Melinda and Paula shovel driveways and sidewalks in the winter as a way to earn extra money. Together they shoveled 450 square feet of sidewalk in 30 minutes. Then Melinda shoveled for 20 minutes while Paula shoveled for 25 minutes to complete 345 square feet of driveway.

30x + 30y = 450
20x + 25y = 345
How much more can Paula shovel in 1 minute than Melinda?

3 square feet per minute
6 square feet per minute
9 square feet per minute
15 square feet per minute

1 answer

We can set up a system of equations based on the information provided.

The first equation is based on the total area they shoveled together in 30 minutes:

\[ 30x + 30y = 450 \]

Dividing through by 30 simplifies to:

\[ x + y = 15 \quad \text{(Equation 1)} \]

where \(x\) is the rate at which Melinda shovels (in square feet per minute) and \(y\) is the rate at which Paula shovels (in square feet per minute).

The second equation is based on the work done individually by Melinda and Paula:

\[ 20x + 25y = 345 \]

Dividing through by 5 simplifies to:

\[ 4x + 5y = 69 \quad \text{(Equation 2)} \]

Now we have the following system of equations:

  1. \(x + y = 15\)
  2. \(4x + 5y = 69\)

We can solve this system using substitution or elimination. Let's use substitution. From Equation 1:

\[ y = 15 - x \]

Now, substitute \(y\) into Equation 2:

\[ 4x + 5(15 - x) = 69 \]

Distributing the \(5\):

\[ 4x + 75 - 5x = 69 \]

Combine like terms:

\[ -x + 75 = 69 \]

Subtracting \(75\) from both sides gives:

\[ -x = 69 - 75 \] \[ -x = -6 \quad \Rightarrow \quad x = 6 \]

Now substitute back to find \(y\):

\[ y = 15 - x = 15 - 6 = 9 \]

So, we have:

  • Melinda's rate \(x = 6\) square feet per minute.
  • Paula's rate \(y = 9\) square feet per minute.

To find out how much more Paula can shovel in one minute than Melinda, we calculate:

\[ y - x = 9 - 6 = 3 \]

Thus, Paula can shovel 3 square feet per minute more than Melinda.

The answer is: 3 square feet per minute.