Water for Schools
Menu

Math problem set Grade 8 Math

Meridian Water Bills: Linear Functions for Two Rate Plans

A grade 8 problem set where students write and compare linear functions for two fictional water-bill rate plans, find the break-even usage, and check affordability.

Objectives

  1. Students will write a linear function for a monthly water bill from a base charge and a per-unit rate.
  2. Students will interpret the slope and initial value of a bill function in terms of price and base charge.
  3. Students will find the usage at which two rate plans cost the same and compare which is cheaper on each side.

Vocabulary

Key words: linear function, rate of change, initial value, slope, base charge

Teacher overview

A monthly water bill usually has two parts: a fixed base charge that every customer pays, and a usage charge that depends on how much water the household used. That structure is a linear function, with the base charge as the initial value and the per-unit price as the rate of change. This problem set uses the fictional Meridian Water District and a neighboring district to have students write, compare, and interpret two such functions. A common misconception is to treat the whole bill as directly proportional to usage, so that doubling the water doubles the bill; the fixed base charge means the bill is not proportional, and low-usage months still carry that base cost.

Meridian Water Bills: Linear Functions for Two Rate Plans — Grade 8 Math
Name: ____________________________ Date: ______________

Problems

Use the following information about the fictional Meridian Water District for every problem. Let g stand for the number of thousands of gallons a household uses in one month.

  • Plan A (Meridian): a base charge of $18 per month plus $6 for each 1,000 gallons used.
  • Plan B (a neighboring district): a base charge of $9 per month plus $9 for each 1,000 gallons used.
  1. Write a linear function for Plan A's monthly bill, A, in dollars, in terms of g.
  2. Write a linear function for Plan B's monthly bill, B, in dollars, in terms of g.
  3. In Plan A's function, name the rate of change and what it represents, and name the initial value and what it represents.
  4. A household uses 5,000 gallons in a month, so g = 5. Find its bill under Plan A.
  5. Find the same household's bill under Plan B.
  6. Find the usage g at which Plan A and Plan B cost exactly the same. Show your work.
  7. Using g = 1 and g = 6, decide which plan is cheaper below the break-even usage and which is cheaper above it.
  8. Compare the slopes of the two functions. Which plan's bill rises faster as usage grows, and what does that mean for a household that uses a lot of water?
  9. A widely used rule of thumb, built from two EPA benchmarks, treats combined water and sewer costs above about 4.5 percent of household income as a high burden. Suppose a Meridian household earns $1,500 per month and pays the Plan A water bill from problem 4. What percent of its monthly income is the water bill alone, and is it below the 4.5 percent threshold?
  10. Meridian adds a conservation tier to Plan A: every 1,000 gallons above 10,000 gallons (above g = 10) costs $12 instead of $6. Find the Plan A bill for a household that uses 12,000 gallons in a month.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key
  1. Answer: A = 18 + 6g.
    Plan A charges a fixed $18 base plus $6 for each thousand gallons, so A = 18 + 6g.
  2. Answer: B = 9 + 9g.
    Plan B charges a fixed $9 base plus $9 for each thousand gallons, so B = 9 + 9g.
  3. Answer: Rate of change $6; initial value $18.
    The rate of change is $6 per 1,000 gallons, the price for each thousand gallons used. The initial value is $18, the fixed base charge a household pays even at zero usage.
  4. Answer: $48.
    A = 18 + 6(5) = 18 + 30 = 48 dollars.
  5. Answer: $54.
    B = 9 + 9(5) = 9 + 45 = 54 dollars.
  6. Answer: g = 3 (3,000 gallons).
    Set 18 + 6g = 9 + 9g. Subtract 6g from both sides: 18 = 9 + 3g. Subtract 9: 9 = 3g. Divide by 3: g = 3, which is 3,000 gallons.
  7. Answer: Plan B is cheaper below, Plan A is cheaper above.
    At g = 1: A = 18 + 6 = 24 and B = 9 + 9 = 18, so Plan B is cheaper. At g = 6: A = 18 + 36 = 54 and B = 9 + 54 = 63, so Plan A is cheaper. Below 3,000 gallons Plan B wins; above it Plan A wins.
  8. Answer: Plan B rises faster (slope 9 versus 6).
    Plan B's slope is $9 per 1,000 gallons and Plan A's is $6 per 1,000 gallons, so Plan B's bill climbs faster; a household that uses a lot of water pays more under Plan B.
  9. Answer: 3.2 percent, below the threshold.
    48 ÷ 1,500 = 0.032, or 3.2 percent of monthly income, which is below the 4.5 percent high-burden threshold. A sewer bill would push the combined share higher.
  10. Answer: $102.
    The first 10,000 gallons cost 18 + 6(10) = 18 + 60 = 78 dollars. The 2,000 gallons above that cost 2 × 12 = 24 dollars. The total is 78 + 24 = 102 dollars.

Differentiation

Support (below grade level)

Give students a two-column table of bills at g = 0, 1, 2, 3, and 4 for both plans so they can see each pattern before writing the functions, and provide the equation for problem 6 already set up as 18 + 6g = 9 + 9g.

Extension (above grade level)

Ask students to graph both plans on the same axes, mark the point where the lines cross, and explain how that crossing point matches the break-even usage they found in problem 6.

ELL supports

Pre-teach "base charge," "rate," "slope," and "break-even" with a simple picture of a bill split into a fixed part and a per-gallon part. Provide sentence starters for problems 3 and 8, such as "The slope is ___, which means ___."

Extensions

Have students design a fair third rate plan for a small town that wants low bills for households that conserve water but still enough revenue to run the system, then write its linear function and explain the base charge and per-unit rate they chose.