Question

Painting a House Three painters (Beth, Dan, and Edie), working together, can paint the exterior of a home in 10 hours (h). Dan and Edie together have painted a similar house in \(15 \mathrm{~h}\). One day, all three worked on this same kind of house for \(4 \mathrm{~h},\) after which Edie left. Beth and Dan required 8 more hours to finish. Assuming no gain or loss in efficiency, how long should it take each person to complete such a job alone?

Short answer

Beth: 30 hours, Dan: 24 hours, Edie: 40 hours

Step-by-step solution

Define the Variables

Let the rate of work for Beth, Dan, and Edie be denoted as follows: \(B\) for Beth, \(D\) for Dan, and \(E\) for Edie. The rates are measured in houses per hour.

Set Up the Equations

Use the given information to set up equations for each scenario. First, find the total rate when all three are working together: \[B + D + E = \frac{1}{10}\]

Equation for Dan and Edie

When Dan and Edie work together, their combined rate is: \[D + E = \frac{1}{15}\]

Calculate Work Done Before Edie Left

All three work together for 4 hours. Let’s calculate the fraction of the house painted: \[4(B + D + E) = 4 \left(\frac{1}{10}\right) = \frac{2}{5}\]

Calculate Remaining Work

After Edie leaves, 3/5 of the house remains to be painted. Beth and Dan take 8 more hours to complete it. Their combined rate for this is: \[8(B + D) = \frac{3}{5}\]

Solve for Beth and Dan's Combined Rate

Solve for \(B + D\) using the remaining work equation: \[\frac{3}{5} = 8(B + D) \rightarrow B + D = \frac{3}{40}\]

Solve the System of Equations

We now have two equations (\(B + D + E = \frac{1}{10}\) and \(D + E = \frac{1}{15}\)). Substitute \(D + E\) from the second equation into the first: \[B + \frac{1}{15} = \frac{1}{10} \implies B = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}\]

Solve for Dan and Edie Individually

Now that we have \(B = \frac{1}{30}\), use \(B + D = \frac{3}{40}\) to find \(D\): \[\frac{1}{30} + D = \frac{3}{40} \implies D = \frac{3}{40} - \frac{1}{30} = \frac{1}{24}\] Finally, use \(D + E = \frac{1}{15}\) to find \(E\): \[\frac{1}{24} + E = \frac{1}{15} \implies E = \frac{1}{15} - \frac{1}{24} = \frac{1}{40}\]

Interpret the Results

Each painter's rate is now known: \(B = \frac{1}{30}\), \(D = \frac{1}{24}\), and \(E = \frac{1}{40}\). To find the time for each to complete one house individually, we simply take the reciprocal of their rates.

Key concepts

Work rate equations

When dealing with work rate problems, it's important to understand the concept of rates. The rate of work refers to how much work can be done per unit of time. To approach these problems, we often use equations to represent the combined work rates of multiple workers. For example, in the painting problem, we denote the work rates of Beth, Dan, and Edie as \(B\), \(D\), and \(E\) respectively. When working together, their combined rate is the sum of their individual rates: \(B + D + E\).

The problem provides that all three painters together can paint the house in 10 hours, which means their combined rate is \(\frac{1}{10}\) house per hour. This relationship is crucial to set up our initial equation: \(B + D + E = \frac{1}{10}\). Understanding how to set up these work rate equations is fundamental, as they lay the groundwork for solving the entire problem.

Solving systems of equations

Once we have our work rate equations set up, the next step is to solve these equations systematically. In our problem, we have two such equations: the combined rate for all three painters and the combined rate for Dan and Edie. The first equation is: \(B + D + E = \frac{1}{10}\) and the second one is: \(D + E = \frac{1}{15}\).

These two equations form a system of linear equations. To solve them, we can use substitution or elimination methods. In this case, we substitute the value from one equation into the other. We know \(D + E = \frac{1}{15}\), so we can replace \(D + E\) in the first equation:
\(B + \frac{1}{15} = \frac{1}{10}\).

By isolating \(B\), we get \(B = \frac{1}{10} - \frac{1}{15}\). Simplifying this gives us the individual work rate of Beth. Solving systems of equations is a crucial algebraic skill that allows us to find individual components from combined information.

Algebraic problem solving

Algebra is the backbone of solving work rate problems. It helps us translate word problems into mathematical expressions and find solutions. In work rate problems, algebra lets us form equations that represent the relationships described in the problem. For instance, once we know the proportions of work done by each painter, algebraic manipulation helps us find their individual rates.

From our equations, after finding \(B\)'s value, we use the second equation to solve for \(D\). If \(B = \frac{1}{30}\), then substituting into \(B + D = \frac{3}{40}\) helps us isolate and solve for \(D\).

When we reach the calculated rate for Dan as \(\frac{1}{24}\), using it in \(D + E = \frac{1}{15}\) makes it possible to solve for \(E\). Algebraic problem-solving involves these iterative steps to simplify and solve for unknowns efficiently.

Combined work rates

Understanding combined work rates is fundamental in work rate problems. It is about knowing how individual rates add up when working together. For example, Beth, Dan, and Edie's combined work rate together can paint the house in 10 hours, denoted as \(B + D + E = \frac{1}{10}\).

When only Dan and Edie are working, their combined rate is recognized as \(D + E = \frac{1}{15}\). These parts are critical to determining how each person's contribution influences the total. Suppose after working together for some time, the remaining work is completed by fewer workers; this scenario also needs understanding of combined rates.

When Edie leaves after 4 hours, Beth and Dan complete the remaining \(\frac{3}{5}\) of the house in 8 hours. Their combined rate thus becomes \(\frac{3}{5} / 8 = \frac{3}{40}\). Recognizing how combined work rates change based on team members and calculating individual contributions help solve such multi-step problems.