Rocco mows lawns in the summer. He charges $10 for every 1/2 hour of mowing. This is represented by the equation y=10x where y is the total amount Rocco charges and x is the number of 1/2 hour sessions of mowing. If Mr. Stevenson lawn takes 2-5 hours to mow and Mr. price lawn takes 4 hours to mow, how much more would Rocco charge Mr. Price than Mr. Stevenson?

1 answer

To determine how much more Rocco would charge Mr. Price than Mr. Stevenson, we first need to find out how many half-hour sessions are in the mowing times given for both lawns.

  1. Mr. Stevenson's lawn: It takes between 2 to 5 hours to mow.

    • In half-hour sessions, 2 hours equals \(2 \times 2 = 4\) sessions.
    • 5 hours equals \(5 \times 2 = 10\) sessions.

    Therefore, the charge for Mr. Stevenson can range from:

    • Minimum: \(y = 10 \times 4 = 40\) dollars
    • Maximum: \(y = 10 \times 10 = 100\) dollars
  2. Mr. Price's lawn: It takes 4 hours to mow.

    • In half-hour sessions, 4 hours equals \(4 \times 2 = 8\) sessions.

    Therefore, the charge for Mr. Price is:

    • \(y = 10 \times 8 = 80\) dollars
  3. Calculating the difference:

    • The minimum charge for Mr. Stevenson is \(40\) dollars and the maximum charge is \(100\) dollars.
    • The charge for Mr. Price is \(80\) dollars.

Now, we check the difference:

  • If Mr. Stevenson is charged the minimum ($40), the difference is: \[ 80 - 40 = 40 \text{ dollars} \]

  • If Mr. Stevenson is charged the maximum ($100), the difference is: \[ 80 - 100 = -20 \text{ dollars} \] (indicating Mr. Stevenson would be charged more.)

So, depending on how long it takes to mow Mr. Stevenson's lawn, Rocco could charge anywhere from $40 less (when Mr. Stevenson takes the maximum time of 5 hours) to $20 more (when Mr. Stevenson takes the minimum time of 2 hours).

To summarize, Rocco charges Mr. Price $40 more than the minimum he could charge Mr. Stevenson and $20 less than the maximum. Therefore, the difference in charges varies depending on the time it takes to mow Mr. Stevenson's lawn.