- Define the variables for Dominic's rate (\(d\)) and Cynthia's rate (\(c\)).
- Use the given information to create a system of two equations: \(\frac{d}{c} = \frac{6}{5}\) and \(d = c + 10\).
- Substitute the first equation into the second to solve for \(c\).
- Conclude that Cynthia can install \(\boxed{50}\) shingles per hour.
Explanation
- Problem Analysis and Data Formulation
We are given the rates of two workers, Dominic and Cynthia. Let \(d\) represent the number of shingles Dominic can install per hour, and \(c\) represent the number of shingles Cynthia can install per hour. We are given two pieces of information:
- Dominic installs 180 shingles in the same time it takes Cynthia to install 150. If \(t\) is the time taken, we have \(d \times t = 180\) and \(c \times t = 150\). Dividing these two equations gives \(\frac{d}{c} = \frac{180}{150} = \frac{6}{5}\).
- Dominic installs 10 more shingles per hour than Cynthia, so \(d = c + 10\).
Our goal is to determine the value of \(c\).
- Solving for Cynthia's Rate
From the ratio \(\frac{d}{c} = \frac{6}{5}\), we can express Dominic's rate in terms of Cynthia's rate as follows:
\[d = \frac{6}{5}c = 1.2c\]
Substituting this expression into the second constraint, \(d = c + 10\), we obtain:
\[1.2c = c + 10\]
Subtracting \(c\) from both sides of the equation, we get:
\[0.2c = 10\]
Solving for \(c\) by dividing both sides by 0.2:
\[c = \frac{10}{0.2} = 50\]
Therefore, Cynthia can install 50 shingles per hour.
- Verification and Conclusion
We have determined that \(c = 50\) shingles per hour. Consequently, Dominic's rate is \(d = 50 + 10 = 60\) shingles per hour. Checking the first condition:
\[\frac{180}{60} = 3 \text{ hours}\]
\[\frac{150}{50} = 3 \text{ hours}\]
Both workers finish their respective tasks in the same amount of time, confirming that our calculated rates are correct. Cynthia can install 50 shingles per hour. The final answer is \(\boxed{50}\).
Examples
In real-life scenarios like construction management or manufacturing, determining the productivity of different workers allows for better scheduling and resource allocation. If you know the relative efficiency of your team members, you can predict project completion times and balance workloads effectively. By applying algebraic ratios to work-rate problems, project managers can ensure that tasks are assigned efficiently, helping to minimize delays and optimize overall project timelines.