the three quantities is unknown and select the form of the equation that isolates that quantity.
When the Entire Job Equals 1
In most work-rate problems, one complete job is represented by
\[ W = 1. \]
If a worker can complete the entire job in \(t\) hours, the worker's rate is
\[ R = \frac{1}{t}. \]
For example, a worker who completes a job in \(5\) hours works at the rate
\[ R = \frac{1}{5} \]
of the job per hour.
This means that after \(1\) hour the worker has completed \(\frac{1}{5}\) of the job, after \(2\) hours the worker has completed \(\frac{2}{5}\), and after \(5\) hours the worker has completed the entire job.
Example 1: Finding the Work Rate
A machine can complete one manufacturing cycle in \(8\) hours. What is the machine's work rate?
The complete job is
\[ W = 1, \]
and the time is
\[ T = 8. \]
Use
\[ R = \frac{W}{T}. \]
Substitute the known values:
\[ R = \frac{1}{8}. \]
Therefore, the machine completes
\[ \boxed{\frac{1}{8}} \]
of the job each hour.
Example 2: Finding the Work Completed
A painter can paint one room in \(6\) hours. How much of the room can the painter complete in \(4\) hours?
First find the painter's rate:
\[ R = \frac{1}{6}. \]
The painter works for
\[ T = 4 \]
hours.
Use
\[ W = RT. \]
Substitute:
\[ W = \frac{1}{6}(4). \]
Simplify:
\[ W = \frac{4}{6} = \frac{2}{3}. \]
Therefore, the painter completes
\[ \boxed{\frac{2}{3}} \]
of the room in \(4\) hours.
A Useful Formula for Partial Work
Suppose a worker can complete an entire job in \(a\) hours. The worker's rate is
\[ R = \frac{1}{a}. \]
If the worker operates for \(t\) hours, then
\[ W = RT. \]
Substituting the rate gives
\[ W = \frac{1}{a}t. \]
Therefore,
\[ \boxed{W = \frac{t}{a}}. \]
This formula states that the fraction of the job completed equals the time worked divided by the total time required to complete the job.
Example 3: Finding the Time Required
A data-processing system completes \(\frac{3}{5}\) of a job at a constant rate of \(\frac{1}{10}\) of the job per hour. How long has the system been operating?
The known values are
\[ W = \frac{3}{5} \]
and
\[ R = \frac{1}{10}. \]
Use
\[ T = \frac{W}{R}. \]
Substitute:
\[ T = \frac{\frac{3}{5}}{\frac{1}{10}}. \]
Dividing by a fraction means multiplying by its reciprocal:
\[ T = \frac{3}{5}\times 10. \]
Therefore,
\[ T = 6. \]
The system has been operating for
\[ \boxed{6\text{ hours}}. \]
Example 4: Finding a Completion Time
A pump fills \(\frac{2}{7}\) of a reservoir in \(3\) hours. Assuming the pump operates at a constant rate, how long would it take to fill the entire reservoir?
First determine the rate:
\[ R = \frac{W}{T}. \]
Substitute:
\[ R = \frac{\frac{2}{7}}{3}. \]
Simplify:
\[ R = \frac{2}{7}\times\frac{1}{3} = \frac{2}{21}. \]
The pump fills \(\frac{2}{21}\) of the reservoir per hour.
To fill the entire reservoir, use
\[ T = \frac{W}{R}. \]
Since a full reservoir is \(W=1\),
\[ T = \frac{1}{\frac{2}{21}} = \frac{21}{2}. \]
Thus,
\[ \boxed{T = 10.5\text{ hours}}. \]
Understanding the Units
Units help explain why the Work Principle works.
Suppose the rate is
\[ \frac{1}{5}\frac{\text{job}}{\text{hour}}. \]
If the worker operates for \(3\) hours, then
\[ W = \frac{1}{5}\frac{\text{job}}{\text{hour}} \times 3\text{ hours}. \]
The hour units cancel:
\[ W = \frac{3}{5}\text{ job}. \]
This provides a useful check. A rate multiplied by time should produce an amount of work.
Work as a Fraction or Percentage
Work may be expressed as a fraction, decimal, or percentage.
For example,
\[ \frac{3}{4} = 0.75 = 75\%. \]
If a worker completes \(\frac{3}{4}\) of a job, the worker has completed \(75\%\) of the job.
Similarly, if a problem states that \(40\%\) of a job has been completed, then
\[ W = 0.40 = \frac{2}{5}. \]
Converting percentages into fractions or decimals may make the Work Principle easier to apply.
Finding the Remaining Work
If the entire job is represented by \(1\), then the remaining work is
\[ W_{\text{remaining}} = 1-W_{\text{completed}}. \]
For example, if a worker has completed
\[ \frac{3}{8} \]
of a job, then the remaining work is
\[ 1-\frac{3}{8} = \frac{5}{8}. \]
Therefore,
\[ \boxed{W_{\text{remaining}}=\frac{5}{8}}. \]
This idea becomes especially important in later problems involving interrupted work and workers who begin or stop at different times.
Method for Solving Single-Rate Problems
- Identify the complete job. If the problem concerns one complete task, represent it by \(1\).
- Identify the known quantities. Determine whether work, rate, or time is given.
- Convert a completion time into a rate if necessary. If the job takes \(t\) hours, use \(R=\frac{1}{t}\).
- Select the appropriate form of the Work Principle. Use \(W=RT\), \(R=\frac{W}{T}\), or \(T=\frac{W}{R}\).
- Substitute the known values. Keep units consistent.
- Solve and simplify. Express the result in a form appropriate to the problem.
- Interpret the answer. State whether the result represents work completed, a work rate, or a completion time.
Common Errors
- Using completion time as though it were the rate. A worker who takes \(5\) hours to complete a job has rate \(\frac{1}{5}\), not \(5\).
- Forgetting that a whole job is represented by \(1\). Fractions such as \(\frac{2}{5}\) represent portions of that whole.
- Multiplying or dividing the wrong quantities. Start from \(W=RT\) rather than relying on memory alone.
- Mixing units. Convert minutes to hours or hours to minutes when necessary before using the equation.
- Reporting a fraction without interpreting it. If \(W=\frac{3}{5}\), state that three-fifths of the job has been completed.
Practice Problems
- A machine can complete one job in \(12\) hours. What is its work rate?
- A worker can complete a project in \(9\) hours. What fraction of the project can the worker complete in \(4\) hours?
- A pump operates at a rate of \(\frac{1}{7}\) of a tank per hour. How long will it take to fill the tank?
- A printer completes \(\frac{5}{8}\) of a print run in \(10\) minutes. What is its rate in jobs per minute?
- A technician completes \(35\%\) of a task in \(2\) hours. Assuming a constant rate, what fraction of the task remains?
- A machine completes \(\frac{3}{10}\) of a job per hour. How much work does it complete in \(2.5\) hours?
- A worker completes \(\frac{7}{12}\) of a job in \(3.5\) hours. What is the worker's rate?
- A process completes \(\frac{4}{9}\) of a job in \(8\) hours. How long would the entire job require at the same rate?
Practice Answers
- \[ \boxed{\frac{1}{12}\text{ job per hour}} \]
- \[ \boxed{\frac{4}{9}} \]
- \[ \boxed{7\text{ hours}} \]
- \[ R = \frac{\frac{5}{8}}{10} = \boxed{\frac{1}{16}\text{ job per minute}} \]
- \[ 1-0.35=0.65 \] so \[ \boxed{65\%} \] remains.
- \[ W = \frac{3}{10}(2.5) = \boxed{\frac{3}{4}} \]
- \[ R = \frac{\frac{7}{12}}{\frac{7}{2}} = \boxed{\frac{1}{6}\text{ job per hour}} \]
- \[ R = \frac{\frac{4}{9}}{8} = \frac{1}{18} \] so \[ \boxed{18\text{ hours}} \]
Summary
Single-rate problems involve one process operating at a constant rate. Every problem can be modeled using
\[ \boxed{W=RT}. \]
If a complete job requires \(t\) units of time, then the rate is
\[ \boxed{R=\frac{1}{t}}. \]
If a worker who requires \(a\) hours to complete the full job works for \(t\) hours, the amount completed is
\[ \boxed{W=\frac{t}{a}}. \]
Once these relationships are understood, the next step is to examine what happens when two or more work rates operate at the same time.
Continue
Continue to 2. Combined-Rate Problems