9.3: Work-rate problems
- Page ID
- 45129
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)If it takes one person \(4\) hours to paint a room and another person \(12\) hours to paint the same room, working together they could paint the room even quicker. As it turns out, they would paint the room in \(3\) hours together. This is reasoned by the following logic. If the first person paints the room in \(4\) hours, she paints \(\dfrac{1}{4}\) of the room each hour. If the second person takes \(12\) hours to paint the room, he paints \(\dfrac{1}{12}\) of the room each hour. So together, each hour they paint \(\dfrac{1}{4}+\dfrac{1}{12}\) of the room. Let’s simplify this sum:
\[\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\nonumber\]
This means each hour, working together, they complete \(\dfrac{1}{3}\) of the room. If \(\dfrac{1}{3}\) of the room is painted each hour, it follows that it will take \(3\) hours to complete the entire room.
If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total). We can use the work-rate equation:
\[\underset{\text{job per time A}}{\underbrace{\dfrac{1}{A}}}+\underset{\text{job per time B}}{\underbrace{\dfrac{1}{B}}}=\underset{\text{job per time T}}{\underbrace{\dfrac{1}{T}}}\nonumber\]
The Egyptians were the first to work with fractions. When the Egyptians wrote fractions, they were all unit fractions (a numerator of one). They used these types of fractions for about 2,000 years. Some believe that this cumbersome style of using fractions was used for so long out of tradition. Others believe the Egyptians had a way of thinking about and working with fractions that has been completely lost in history.
One Unknown Time
Adam can clean a room in 3 hours. If his sister Maria helps, they can clean it in \(2\dfrac{2}{5}\) hours. How long will it take Maria to do the job alone?
Solution
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table:
| time | job per hour | |
|---|---|---|
| Adam | \(3\) | \(\dfrac{1}{3}\) |
| Maria | \(t\) | \(\dfrac{1}{t}\) |
| Together | \(2\dfrac{2}{5}\) | \(\dfrac{1}{2\dfrac{2}{5}}\) |
Now, let’s set up the equation and solve. Notice, \(\dfrac{1}{2\dfrac{2}{5}}\) is an improper fraction and we can rewrite this as \(\dfrac{1}{\dfrac{12}{5}}=\dfrac{5}{12}\). We first clear denominators, then solve the linear equation as usual.
\[\begin{aligned}\dfrac{1}{3}+\dfrac{1}{t}&=\dfrac{5}{12} \\ \color{blue}{12t}\color{black}{}\cdot\dfrac{1}{3}+\color{blue}{12t}\color{black}{}\cdot\dfrac{1}{t}&=\color{blue}{12t}\color{black}{}\cdot\dfrac{5}{12}\\ 4t+12&=5t \\ 12&=t \\ t&=12\end{aligned}\]
Thus, it would take Maria \(12\) hours to clean the room by herself.
A sink can be filled by a pipe in \(5\) minutes, but it takes \(7\) minutes to drain a full sink. If both the pipe and the drain are open, how long will it take to fill the sink?
Solution
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table:
| time | fill per minute | |
|---|---|---|
| Fill the sink | \(5\) | \(\dfrac{1}{5}\) |
| Drain the sink | \(7\) | \(\dfrac{1}{7}\) |
| Together | \(t\) | \(\dfrac{1}{t}\) |
Now, let’s set up the equation and solve. Notice, were are filling the sink and draining it. Since we are draining the sink, we are losing water as the sink fills. Hence, we will subtract the rate in which the sink drains. We first clear denominators, then solve the linear equation as usual.
\[\begin{aligned}\dfrac{1}{5}-\dfrac{1}{7}&=\dfrac{1}{t} \\ \color{blue}{35t}\color{black}{}\cdot\dfrac{1}{5}-\color{blue}{35t}\color{black}{}\cdot\dfrac{1}{7}&=\color{blue}{35t}\color{black}{}\cdot\dfrac{1}{t} \\ 7t-5t&=35 \\ 2t&=35 \\ t&=\dfrac{35}{2}\end{aligned}\]
Thus, it would take \(\dfrac{35}{2}\) minutes to fill the sink, i.e., \(17\dfrac{1}{2}\) minutes.
Two Unknown Times
Mike takes twice as long as Rachel to complete a project. Together they can complete a project in 10 hours. How long will it take each of them to complete a project alone?
Solution
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table:
| time | project per hour | |
|---|---|---|
| Mike | \(2t\) | \(\dfrac{1}{2t}\) |
| Rachel | \(t\) | \(\dfrac{1}{t}\) |
| Together | \(10\) | \(\dfrac{1}{10}\) |
Now, let’s set up the equation and solve. We first clear denominators, then solve the linear equation as usual.
\[\begin{aligned}\dfrac{1}{2t}+\dfrac{1}{t}&=\dfrac{1}{10} \\ \color{blue}{10t}\color{black}{}\cdot\dfrac{1}{2t}+\color{blue}{10t}\color{black}{}\cdot\dfrac{1}{t}&=\color{blue}{10t}\color{black}{}\cdot\dfrac{1}{10} \\5+10&=t \\ 15&=t \\ t&=15\end{aligned}\]
Thus, it would take Rachel \(15\) hours to complete a project and Mike twice as long, \(30\) hours.
Brittney can build a large shed in \(10\) days less than Cosmo. If they built it together, it would take them \(12\) days. How long would it take each of them working alone?
Solution
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table:
| time | build per day | |
|---|---|---|
| Cosmo | \(t\) | \(\dfrac{1}{t}\) |
| Brittney | \(t-10\) | \(\dfrac{1}{(t-10)}\) |
| Together | \(12\) | \(\dfrac{1}{12}\) |
Now, let’s set up the equation and solve. We first clear denominators, then solve the equation as usual.
\[\begin{array}{rl}\dfrac{1}{t}+\dfrac{1}{t-10}=\dfrac{1}{12}&\text{Apply the work-rate equation} \\ \color{blue}{12t(t-10)}\color{black}{}\cdot\dfrac{1}{t}+\color{blue}{12t(t-10)}\color{black}{}\cdot\dfrac{1}{t-10}=\color{blue}{12t(t-10)}\color{black}{}\cdot\dfrac{1}{12}&\text{Clear denominators} \\ 12(t-10)+12t=t(t-10)&\text{Distribute} \\ 12t-120+12t=t^2-10t &\text{Combine like terms} \\ 24t-120=t^2-10t&\text{Notice the }t^2\text{ term; solve by factoring} \\ t^2-34t+120=0&\text{Factor} \\ (t-4)(t-30)=0&\text{Apply zero product rule} \\ t-4=0\text{ or }t-30=0&\text{Isolate variable terms} \\ t=4\text{ or }t=30&\text{Solutions}\end{array}\nonumber\]
We obtained \(t = 4\) and \(t = 30\) for the solutions. However, we need to verify these solutions with Cosmo and Brittney’s times. If \(t = 4\), then Brittney’s time would be \(4 − 10 = −6\) days. This makes no sense since days are always positive. Thus, it would take Cosmo \(30\) days to build a shed and Brittney \(10\) less days, \(20\) days.
An electrician can complete a job in one hour less than his apprentice. Together they do the job in \(1\) hour and \(12\) minutes. How long would it take each of them working alone?
Solution
We use the work-rate equation to model the problem, but before doing this, we can display the information on a table. Notice the time given doing the job together: \(1\) hour and \(12\) minutes. Unfortunately, we cannot use this format in the work-rate equation. Hence, we need to convert this to the same time units: \(1\) hour and \(12\) minutes \(= 1\dfrac{12}{60}\) hours \(= 1.2\) hours \(= \dfrac{6}{5}\) hours.
| time | job per hour | |
|---|---|---|
| Electrician | \(t-1\) | \(\dfrac{1}{(t-1)}\) |
| Apprentice | \(t\) | \(\dfrac{1}{t}\) |
| Together | \(\dfrac{6}{5}\) | \(\dfrac{5}{6}\) |
Note, \(\dfrac{1}{\dfrac{6}{5}} = \dfrac{5}{6}\). Now, let’s set up the equation and solve. We first clear denominators, then solve the equation as usual.
\[\begin{array}{rl}\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{5}{6}&\text{Apply the work-rate equation} \\ \color{blue}{6t(t-1)}\color{black}{}\cdot\dfrac{1}{t-1}+\color{blue}{6t(t-1)}\color{black}{}\cdot\dfrac{1}{t}=\color{blue}{6t(t-1)}\color{black}{}\cdot\dfrac{5}{6}&\text{Clear denominators} \\ 6t+6(t-1)=5t(t-1)&\text{Distribute} \\ 6t+6t-6=5t^2-5t&\text{Combine like terms} \\ 12t-6=5t^2-5t&\text{Notice the }5t^2\text{ term; solve by factoring} \\ 5t^2-17t+6=0&\text{Factor} \\ (5t-2)(t-3)=0&\text{Apply zero product rule} \\ 5t-2=0\text{ or }t-3=0&\text{Isolate variable terms} \\ t=\dfrac{2}{5}\text{ or }t=3&\text{Solutions}\end{array}\nonumber\]
We obtained \(t = \dfrac{2}{5}\) and \(t = 3\) for the solutions. However, we need to verify these solutions with the electrician and apprentice’s times. If \(t =\dfrac{2}{5}\), then the electrician’s time would be \(\dfrac{2}{5} −1 = −\dfrac{3}{5}\) hours. This makes no sense since hours are always positive. Thus, it would take the apprentice \(3\) hours to complete a job and the electrician \(1\) less hour, \(2\) hours.
Work-Rate Problems Homework
Bill’s father can paint a room in two hours less than Bill can paint it. Working together they can complete the job in two hours and \(24\) minutes. How much time would each require working alone?
Of two inlet pipes, the smaller pipe takes four hours longer than the larger pipe to fill a pool. When both pipes are open, the pool is filled in three hours and forty-five minutes. If only the larger pipe is open, how many hours are required to fill the pool?
Jack can wash and wax the family car in one hour less than Bob can. The two working together can complete the job in \(1\dfrac{1}{5}\) hours. How much time would each require if they worked alone?
If A can do a piece of work alone in \(6\) days and B can do it alone in \(4\) days, how long will it take the two working together to complete the job?
Working alone it takes John \(8\) hours longer than Carlos to do a job. Working together they can do the job in \(3\) hours. How long will it take each to do the job working alone?
A can do a piece of work in \(3\) days, B in \(4\) days, and C in \(5\) days each working alone. How long will it take them to do it working together?
A can do a piece of work in \(4\) days and B can do it in half the time. How long will it take them to do the work together?
A cistern can be filled by one pipe in \(20\) minutes and by another in \(30\) minutes. How long will it take both pipes together to fill the tank?
If A can do a piece of work in \(24\) days and A and B together can do it in \(6\) days, how long would it take B to do the work alone?
A carpenter and his assistant can do a piece of work in \(3\dfrac{3}{4}\) days. If the carpenter himself could do the work alone in \(5\) days, how long would the assistant take to do the work alone?
If Sam can do a certain job in \(3\) days, while it takes Fred \(6\) days to do the same job, how long will it take them, working together, to complete the job?
Tim can finish a certain job in \(10\) hours. It take his wife JoAnn only \(8\) hours to do the same job. If they work together, how long will it take them to complete the job?
Two people working together can complete a job in \(6\) hours. If one of them works twice as fast as the other, how long would it take the faster person, working alone, to do the job?
If two people working together can do a job in \(3\) hours, how long will it take the slower person to do the same job if one of them is \(3\) times as fast as the other?
A water tank can be filled by an inlet pipe in \(8\) hours. It takes twice that long for the outlet pipe to empty the tank. How long will it take to fill the tank if both pipes are open?
A sink can be filled from the faucet in \(5\) minutes. It takes only \(3\) minutes to empty the sink when the drain is open. If the sink is full and both the faucet and the drain are open, how long will it take to empty the sink?
It takes \(10\) hours to fill a pool with the inlet pipe. It can be emptied in \(15\) hrs with the outlet pipe. If the pool is half full to begin with, how long will it take to fill it from there if both pipes are open?
A sink is \(\dfrac{1}{4}\) full when both the faucet and the drain are opened. The faucet alone can fill the sink in \(6\) minutes, while it takes \(8\) minutes to empty it with the drain. How long will it take to fill the remaining \(\dfrac{3}{4}\) of the sink?
A sink has two faucets, one for hot water and one for cold water. The sink can be filled by a cold-water faucet in \(3.5\) minutes. If both faucets are open, the sink is filled in \(2.1\) minutes. How long does it take to fill the sink with just the hot-water faucet open?
A water tank is being filled by two inlet pipes. Pipe A can fill the tank in \(4\dfrac{1}{2}\) hrs, while both pipes together can fill the tank in \(2\) hours. How long does it take to fill the tank using only pipe B?
A tank can be emptied by any one of three caps. The first can empty the tank in \(20\) minutes while the second takes \(32\) minutes. If all three working together could empty the tank in \(8\dfrac{8}{59}\) minutes, how long would the third take to empty the tank?
One pipe can fill a cistern in \(1\dfrac{1}{2}\) hours while a second pipe can fill it in \(2\dfrac{1}{3}\) hrs. Three pipes working together fill the cistern in \(42\) minutes. How long would it take the third pipe alone to fill the tank?
Sam takes \(6\) hours longer than Susan to wax a floor. Working together they can wax the floor in \(4\) hours. How long will it take each of them working alone to wax the floor?
It takes Robert \(9\) hours longer than Paul to rapair a transmission. If it takes them \(2 \dfrac{2}{5}\) hours to do the job if they work together, how long will it take each of them working alone?
It takes Sally \(10\dfrac{1}{2}\) minutes longer than Patricia to clean up their dorm room. If they work together they can clean it in \(5\) minutes. How long will it take each of them if they work alone?
A takes \(7 \dfrac{1}{2}\) minutes longer than B to do a job. Working together they can do the job in \(9\) minutes. How long does it take each working alone?
Secretary A takes \(6\) minutes longer than Secretary B to type \(10\) pages of manuscript. If they divide the job and work together it will take them \(8 \dfrac{3}{4}\) minutes to type \(10\) pages. How long will it take each working alone to type the \(10\) pages?
It takes John \(24\) minutes longer than Sally to mow the lawn. If they work together they can mow the lawn in \(9\) minutes. How long will it take each to mow the lawn if they work alone?