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Worksheet Grade 8 Careers & CTE

Cottersby's Pelham Row Main: Repair Every Year or Replace Once?

A grade 8 worksheet where students model a village's repair and replacement costs as linear equations, solve for the crossover year, and meet a no-solution case.

Objectives

  1. Students will write linear cost functions for repairing and for replacing an aging water main.
  2. Students will solve an equation with the variable on both sides and interpret the solution as a crossover year.
  3. Students will recognize an equation with no solution and explain what it means for the decision.

Vocabulary

Water main
A big underground pipe that carries drinking water beneath the streets to a neighborhood. Smaller service lines branch off it to reach each building.
Asset management
The practice of keeping track of everything a utility owns, like pipes and pumps, and planning the smartest times to fix or replace each one. It helps spend repair money wisely.

Also: crossover year, rate of change, no solution

Teacher overview

The ASCE 2025 Infrastructure Report Card gave US drinking water a C-, and much of that grade comes from pipe that has outlived its design life. Deciding which mains to replace, and when, is the daily work of an asset management analyst, who compares what a utility is already spending on emergency repairs against what a planned replacement would cost. This worksheet walks students through that comparison for one invented street. Two misconceptions show up here: students often assume the option with the large up-front price is automatically the expensive one, and when an equation reduces to a false statement they usually hunt for an arithmetic error instead of recognizing that two parallel lines never meet.

Cottersby's Pelham Row Main: Repair Every Year or Replace Once? — Grade 8 Careers & CTE
Name: ____________________________ Date: ______________

Student Worksheet

Suppose you are an asset management analyst for the village of Cottersby, a fictional water system. The cast-iron main under Pelham Row keeps breaking. Every number below was invented for this worksheet. Let t stand for the number of years from today.

  • Keep repairing. Keeping the old main in service requires relining its worst section now for $54,000, and emergency repairs then run about $18,000 per year.
  • Replace it now. A new main costs $310,000 to install, and routine maintenance on new pipe runs about $2,000 per year.
  1. Write a function for the total repair cost C after t more years. State the initial value and the rate of change.
  2. Write a function for the total replacement cost R after t years.
  3. What is the total repair cost after 5 years? Show your substitution.
  4. What is the total replacement cost after 5 years? Show your substitution.
  5. At 5 years, how much more does the replacement option cost than the repair option?
  6. Solve for the year t at which the two totals are equal. Show every step.
  7. What is the total cost of each option at that crossover year?
  8. Cottersby expects Pelham Row to stay in service for 25 more years. What is the total repair cost over 25 years?
  9. What is the total replacement cost over the same 25 years?
  10. Over 25 years, how much less does the replacement option cost?
  11. A second contractor offers to install the new main for the same $310,000, but its maintenance plan would cost $18,000 per year. Write and solve the equation comparing that offer to the repair option, and state what your result means.
  12. Explain to the village board, in your own words, why the answer to item 11 does not mean you made a mistake.

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: C = 54,000 + 18,000t.
    The initial value is the $54,000 relining cost due now, and the rate of change is the $18,000 spent on repairs each year.
  2. Answer: R = 310,000 + 2,000t.
    The initial value is the $310,000 installation cost, and the rate of change is the $2,000 of yearly maintenance.
  3. Answer: $144,000.
    C = 54,000 + 18,000(5) = 54,000 + 90,000 = 144,000.
  4. Answer: $320,000.
    R = 310,000 + 2,000(5) = 310,000 + 10,000 = 320,000.
  5. Answer: $176,000.
    320,000 - 144,000 = 176,000, so replacing costs $176,000 more at the five-year mark.
  6. Answer: 16 years.
    Set 54,000 + 18,000t = 310,000 + 2,000t. Subtract 2,000t from both sides: 54,000 + 16,000t = 310,000. Subtract 54,000 from both sides: 16,000t = 256,000. Divide both sides by 16,000: t = 16.
  7. Answer: $342,000.
    Repair: 54,000 + 18,000(16) = 54,000 + 288,000 = 342,000. Replace: 310,000 + 2,000(16) = 310,000 + 32,000 = 342,000. The totals agree, which checks item 6.
  8. Answer: $504,000.
    C = 54,000 + 18,000(25) = 54,000 + 450,000 = 504,000.
  9. Answer: $360,000.
    R = 310,000 + 2,000(25) = 310,000 + 50,000 = 360,000.
  10. Answer: $144,000.
    504,000 - 360,000 = 144,000, so over 25 years the replacement option saves the village $144,000.
  11. Answer: No solution.
    The equation is 54,000 + 18,000t = 310,000 + 18,000t. Subtracting 18,000t from both sides leaves 54,000 = 310,000, which is false for every value of t, so the equation has no solution.
  12. Answer: The two costs rise at the same rate, so they never become equal.
    Both options in item 11 add $18,000 every year, which means the gap between them stays fixed at $256,000 forever. On a graph the lines are parallel and never cross, so there is no crossover year to find. For the board, the practical meaning is that this second offer never pays for itself.

Differentiation

Support

Give students a table for this invented scenario with columns for 5, 10, 15, 20, and 25 years and the first row filled in, so the crossover appears in the numbers before students solve for it algebraically. Provide the function frame "total = (up-front cost) + (cost per year) times t" for items 1 and 2.

Extension

Ask students to find the yearly repair cost that would move the crossover year in this invented comparison to exactly 10 years, then verify the answer by substituting it into both functions.

ELL supports

Pre-teach "water main," "crossover year," and "no solution" with a graph showing two lines that cross and two lines that are parallel. Allow item 12 to be answered with the parallel-lines sketch plus one or two sentences.

Extensions

Have students graph both options from items 1 and 2 on the same axes and shade the region where repairing costs less, then repeat with the second contractor's offer from item 11 to see parallel lines. Students can also find out how their own utility reports water main breaks, and which job title decides the replacement order.