Lesson plan Grade 7 Math
Budgeting Pump Hours at Elm Ridge Well: A Grade 7 Inequality Lesson
A 45-minute grade 7 lesson where students write and solve an inequality to find the maximum daily hours a well pump can run within a fixed water budget.
Objectives
- Students will write a one-variable inequality representing a real water-budget constraint with a rate and a fixed amount.
- Students will solve a one-variable inequality and interpret the solution set in context.
Vocabulary
Key words: inequality, fixed amount, variable, solution set
Teacher overview
The Elm Ridge Well Station (a fictional small water system used for this lesson) draws water with a pump that delivers 250 gallons every hour it runs. Because the well's groundwater permit limits how much water the station can withdraw from this well each day, the operator has to know exactly how many hours the pump can safely run, especially since a fixed 150 gallons is already used each day for an automated system flush before the pump's variable run-time even starts. This lesson has students set up and solve an inequality that answers exactly that question. A common misconception in this lesson is solving the inequality as if it were an equation and stopping at a single value, rather than reporting a full range of hours that keeps the total at or under budget.
Procedure
Hook (5 min): Ask students about Elm Ridge Well Station (a fictional system): if a pump adds 250 gallons for every hour it runs, and the well already has 150 gallons "spoken for" before the pump starts, how would you find the most hours the pump could run without going over a 2,000-gallon daily limit? Let a few students describe their thinking in words before introducing the inequality.
Direct instruction (12 min): Model writing the inequality with a variable h for hours: 250h + 150 ≤ 2,000. Solve it step by step: subtract 150 from both sides (250h ≤ 1,850), then divide both sides by 250 (h ≤ 7.4). Discuss what h ≤ 7.4 means in context: the pump can run for any number of hours up to and including 7.4 hours without exceeding the daily 2,000-gallon budget.
Guided practice (15 min): In pairs, students write and solve a new inequality for a day when the fixed flush uses 300 gallons instead of 150 gallons, with the same 250 gallons-per-hour pump rate and the same 2,000-gallon daily budget.
Independent practice (8 min): Each student writes and solves a third inequality on the handout: the fixed use is 150 gallons, the pump rate is 200 gallons per hour, and the daily budget is 1,750 gallons.
Closure / exit ticket (5 min): Students answer one question on an index card: "If Elm Ridge's daily budget were raised to 2,250 gallons, with the fixed 150-gallon flush and the 250-gallon-per-hour pump rate, what is the new maximum number of hours the pump could run?" Collect and check for the correct inequality setup before dismissal.
Student Handout
Elm Ridge Well Station is a fictional small water system. For each situation, write an inequality using h for the number of hours the pump runs, solve it, and report the greatest number of hours the pump can run without going over the daily budget.
The Elm Ridge pump delivers 250 gallons for every hour it runs. On this day the fixed flush uses 300 gallons before the pump starts, and the daily budget is 2,000 gallons. What is the greatest number of hours the pump can run?
On a different day the fixed flush uses 150 gallons, the pump delivers 200 gallons per hour, and the daily budget is 1,750 gallons. What is the greatest number of hours the pump can run?
Exit ticket: Elm Ridge's daily budget is raised to 2,250 gallons. The fixed flush still uses 150 gallons and the pump still delivers 250 gallons per hour. What is the greatest number of hours the pump can run?
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: 6.8 hr.
250h + 300 ≤ 2,000. Subtract 300: 250h ≤ 1,700. Divide by 250: h ≤ 6.8. The pump can run up to 6.8 hours. - Answer: 8 hr.
200h + 150 ≤ 1,750. Subtract 150: 200h ≤ 1,600. Divide by 200: h ≤ 8. The pump can run up to 8 hours. - Answer: 8.4 hr.
This is the exit-ticket answer. 250h + 150 ≤ 2,250. Subtract 150: 250h ≤ 2,100. Divide by 250: h ≤ 8.4. The pump can run up to 8.4 hours.
Differentiation
Support (below grade level)
Provide the inequality already set up (250h + 150 ≤ 2,000) with only the two solving steps left blank, and give a printed reminder of the two-step "subtract, then divide" solving order used for one-variable inequalities like this one.
Extension (above grade level)
Ask students to write a single inequality using a second variable for the fixed amount (250h + f ≤ 2,000) and solve for h in terms of f. Then have them find, for the original 2,000-gallon budget, the largest fixed amount f could be while still allowing the pump to run at least 5 hours.
ELL supports
Pre-teach "budget," "fixed," and "at most" using a simple visual of a bucket filling partway (the fixed amount) with a hose adding more (the pump's hourly rate) up to a marked line (the daily limit). Provide a sentence starter for the exit ticket: "The pump can run up to ___ hours because ___."