Overview
Combined-rate problems involve two or more workers, machines, or processes contributing toward the same task at the same time. Each participant performs some fraction of the total job during each unit of time, and those contributions are added to determine the total rate of work.
The key principle is:
\[ \boxed{R_{\text{combined}} = R_1 + R_2 + R_3 + \cdots} \]
Once the combined rate is known, the ordinary Work Principle applies:
\[ \boxed{W = RT}. \]
Why the Rates Add
Suppose Worker A completes \(\frac{1}{6}\) of a job each hour and Worker B completes \(\frac{1}{4}\) of the same job each hour. If they work together for one hour, Worker A contributes \(\frac{1}{6}\) of the job while Worker B contributes \(\frac{1}{4}\).
The total amount completed during that hour is therefore
\[ \frac{1}{6}+\frac{1}{4}. \]
Using a common denominator,
\[ \frac{1}{6}+\frac{1}{4} = \frac{2}{12}+\frac{3}{12} = \frac{5}{12}. \]
Together, the workers complete \(\frac{5}{12}\) of the job each hour.
The completion times themselves are not added. The quantities that describe simultaneous productivity are the rates.
Convert Completion Times to Rates First
Combined-rate problems often provide the time each worker would require to complete the job alone. Those times must first be converted into rates.
If Worker A completes the job in \(a\) hours, then
\[ R_A = \frac{1}{a}. \]
If Worker B completes the job in \(b\) hours, then
\[ R_B = \frac{1}{b}. \]
Their combined rate is
\[ R_{\text{combined}} = \frac{1}{a}+\frac{1}{b}. \]
This conversion from time per job to jobs per unit of time is the essential first step.
The Standard Combined-Rate Formula
Suppose Worker A can complete a job in \(a\) hours, Worker B can complete the same job in \(b\) hours, and together they complete it in \(t\) hours.
Their combined rate is
\[ \frac{1}{a}+\frac{1}{b}. \]
Since the entire job is completed in \(t\) hours,
\[ 1 = t\left(\frac{1}{a}+\frac{1}{b}\right). \]
Dividing by \(t\) gives
\[ \boxed{\frac{1}{a}+\frac{1}{b}=\frac{1}{t}}. \]
This is the standard two-worker combined-rate equation.
Example 1: Two Workers Together
Worker A can complete a job in \(6\) hours. Worker B can complete the same job in \(4\) hours. How long will they take if they work together?
Worker A's rate is
\[ R_A = \frac{1}{6}. \]
Worker B's rate is
\[ R_B = \frac{1}{4}. \]
Add the rates:
\[ R_{\text{combined}} = \frac{1}{6}+\frac{1}{4}. \]
Using a common denominator,
\[ R_{\text{combined}} = \frac{2}{12}+\frac{3}{12} = \frac{5}{12}. \]
Now use
\[ T=\frac{W}{R}. \]
Since \(W=1\),
\[ T = \frac{1}{\frac{5}{12}} = \frac{12}{5}. \]
Therefore,
\[ \boxed{T=2.4\text{ hours}}. \]
The two workers complete the job together in \(2\) hours and \(24\) minutes.
A Useful Reasonableness Check
When two workers contribute positively toward the same job, their combined completion time must be less than the completion time of the faster worker working alone.
In the previous example, the faster worker requires \(4\) hours. Therefore, the combined time must satisfy
\[ T<4. \]
The result
\[ T=2.4 \]
passes this test.
If a calculation had produced \(5\) hours, \(6\) hours, or any larger time, that would signal an error.
Example 2: Two Machines
Machine A can complete a production batch in \(10\) hours. Machine B can complete the same batch in \(15\) hours. How long will the batch take if both machines operate together?
The rates are
\[ R_A=\frac{1}{10} \]
and
\[ R_B=\frac{1}{15}. \]
Add them:
\[ R_{\text{combined}} = \frac{1}{10}+\frac{1}{15}. \]
The least common denominator is \(30\):
\[ R_{\text{combined}} = \frac{3}{30}+\frac{2}{30} = \frac{5}{30} = \frac{1}{6}. \]
The combined rate is therefore one-sixth of the batch per hour.
The completion time is
\[ T = \frac{1}{\frac{1}{6}} = 6. \]
Thus,
\[ \boxed{T=6\text{ hours}}. \]
Three or More Combined Rates
The same principle extends to any number of workers or machines.
If three workers have rates \(R_1\), \(R_2\), and \(R_3\), then
\[ R_{\text{combined}} = R_1+R_2+R_3. \]
If their individual completion times are \(a\), \(b\), and \(c\), then
\[ R_{\text{combined}} = \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \]
For a complete job,
\[ 1 = T\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right). \]
Example 3: Three Workers
Three workers can complete a job individually in \(6\), \(8\), and \(12\) hours. How long will they take working together?
Their rates are
\[ \frac{1}{6}, \qquad \frac{1}{8}, \qquad \frac{1}{12}. \]
The combined rate is
\[ R = \frac{1}{6}+\frac{1}{8}+\frac{1}{12}. \]
The least common denominator is \(24\):
\[ R = \frac{4}{24}+\frac{3}{24}+\frac{2}{24} = \frac{9}{24} = \frac{3}{8}. \]
Therefore,
\[ T = \frac{1}{\frac{3}{8}} = \frac{8}{3}. \]
So the workers complete the job in
\[ \boxed{\frac{8}{3}\text{ hours}} \]
or \(2\) hours and \(40\) minutes.
Finding an Unknown Work Rate
Sometimes the combined completion time and one worker's completion time are known, while the other worker's rate or completion time must be found.
For two workers,
\[ R_A+R_B=R_{\text{combined}}. \]
Therefore,
\[ R_B = R_{\text{combined}}-R_A. \]
Once the unknown rate is found, its reciprocal gives the worker's individual completion time.
Example 4: Finding an Unknown Completion Time
Worker A can complete a job in \(6\) hours. Working together, Workers A and B complete the job in \(4\) hours. How long would Worker B take to complete the job alone?
Worker A's rate is
\[ R_A=\frac{1}{6}. \]
The combined rate is
\[ R_{\text{combined}}=\frac{1}{4}. \]
Subtract Worker A's rate:
\[ R_B = \frac{1}{4}-\frac{1}{6}. \]
Using a common denominator,
\[ R_B = \frac{3}{12}-\frac{2}{12} = \frac{1}{12}. \]
Therefore Worker B's completion time is
\[ T_B = \frac{1}{\frac{1}{12}} = 12. \]
Thus,
\[ \boxed{T_B=12\text{ hours}}. \]
Combined Rates for Part of a Job
Workers do not always continue until the entire job is complete. If several workers operate together for only a fixed period of time, use
\[ W=RT \]
with the combined rate.
For example, suppose two workers have rates
\[ \frac{1}{8} \qquad\text{and}\qquad \frac{1}{12}. \]
Their combined rate is
\[ \frac{1}{8}+\frac{1}{12} = \frac{3}{24}+\frac{2}{24} = \frac{5}{24}. \]
If they work together for \(3\) hours, then
\[ W = \frac{5}{24}(3) = \frac{15}{24} = \frac{5}{8}. \]
They complete
\[ \boxed{\frac{5}{8}} \]
of the job.
Why You Cannot Average Completion Times
A common mistake is to average the individual completion times. Suppose one worker requires \(6\) hours and another requires \(4\) hours. Their arithmetic mean is
\[ \frac{6+4}{2}=5. \]
But \(5\) hours cannot be the combined completion time because one worker alone can already complete the job in \(4\) hours.
The error occurs because completion times measure hours per job, while combined work requires adding jobs per hour.
The correct rates are
\[ \frac{1}{6} \qquad\text{and}\qquad \frac{1}{4}, \]
and these rates, not the completion times, must be combined.
Method for Solving Combined-Rate Problems
- Represent the entire job by \(1\).
- Identify each individual completion time.
- Convert each completion time into a work rate. Use \(R=\frac{1}{t}\).
- Add the contributing rates.
- Use the Work Principle. If the entire job is completed, use \(1=R_{\text{combined}}T\).
- Solve for the unknown quantity.
- Check whether the result is reasonable. The combined completion time should be less than the fastest individual completion time when all workers contribute positively for the entire duration.
Common Errors
- Adding completion times instead of rates. Convert each completion time into its reciprocal before combining them.
- Averaging the completion times. Combined productivity is determined by rate addition, not by averaging hours.
- Adding fractions without a common denominator. Rates such as \(\frac{1}{6}\) and \(\frac{1}{4}\) must be rewritten with a common denominator before addition.
- Forgetting to take the reciprocal at the end. A combined rate such as \(\frac{5}{12}\) means \(\frac{5}{12}\) of a job per hour; it does not mean the job takes \(\frac{5}{12}\) of an hour.
- Failing to check the answer. A combined time greater than the faster individual time usually indicates an error when all rates are positive.
Practice Problems
- Worker A can complete a job in \(8\) hours and Worker B can complete it in \(12\) hours. How long will they take working together?
- Two machines can complete a production run individually in \(9\) hours and \(18\) hours. How long will the run take if both machines operate together?
- Three workers can complete a task individually in \(6\), \(10\), and \(15\) hours. What is their combined work rate?
- Worker A can complete a job in \(5\) hours. Worker B can complete it in \(20\) hours. What fraction of the job will they complete together in \(2\) hours?
- A machine completes a job in \(8\) hours. When another machine operates with it, the job takes \(4\) hours. How long would the second machine take to complete the job alone?
- Worker A completes a job in \(10\) hours and Worker B completes it in \(15\) hours. If they work together for \(3\) hours, what fraction of the job remains?
- Three pumps can fill a reservoir individually in \(12\), \(18\), and \(36\) hours. How long will the reservoir take to fill if all three operate together?
- Two workers together complete a job in \(3\) hours. One worker alone requires \(5\) hours. How long would the other worker require alone?
Practice Answers
- \[ R = \frac{1}{8}+\frac{1}{12} = \frac{5}{24} \] so \[ T = \frac{24}{5} = \boxed{4.8\text{ hours}}. \]
- \[ R = \frac{1}{9}+\frac{1}{18} = \frac{1}{6} \] so \[ \boxed{6\text{ hours}}. \]
- \[ R = \frac{1}{6}+\frac{1}{10}+\frac{1}{15} = \frac{5}{30}+\frac{3}{30}+\frac{2}{30} = \boxed{\frac{1}{3}}. \]
- \[ R = \frac{1}{5}+\frac{1}{20} = \frac{1}{4} \] so \[ W = \frac{1}{4}(2) = \boxed{\frac{1}{2}}. \]
- \[ \frac{1}{8}+R_B=\frac{1}{4} \] so \[ R_B=\frac{1}{8}. \] Therefore, \[ \boxed{8\text{ hours}}. \]
- \[ R = \frac{1}{10}+\frac{1}{15} = \frac{1}{6}. \] In \(3\) hours, \[ W=\frac{1}{6}(3)=\frac{1}{2}. \] Therefore, \[ \boxed{\frac{1}{2}} \] of the job remains.
- \[ R = \frac{1}{12}+\frac{1}{18}+\frac{1}{36} = \frac{3}{36}+\frac{2}{36}+\frac{1}{36} = \frac{1}{6}. \] Therefore, \[ \boxed{6\text{ hours}}. \]
- \[ R_B = \frac{1}{3}-\frac{1}{5} = \frac{2}{15}. \] Therefore, \[ T_B = \frac{1}{\frac{2}{15}} = \boxed{\frac{15}{2}\text{ hours}} \] or \(7.5\) hours.
Summary
Combined-rate problems involve several positive work rates operating toward the same goal. The fundamental relationship is
\[ \boxed{R_{\text{combined}}=R_1+R_2+\cdots}. \]
If the workers have individual completion times \(a\), \(b\), and \(c\), then
\[ R_{\text{combined}} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. \]
For two workers completing one full job together in \(t\) hours,
\[ \boxed{\frac{1}{a}+\frac{1}{b}=\frac{1}{t}}. \]
The central idea is simple: completion times are converted into rates, and simultaneous contributing rates are added.
The next problem type modifies this idea by introducing processes that work in opposite directions.
Continue
Continue to 3. Opposing-Rate Problems
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