Lesson plan Grade 11 Math

Calverhill's Lead Service Lines: A Fixed Count or a Fixed Percent

A 50-minute grade 11 lesson comparing a linear and an exponential lead service line replacement plan, with logarithms used to find when each plan reaches its target.

Grade: Grade 11
Subject: Math
Material: Lesson plan
Time: 50 min
Prep: Low prep
Materials: Printed handout, Scientific calculators

Standards

HSF-LE.A.1 Common Core Math

Distinguish between situations that can be modeled with linear functions and with exponential functions, recognizing that linear functions grow by equal differences over equal intervals and exponential functions grow by equal factors over equal intervals.

How this material aligns: Students decide which replacement plan changes by equal differences over equal intervals and which changes by equal factors, then justify the choice from the year-by-year counts in questions 3 through 5.

All materials for HSF-LE.A.1 →

HSA-CED.A.1 Common Core Math

Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

How this material aligns: Students create and solve an exponential equation in one variable with logarithms to find the year the remaining inventory falls below a stated target.

All materials for HSA-CED.A.1 →

Download Word file

This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-09-04.

Objectives

Students will distinguish a constant-count replacement plan from a constant-percent plan using successive differences and ratios.
Students will solve an exponential equation for time using logarithms.
Students will compare two plans against a stated ten-year deadline.

Vocabulary

Service line: A service line is the small pipe that connects the water main under the street to the plumbing inside one home or building. It is the last leg of the trip water makes to your tap.

Also: replacement plan, exponential decay, logarithm, inventory

Teacher overview

EPA estimates that about 4 million lead service lines remain in United States water systems, a November 2025 figure built from the inventories utilities reported that year, and the Lead and Copper Rule Improvements require utilities to replace nearly all of theirs within ten years. A replacement program could be written in one of two ways, either a fixed number of lines each year or a fixed share of whatever remains, and the two produce very different finish dates. This lesson gives students both plans for the fictional city of Calverhill and asks them to decide which one meets the deadline. The misconception it targets is the belief that a percentage plan is always faster because it starts faster, which the first year's numbers appear to confirm and the tenth year's numbers contradict.

Procedure

Launch (5 min): Give the class the fictional Calverhill inventory of 6,400 lead service lines and the two plans, one replacing 640 lines a year and the other replacing 15 percent of the lines still in the ground. Ask for a show of hands on which plan finishes first, and record the tally on the board without comment.

Direct instruction (12 min): Build a five-row table for each plan on the board, showing the count replaced during each year and the count remaining after it. Name the pattern in each column: equal differences for the first plan and equal ratios for the second. Write the two models, one as a linear function and one as 6,400 times 0.85 raised to the power t.

Guided practice (10 min): Working in pairs, students find the year the linear plan reaches zero and then attempt the same question for the exponential plan. When pairs report that the exponential plan never reaches zero, introduce the target of fewer than 100 lines remaining and show how taking the logarithm of both sides turns the exponential equation into a linear one in t.

Independent practice (18 min): Students complete questions 1 through 11 on the handout, including the modified plan in question 11.

Closure and formative check (5 min): Students answer handout question 12 on an index card, naming the plan they would recommend and the number from their own work that supports it. Collect the cards and count how many cite a specific figure rather than a general preference, with fewer than three-quarters citing a figure signaling that the next session should open with evidence-based justification.

Calverhill's Lead Service Lines: A Fixed Count or a Fixed Percent — Grade 11 Math

Name: ____________________________ Date: ______________

Student Handout

The fictional city of Calverhill has 6,400 lead service lines still in the ground. Its utility is choosing between two replacement plans.

Plan A replaces exactly 640 lines each year.
Plan B replaces 15 percent of the lines still remaining at the start of each year.

Let t stand for the number of years since the program began.

Under Plan A, find the number of lines remaining after 5 years.
Under Plan B, find the number of lines remaining after 5 years, to the nearest whole line.
Under Plan B, find the number of lines replaced during year 1.
Under Plan B, find the number of lines replaced during year 5, to the nearest whole line.
Explain in one or two sentences why Plan A replaces the same count every year while Plan B does not.
Find the number of years Plan A needs to reach zero lines remaining.
Write the function giving the number of lines remaining under Plan B after t years.
Solve your equation from question 7 to find the number of whole years until fewer than 100 lines remain under Plan B.
The Lead and Copper Rule Improvements require replacing nearly all lead service lines within 10 years. Find the number of lines still in the ground after 10 years under Plan B, to the nearest whole line.
Find what percent of the starting inventory your answer to question 9 represents, to one decimal place.
Calverhill wants a percentage plan that leaves fewer than 100 lines after 10 years. Find the yearly replacement percent that leaves exactly 100 lines at 10 years, to the nearest whole percent.
Name the plan you would recommend to the Calverhill council, and support your choice with one number from your own work.

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

Answer Key

Answer: 3,200 lines.
Subtracting gives 6,400 - 640(5) = 6,400 - 3,200 = 3,200 lines remaining.
Answer: 2,840 lines.
Raising 0.85 to the fifth power gives 0.443705, and 6,400 x 0.443705 = 2,839.7, which rounds to 2,840 lines.
Answer: 960 lines.
Fifteen percent of the starting inventory is 0.15 x 6,400 = 960 lines.
Answer: 501 lines.
At the start of year 5 there are 6,400 x 0.85 to the fourth power = 3,340.8 lines, and 0.15 x 3,340.8 = 501.1, which rounds to 501 lines.
Answer: Plan A removes a fixed count while Plan B removes a fixed share of a shrinking inventory.
Plan A subtracts the same 640 every year, which gives equal differences, while Plan B multiplies by 0.85 every year, so each year's replacement count is 15 percent of a smaller number than the year before.
Answer: 10 years.
Dividing gives 6,400 divided by 640 = 10 years, at which point no lines remain.
Answer: L(t) = 6,400(0.85) raised to the power t.
The starting value is 6,400 lines and each year retains 85 percent of the lines that were there, so the retention factor is 0.85.
Answer: 26 years.
Setting 6,400(0.85) to the power t = 100 gives 0.85 to the power t = 0.015625. Taking logarithms gives t = log(0.015625) divided by log(0.85) = 25.59, so the count first falls below 100 during year 26.
Answer: 1,260 lines.
Raising 0.85 to the tenth power gives 0.196874, and 6,400 x 0.196874 = 1,260.0 lines.
Answer: 19.7%.
Dividing gives 1,260 divided by 6,400 = 0.1969, which is 19.7 percent of the starting inventory.
Answer: 34%.
Setting 6,400(1 - p) to the tenth power = 100 gives (1 - p) to the tenth power = 0.015625, so 1 - p = 0.015625 raised to the power 0.1 = 0.6597 and p = 0.3403, which rounds to 34 percent.
Answer: Plan A.
Plan A clears the inventory in 10 years, matching the rule's deadline, while Plan B still leaves 1,260 lines at year 10 and needs 26 years to fall below 100. A recommendation of Plan B is defensible only if the replacement percent is raised to about 34 percent as in question 11, and an answer along those lines earns credit when it cites that figure.

Differentiation

Support

Give these students the two five-row tables for the fictional Calverhill inventory already started, with the year-1 and year-2 rows filled in for both plans, so the pattern-naming work in question 5 comes before any modeling. Provide the logarithm step for question 8 as a worked template with the numbers left blank, and allow a table of values as an alternative route to the answer.

Extension

Ask these students to find, for the fictional Calverhill inventory, the constant yearly count that would leave the same number of lines at year 10 as the 15 percent plan does, and then to explain which of the two plans costs a utility more in the first three years if each replacement costs the same amount. Have them describe how a crew hiring limit would change their recommendation.

ELL supports

Pre-teach "service line," "inventory," and "replacement" with a labeled diagram showing the pipe running from the water main under the street to a house. Provide a two-column comparison frame with the headings "same amount each year" and "same percent each year," and sentence starters for question 12 such as "I recommend Plan ___ because after 10 years it leaves ___ lines."

Extensions

Students can look up whether their own state publishes a lead service line inventory and describe, without adding numbers to their Calverhill work, how a utility would know the count in the first place. A second task is to write the two Calverhill plans as recursive rules rather than closed-form functions, and to explain which form makes the year-by-year replacement counts easier to read.