Math problem set Grade 7 Math

Ardsley Desalination Energy Use: A Grade 7 Proportional Reasoning Set

A grade 7 problem set on desalination's energy cost: students use per-1,000-gallon kWh rates to compare seawater and conventional treatment and find unit rates.

Grade: Grade 7
Subject: Math
Material: Math problem set
Time: 40 min
Prep: No prep
Materials: Calculator

Standards

7.RP.A.2 Common Core Math

Recognize and represent proportional relationships between quantities, including identifying the constant of proportionality and representing proportional relationships by equations.

How this material aligns: Problems 1 through 4 and 7 use the fixed rate of kWh per 1,000 gallons as a proportional relationship, multiplying it by the number of thousand-gallon units to find total energy, which is the constant of proportionality this standard names.

All materials for 7.RP.A.2 →

7.RP.A.3 Common Core Math

Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

How this material aligns: Problem 6 asks students to compute how many percent more energy desalination uses than conventional treatment, the multistep percent reasoning this standard names.

All materials for 7.RP.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will use a per-1,000-gallon energy rate to find total energy for a given volume.
Students will find the unit rate of extra energy desalination uses over conventional treatment.
Students will compute the percent more energy desalination requires.

Vocabulary

Desalination: Turning salty water, like ocean water, into fresh water people can use. It works, but it takes a lot of energy and leaves behind very salty water that must be handled carefully.
Reverse osmosis: Reverse osmosis cleans water by pushing it through a membrane with holes so tiny that salt and most other dissolved things cannot fit through. It can even turn seawater into fresh water.

Also: kilowatt-hour, unit rate

Teacher overview

Turning seawater into drinking water with reverse osmosis takes a lot of electricity, more than treating fresh river water does. This set uses the fictional coastal town of Ardsley to have students compare the two using rates measured in kilowatt-hours (kWh) per 1,000 gallons. A common misconception to watch for is multiplying the energy rate by the full gallon count without first dividing by 1,000; the rate is given per 1,000 gallons, so the gallons must be scaled to thousands before multiplying. Watch also for students who try to add the two rates to the gallon amount instead of multiplying.

Ardsley Desalination Energy Use: A Grade 7 Proportional Reasoning Set — Grade 7 Math

Name: ____________________________ Date: ______________

Problems

Ardsley is a fictional coastal town. Suppose its desalination plant uses 12 kWh of electricity for every 1,000 gallons it produces, and a conventional plant nearby uses 1.5 kWh for every 1,000 gallons.

How much energy does the desalination plant use to produce 10,000 gallons, at 12 kWh per 1,000 gallons?
The desalination plant produces 500,000 gallons in one day. How many kWh does that take at 12 kWh per 1,000 gallons?
The conventional plant treats river water at 1.5 kWh per 1,000 gallons. How many kWh would it use for the same 500,000 gallons?
How many more kWh does desalination use than conventional treatment for that 500,000 gallons?
Find the unit rate of extra energy, in kWh per 1,000 gallons, that desalination uses compared with conventional treatment.
Desalination's 12 kWh is what percent more than conventional's 1.5 kWh per 1,000 gallons? Round to the nearest whole percent.
A new membrane lowers Ardsley's use to 11 kWh per 1,000 gallons. How many kWh would it then use for 500,000 gallons, and how many kWh per day does that save compared with 12 kWh?

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: 120 kWh.
10,000 gallons is 10 thousand-gallon units. 10 times 12 kWh equals 120 kWh.
Answer: 6,000 kWh.
500,000 gallons is 500 thousand-gallon units. 500 times 12 kWh equals 6,000 kWh.
Answer: 750 kWh.
500 thousand-gallon units times 1.5 kWh equals 750 kWh.
Answer: 5,250 kWh.
6,000 kWh minus 750 kWh equals 5,250 kWh more for desalination.
Answer: 10.5 kWh per 1,000 gallons.
12 kWh minus 1.5 kWh equals 10.5 kWh of extra energy for every 1,000 gallons.
Answer: 700 percent.
The extra energy is 12 minus 1.5, which is 10.5 kWh. Then 10.5 divided by 1.5 equals 7.0, or 700 percent more.
Answer: 5,500 kWh, saving 500 kWh.
500 units times 11 kWh equals 5,500 kWh. The old use was 6,000 kWh, so 6,000 minus 5,500 saves 500 kWh per day.

Differentiation

Support

Give students a first step reminder for each problem: divide the gallons by 1,000 to get the number of thousand-gallon units, then multiply by the rate. Provide the two rates already labeled so students do not mix them up.

Extension

Suppose Ardsley runs its plant every day for a month. Ask students to find the monthly energy use at the 12 kWh rate for 500,000 gallons per day, then compare it with the 11 kWh membrane, and explain how a small change in the rate adds up over 30 days.

ELL supports

Pre-teach "energy," "rate," and "per" with a simple example (miles per hour). Provide a two-column table labeled "gallons" and "kWh" so students can line up each calculation before solving.