If 3 painters can paint a house in 10 days, how long would 5 painters take at the same rate?
The problem describes a task (painting a house) that can be completed by a group of people working at a constant rate. We know that 3 painters take 10 days to finish this job. To find the time taken by 5 painters, we assume that each painter performs work at the same constant rate, which we denote as \(r\), measured in houses per day per painter. The total amount of work is defined as one house. The relationship between the number of painters \(N\), time \(T\), and the rate of work \(r\) is given by the formula:
\[\text{Work} = N \times T \times r\]
Using the information provided for 3 painters, we can set up the equation for the total work:
\[1 \text{ (house)} = 3 \text{ (painters)} \times 10 \text{ (days)} \times r\]
\[1 = 30 \times r\]
From this, we solve for the rate \(r\):
\[r = \frac{1}{30} \text{ houses per day per painter}\]
This means that a single painter completes \(\frac{1}{30}\) of a house in one day.
Now we use this rate to calculate the time \(T\) it would take for 5 painters to complete the same amount of work:
\[1 \text{ (house)} = 5 \text{ (painters)} \times T \text{ (days)} \times \left(\frac{1}{30} \text{ houses per day per painter}\right)\]
\[1 = 5 \times T \times \frac{1}{30}\]
\[1 = \frac{5T}{30}\]
\[1 = \frac{T}{6}\]
Solving for \(T\), we multiply both sides by 6:
\[T = 6\]
Thus, 5 painters will take 6 days to finish the house. This corresponds to option B.
This concept of inverse variation is widely used in project management to estimate timelines for teams. For example, if a team of 4 software developers takes 20 days to build a specific software feature, you can estimate that doubling the team size to 8 developers might theoretically halve the time needed, assuming the workload is divisible and communication overhead is minimal. Understanding these relationships helps in efficient resource allocation and scheduling for any collaborative task.