Math problem set Grade 6 Math

Tern Island Desalination Skid: Recovery and Rejection

A grade 6 math problem set on a fictional seawater plant, where students find recovery percent, salt rejection percent, and an energy unit rate.

Grade: Grade 6
Subject: Math
Material: Math problem set
Time: 40 min
Prep: No prep
Materials: Calculator

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: Problems 1 and 4 have students find recovery percent and salt rejection percent from stated feed and product amounts, the percent reasoning this standard targets.

All materials for 6.RP.A.3 →

6.RP.A.2 Common Core Math

Understand the concept of a unit rate a/b associated with a ratio a:b with b not equal to 0, and use rate language in the context of a ratio relationship.

How this material aligns: Problems 5 and 6 have students apply a fixed energy rate per 1,000 gallons to new volumes, the unit-rate reasoning this standard requires.

All materials for 6.RP.A.2 →

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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 distinguish recovery percent from salt rejection percent and calculate each from stated amounts.
Students will apply a fixed energy rate per 1,000 gallons and reverse a recovery percent to find a needed feed amount.

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, rejection, unit rate

Teacher overview

A reverse osmosis plant makes fresh water from seawater by pushing it through a membrane at high pressure. Part of the seawater passes through as clean water, and the rest leaves as salty brine. Two different percents describe how the plant performs: recovery, which is the share of the feed water that becomes clean water, and salt rejection, which is the share of the salt that the membrane holds back. Tern Island is a fictional plant created for this problem set, so its numbers are made up for practice. A common misconception is treating recovery and rejection as the same percent; they measure two different things and are usually far apart.

Tern Island Desalination Skid: Recovery and Rejection — Grade 6 Math

Name: ____________________________ Date: ______________

Problems

Suppose the fictional Tern Island plant makes fresh water from seawater. It pushes seawater through a membrane. Some seawater becomes clean water. The rest leaves as salty brine. Use the numbers in each problem.

The plant pushes in 1,000 gallons of seawater and makes 450 gallons of clean water. Find the recovery percent.
How many gallons leave as brine from that 1,000 gallons?
The seawater at Tern Island holds 35,000 mg/L of salt. The clean water holds 350 mg/L. How much salt was held back per liter?
Find the salt rejection percent: salt held back, divided by salt in, times 100.
The plant uses 12 kWh of energy for every 1,000 gallons of clean water. How much energy does 2,000 gallons take?
At that same rate, how much energy does 5,000 gallons of clean water take?
Recovery stays at 45 percent. The plant needs 900 gallons of clean water. How many gallons of seawater must it push in?
Explain why the brine leaving the plant is saltier than the seawater that came in.

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: 45%. Recovery is clean water divided by feed: 450 ÷ 1,000 = 0.45, or 45 percent.
Answer: 550 gallons. The feed was 1,000 gallons and 450 became clean water, so 1,000 − 450 = 550 gallons of brine.
Answer: 34,650 mg/L. The salt dropped from 35,000 to 350, and 35,000 − 350 = 34,650 mg/L held back.
Answer: 99%. Rejection is salt held back divided by salt in: 34,650 ÷ 35,000 = 0.99, or 99 percent.
Answer: 24 kWh. The rate is 12 kWh per 1,000 gallons, and 2,000 gallons is two such units: 2 × 12 = 24 kWh.
Answer: 60 kWh. 5,000 gallons is five units of 1,000, and 5 × 12 = 60 kWh.
Answer: 2,000 gallons. At 45 percent recovery, 900 is 0.45 of the feed, so feed = 900 ÷ 0.45 = 2,000 gallons.
Answer: The salt stays behind. Clean water leaves through the membrane while nearly all the salt remains in the brine, so the brine holds the same salt in less water.

Differentiation

Support

Provide a labeled diagram showing seawater entering, clean water leaving one side, and brine leaving the other, so students can see where each number belongs. Give the recovery formula and the rejection formula on a printed card, and let students use a calculator for the division steps.

Extension

Ask students to find the recovery percent the plant would need to make 900 gallons of clean water from only 1,500 gallons of seawater, then explain whether that is higher or lower than the plant's current 45 percent and why.

ELL supports

Pre-teach seawater, membrane, brine, recovery, and rejection with a simple picture and a short home-language phrase for each. Allow students to record answers as labeled number sentences. Provide a sentence frame for problem 8: "The brine is saltier because ___."