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Worksheet Grade 11 Math

Thornleigh's Aging Water Mains: A Replacement Plan Under Two Limits

A grade 11 worksheet writing budget and crew-capacity inequalities for a fictional city's main replacement program, then solving the system they form.

Objectives

  1. Students will write linear inequalities that represent a budget limit and a labor limit on a replacement program.
  2. Students will solve a two-variable linear system and interpret the solution as a yearly work plan.
  3. Students will compare the pace of two replacement schedules against the length of pipe a system holds.

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.

Also: constraint, system of equations, break rate, capital budget

Teacher overview

Utilities replace water mains on a schedule set by two separate limits, the money available to buy pipe and the crew time available to install it, and either one can run out first. The ASCE 2025 Infrastructure Report Card gave United States drinking water a grade of C-, and buried pipe is a large part of that grade. This worksheet has grade 11 students write both limits as inequalities for the fictional city of Thornleigh, solve the system that forms when both limits are reached at once, and compare the resulting pace against how much old pipe the city owns. A frequent error is to assume the budget always sets the ceiling, when a plan built from inexpensive small-diameter pipe runs into the crew limit long before the money is gone.

Thornleigh's Aging Water Mains: A Replacement Plan Under Two Limits — Grade 11 Math
Name: ____________________________ Date: ______________

Student Worksheet

The city of Thornleigh, its costs, and its break records are all fictional. Use these values throughout.

  • Thornleigh owns 96 miles of unlined iron main that is due for replacement.
  • Replacing 6-inch main costs $310,000 per mile.
  • Replacing 12-inch main costs $520,000 per mile.
  • The annual capital budget for main replacement is $4,500,000.
  • The city's crews can complete at most 12 miles of replacement in a year, whatever the pipe size.
  • Let x be the miles of 6-inch main replaced in one year and let y be the miles of 12-inch main replaced in that same year.
  • Last year the city recorded 68 main breaks across its 96 miles of unlined iron main.
  • Each break costs the city about $8,600 in repair labor, materials, and lost water.
  • The city currently replaces 3.5 miles of main per year.
  1. Write an inequality in x and y for the annual budget limit.
  2. Write an inequality in x and y for the crew capacity limit.
  3. If Thornleigh spent the whole budget on 6-inch main and ignored the crew limit, how many miles could it buy? Round to one decimal place.
  4. If it spent the whole budget on 12-inch main instead, how many miles could it buy? Round to two decimal places.
  5. Solve the system in which the crews work all 12 miles and the budget is spent exactly. Give the miles of 6-inch main, rounded to two decimal places.
  6. Give the miles of 12-inch main in that same plan, rounded to two decimal places.
  7. Find the amount spent on 6-inch main in that plan, to the nearest dollar.
  8. At 12 miles per year, how many years would Thornleigh need to replace all 96 miles?
  9. At its current pace of 3.5 miles per year, how many years would the same job take? Round to one decimal place.
  10. Find last year's break rate as breaks per 100 miles of main. Round to one decimal place.
  11. Find the total cost of last year's breaks.
  12. Suppose the replacement program removes 9 breaks per year from the annual total. After how many full years does the annual break cost first fall below $200,000?
  13. Compare your answers to items 3 and 4 against the crew limit. Which limit controls a plan made almost entirely of 6-inch main, and which controls a plan made almost entirely of 12-inch main? Explain how you can tell.

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: 310,000x + 520,000y is at most 4,500,000.
    Each mile of 6-inch main costs 310,000 dollars and each mile of 12-inch main costs 520,000 dollars, so the total spent is 310,000x + 520,000y, and that total cannot exceed the 4,500,000 dollar budget.
  2. Answer: x + y is at most 12.
    The crews install at most 12 miles in total, and the pipe size does not change that limit, so the sum of the two mileages is capped at 12.
  3. Answer: 14.5 mi.
    4,500,000 divided by 310,000 = 14.516 miles, which rounds to 14.5 miles.
  4. Answer: 8.65 mi.
    4,500,000 divided by 520,000 = 8.6538 miles, which rounds to 8.65 miles.
  5. Answer: 8.29 mi.
    Substitute y = 12 - x into the budget equation: 310,000x + 520,000(12 - x) = 4,500,000. That gives 310,000x + 6,240,000 - 520,000x = 4,500,000, so -210,000x = -1,740,000 and x = 8.2857 miles, which rounds to 8.29 miles.
  6. Answer: 3.71 mi. y = 12 - 8.2857 = 3.7143 miles, which rounds to 3.71 miles.
  7. Answer: $2,568,571.
    310,000 x 8.2857 = 2,568,571.43 dollars. The remaining 520,000 x 3.7143 = 1,931,428.57 dollars, and the two amounts add to exactly the 4,500,000 dollar budget, which confirms the solution.
  8. Answer: 8 yr.
    96 miles divided by 12 miles per year = 8 years.
  9. Answer: 27.4 yr.
    96 divided by 3.5 = 27.43 years, which rounds to 27.4 years.
  10. Answer: 70.8 breaks per 100 mi.
    68 breaks divided by 96 miles = 0.7083 breaks per mile, and multiplying by 100 gives 70.8 breaks per 100 miles.
  11. Answer: $584,800.
    68 breaks x 8,600 dollars per break = 584,800 dollars.
  12. Answer: 5 yr.
    The cost falls below 200,000 dollars once the annual break count drops below 200,000 divided by 8,600 = 23.26 breaks. Starting from 68 and removing 9 per year gives 68 - 9t, and 68 - 9t is less than 23.26 when t is greater than 4.97. After 4 years there are 32 breaks costing 275,200 dollars, and after 5 years there are 23 breaks costing 197,800 dollars, so the answer is 5 years.
  13. Answer: The crew limit controls the small-diameter plan and the budget controls the large-diameter plan.
    Item 3 shows the budget alone would buy 14.5 miles of 6-inch main, which is more than the crews can install, so the crew limit of 12 miles is the one that binds. Item 4 shows the budget alone would buy only 8.65 miles of 12-inch main, which is less than 12, so there the budget binds and the crews finish early. Whichever limit produces the smaller number is the one that controls the plan.

Differentiation

Support

Give students a partially completed table with the two column headings "6-inch main" and "12-inch main" and the row headings "miles," "cost per mile," and "total cost," so the structure of the budget inequality is visible before they write it. Work items 1 and 2 as a class, and provide the substitution step for item 5 with the blanks left only for the arithmetic.

Extension

Ask students to graph the feasible region for the two inequalities in the first quadrant, shade it, and identify each corner point as a work plan the fictional city could actually carry out. Then have them decide which corner replaces the most total miles and which one replaces the most large-diameter miles, and explain why those are not the same plan.

ELL supports

Pre-teach "main," "capital budget," "constraint," "binds," and "break rate" with a labeled sketch of a street cross-section showing a buried pipe. Provide sentence frames for item 13, such as "The limit that controls this plan is ___ because ___," so students can report their reasoning without composing an unfamiliar structure from scratch.

Extensions

Have students find the total miles of water main and the annual replacement mileage reported by their own water utility, either in a consumer confidence report or in a capital improvement plan, and compute the replacement cycle in years the way item 9 does. A utility that replaces 1 percent of its mains each year is on a 100-year cycle, and comparing that figure with the expected life of the pipe material makes a direct connection to the fictional Thornleigh problem.