Math problem set Grade 6 Engineering & Design

Sunmeadow Greenhouse Reverse Osmosis Recovery Problems

A grade 6 engineering set where students size a greenhouse reverse osmosis unit using percent recovery, unit rates, and a ratio table of feed water to clean permeate.

Grade: Grade 6
Subject: Engineering & Design
Material: Math problem set
Time: 40 min
Prep: No prep
Materials: Calculator (optional)

Standards

6.RP.A.3 Common Core Math

Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations, including finding a percent of a quantity and converting measurement units.

How this material aligns: Students find percent recovery in problems 2 and 3, complete the feed-to-permeate ratio table in problem 8, and use a unit price in problem 5, the ratio and percent reasoning this standard targets.

All materials for 6.RP.A.3 →

6.NS.B.3 Common Core Math

Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.

How this material aligns: Problems 5 and 9 require multiplying and dividing multi-digit decimals, such as 0.006 dollars per gallon and dividing by 0.6, exercising the decimal fluency this standard expects.

All materials for 6.NS.B.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 calculate the percent recovery of a reverse osmosis unit from feed and permeate volumes.
Students will use unit rates and a ratio table to size an RO unit for a stated daily water need.

Vocabulary

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: recovery, permeate, feed water

Teacher overview

Sunmeadow Greenhouse, in the fictional town of Halvard, uses a small reverse osmosis (RO) unit to make clean water for its seedlings. An RO unit pushes feed water through a membrane. Some water passes through as clean permeate, and the rest leaves as salty concentrate. This set has students find the unit's recovery and size it for a daily need.

Recovery tells you what share of the feed water becomes clean water. A common mistake is dividing feed by permeate instead of permeate by feed. That flips the ratio and makes recovery look larger than 100 percent, which is impossible. Recovery is always less than 100 percent because some water always leaves as concentrate.

Sunmeadow Greenhouse Reverse Osmosis Recovery Problems — Grade 6 Engineering & Design

Name: ____________________________ Date: ______________

Problems

Use this scenario for every problem. Sunmeadow Greenhouse is a fictional greenhouse in the town of Halvard. Its RO unit takes in 500 gallons of feed water each day. It makes 300 gallons of clean permeate. Recovery (%) = permeate divided by feed, times 100.

How many gallons of concentrate does the unit make each day?
Find the unit's recovery as a percent.
What percent of the feed water leaves as concentrate?
Each membrane makes 75 gallons of permeate per day. How many membranes are needed to make 300 gallons?
The greenhouse pays $6 for every 1,000 gallons of feed water. The unit uses 500 gallons of feed each day. What is the cost of one day of feed water?
How many gallons of permeate does the unit make in a 30-day month?
The seedling benches need 250 gallons of clean water each day. Does 300 gallons of permeate meet that need, and by how much?
Recovery stays at 60 percent. Complete the ratio table of feed water to permeate.

Feed water (gallons) Permeate (gallons)

100

250

400

750
The greenhouse wants 360 gallons of permeate each day at the same 60 percent recovery. How many gallons of feed water are needed?
In one or two sentences, explain why the recovery of an RO unit is always less than 100 percent.

Feed water (gallons)	Permeate (gallons)
100
250
400
750

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: 200 gallons. Feed minus permeate: 500 - 300 = 200 gallons of concentrate.
Answer: 60 percent. Recovery = 300 / 500 = 0.6, and 0.6 times 100 = 60 percent.
Answer: 40 percent. Concentrate is 200 / 500 = 0.4, or 40 percent. This also equals 100 - 60.
Answer: 4 membranes. 300 gallons / 75 gallons per membrane = 4 membranes.
Answer: $3.00. The price is $6 / 1,000 gallons = $0.006 per gallon. Then 500 x 0.006 = $3.00 for one day.
Answer: 9,000 gallons. 300 gallons per day x 30 days = 9,000 gallons.
Answer: 50 gallons. Yes, 300 gallons is enough. 300 - 250 = 50 gallons to spare.
Answer: 60, 150, 240, 450 gallons. Multiply each feed value by 0.6: 100 gives 60; 250 gives 150; 400 gives 240; 750 gives 450.
Answer: 600 gallons. Feed = permeate / recovery = 360 / 0.6 = 600 gallons of feed water.
Answer: Some water always leaves as concentrate. Because concentrate carries away part of the feed, the permeate is only a share of the feed, so recovery stays below 100 percent.

Differentiation

Support

Give students the recovery formula on a printed strip with the words in place: recovery equals permeate divided by feed, times 100. Work problem 8's ratio table first so the steady 60 percent pattern is visible before they tackle problems 2 and 9. Allow a calculator for every division step.

Extension

Ask students to find the recovery of a second unit that takes in 800 gallons of feed and makes 520 gallons of permeate, then explain which unit wastes less water as concentrate and why a greenhouse might still choose the lower-recovery unit.

ELL supports

Provide a labeled diagram of an RO unit showing three arrows: feed water in, permeate out, and concentrate out. Pair each term with a short definition card. Let students point to the diagram to explain problem 10 before writing their sentence.