Math Problem Statement

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You need to rent a moving truck for a day. You have identified two companies that rent trucks. Company A charges $40 per day plus $0.10 per mile K Company B charges $30 per day plus $0.15 per mile. For how many miles will the cost of renting be the same?

Solution

To determine the number of miles where the cost of renting from both companies is the same, we can set up an equation that represents the total cost for each company and solve for the number of miles.

Let:

  • xx represent the number of miles driven.
  • Company A charges $40 per day plus $0.10 per mile, so the total cost for Company A is: CA=40+0.10xC_A = 40 + 0.10x
  • Company B charges $30 per day plus $0.15 per mile, so the total cost for Company B is: CB=30+0.15xC_B = 30 + 0.15x

Now, set the two cost equations equal to find when the costs are the same: 40+0.10x=30+0.15x40 + 0.10x = 30 + 0.15x

Solve for xx:

  1. Subtract 3030 from both sides: 4030+0.10x=0.15x40 - 30 + 0.10x = 0.15x 10+0.10x=0.15x10 + 0.10x = 0.15x

  2. Subtract 0.10x0.10x from both sides: 10=0.05x10 = 0.05x

  3. Divide both sides by 0.050.05: x=100.05=200x = \frac{10}{0.05} = 200

So, the cost of renting the truck will be the same when you drive 200 miles.

Would you like further details or have any questions?


Here are 5 related questions you might find interesting:

  1. What is the total cost for each company if you drive 150 miles?
  2. At how many miles would Company A be cheaper than Company B?
  3. How would the equation change if Company B charged $0.20 per mile instead?
  4. What if both companies charged a higher flat fee but the same per-mile rate?
  5. How can you minimize your total cost if you plan to drive less than 100 miles?

Tip: When comparing two pricing models, set their equations equal to find the breakeven point.

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Math Problem Analysis

Mathematical Concepts

Algebra
Linear Equations

Formulas

Cost Equation for Company A: C_A = 40 + 0.10x
Cost Equation for Company B: C_B = 30 + 0.15x

Theorems

Linear Equation Solving

Suitable Grade Level

Grades 6-8