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Math problem set Grade 10 Math

Kettlebrook's Aging Water Main: A Leak-Loss Problem Set

An 8-problem grade 10 set where students track gallons and cost lost to a leaking water main, chaining unit conversions from gallons per minute to acre-feet.

Objectives

  1. Students will convert a leak's flow rate between gallons per minute, million gallons per day, and acre-feet.
  2. Students will calculate what percent of a water main's total flow is lost to a leak.
  3. Students will calculate the cost of repairing a section of aging pipe from a per-foot rate.

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.
Acre-foot
The amount of water it takes to cover one acre of land, about the size of a football field, one foot deep. Western states use acre-feet to measure large amounts of water.

Also: leak loss, percent of total flow

Teacher overview

Aging pipes fail more often than new ones, and in the United States a water main breaks somewhere roughly every two minutes. This problem set follows a single leak on a fictional aging main in Kettlebrook to build a scenario arc: students calculate the leak's size relative to the main's total flow, the total volume lost before repair, and the cost of fixing the pipe section. A common misconception is treating a "small" leak in gallons per minute as insignificant; multiplying a small rate by enough time, as this set does, shows how quickly a modest leak adds up to a large total loss.

Kettlebrook's Aging Water Main: A Leak-Loss Problem Set — Grade 10 Math
Name: ____________________________ Date: ______________

Problems

Kettlebrook's water utility (a fictional utility used for this problem set) operates a 60-year-old main that normally carries 3.5 million gallons per day (MGD). A crack in the pipe is losing water at a steady rate of 250 gallons per minute (gpm). Use the conversion that 1 MGD equals 694 gpm, and that 1 acre-foot equals 325,851 gallons, for all problems below.

  1. Convert the leak's flow rate (250 gpm) to million gallons per day (MGD). Round to the nearest hundredth.
  2. What percent of the main's total flow (3.5 MGD) is being lost to the leak? Round to the nearest tenth of a percent.
  3. If the leak runs continuously for 30 days before repair crews find and fix it, how many total gallons are lost?
  4. Convert your answer from problem 3 to acre-feet. Round to the nearest tenth.
  5. The utility hires a crew to replace the leaking 40-foot section of pipe at a hypothetical cost of $185 per foot. What is the total repair cost?
  6. If the utility instead takes 60 days to find and repair the leak, how many total gallons would be lost, and how many more gallons is that compared to problem 3's 30-day total?
  7. Using your answer from problem 3, explain in one or two sentences why aging infrastructure like Kettlebrook's 60-year-old main increases both water loss and repair costs the longer a leak goes unrepaired.
  8. Once repaired, the main delivers its full 3.5 MGD instead of losing water to the leak. Using your answer from problem 1, what percent increase in delivered flow does the repair represent, compared to the reduced flow while the main was leaking? Round to the nearest tenth of a percent.

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: 0.36 MGD.
    250 gpm ÷ 694 gpm/MGD = 0.3602, which rounds to 0.36 MGD.

  2. Answer: 10.3 percent.
    The main's total flow in gpm is 3.5 × 694 = 2,429 gpm. 250 ÷ 2,429 × 100 = 10.293, which rounds to 10.3%.

  3. Answer: 10,800,000 gal.
    250 gpm × 60 min/hr × 24 hr/day × 30 days = 10,800,000 gallons.

  4. Answer: 33.1 acre-ft.
    10,800,000 gal ÷ 325,851 gal/acre-ft = 33.144, which rounds to 33.1 acre-feet.

  5. Answer: $7,400.
    40 ft × $185/ft = $7,400.

  6. Answer: 21,600,000 gal.
    250 gpm × 60 × 24 × 60 days = 21,600,000 gallons, which is 10,800,000 gallons more than the 30-day total in problem 3 (twice as long a delay produces exactly twice the loss).

  7. Answer: Longer delays multiply both loss and cost.
    Because the leak loses water at a constant rate, every extra day it runs adds the same 360,000 gallons (250 gpm × 1,440 min/day) to the total, so an old main left unrepaired for months, not weeks, can lose millions of additional gallons and require a larger, costlier repair once the surrounding pipe has degraded further.

  8. Answer: 11.5 percent.
    While leaking, delivered flow was 3.5 − 0.36 = 3.14 MGD. The repair restores flow to 3.5 MGD, an increase of 0.36 MGD. 0.36 ÷ 3.14 × 100 = 11.46, which rounds to 11.5%.

Differentiation

Support (below grade level)

Provide the leak's flow rate for the fictional Kettlebrook main already converted to MGD (0.36 MGD) before problem 2, and give students a printed conversion reference card listing "1 MGD = 694 gpm" and "1 acre-ft = 325,851 gal" so they do not need to recall the factors from memory.

Extension (above grade level)

Ask students to find about how many days it would take the leak, at its original 250 gpm rate, to lose a total of 100 acre-feet of water, working backward from the acre-foot conversion used in problem 4.

ELL supports

Provide a labeled diagram of the water main showing the crack and an arrow representing the leak, alongside a simple flowchart of the unit-conversion chain (gpm to MGD to gallons to acre-feet) that students can reference while solving.