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Math problem set Grade 7 Math

Saltcedar Reverse Osmosis: A Grade 7 Recovery-Percent Set

A grade 7 set where students find recovery and salt-removal percents, percent increase, and energy cost for a fictional reverse-osmosis plant using decimal chains.

Objectives

  1. Students will find recovery and salt-removal as percents of a feed stream.
  2. Students will find a percent increase in product water and a percent decrease in price.
  3. Students will chain decimal steps to compute daily energy use and cost.

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, percent decrease

Teacher overview

A reverse-osmosis plant pushes water through a membrane that lets clean water pass while holding salt back. The clean water is called permeate, and the share of the incoming feed that becomes permeate is the recovery. This set uses a fictional plant to practice recovery percent, salt-removal percent, and a short energy-cost chain. A common misconception is that "60 percent recovery" means the salt is 60 percent removed; recovery and salt removal are two different percents of two different quantities, and problems 1 and 7 keep them apart on purpose.

Saltcedar Reverse Osmosis: A Grade 7 Recovery-Percent Set — Grade 7 Math
Name: ____________________________ Date: ______________

Problems

Saltcedar is a fictional reverse-osmosis plant used for this set. Each day, 500,000 gallons of feed water enter the plant. The plant recovers 60 percent of that feed as clean permeate.

  1. How many gallons of permeate does the plant make each day at 60 percent recovery?
  2. The rest of the feed leaves as salty reject water. How many gallons is the reject, and what percent of the feed is it?
  3. The plant tunes the membranes and raises recovery to 65 percent, so it now makes 325,000 gallons of permeate. Find the percent increase in permeate from the original amount in problem 1.
  4. The plant uses 3.6 kWh of energy for every 1,000 gallons of permeate. How much energy does it use to make the original 300,000 gallons from problem 1?
  5. Electricity costs $0.14 per kWh. What is the daily energy cost to make that permeate?
  6. A power rebate cuts the electricity price by 15 percent. Find the new price per kWh, then the new daily energy cost at the same energy use from problem 4.
  7. The membranes remove 98 percent of the salt. The feed water has 12,000 mg/L of salt. How many mg/L of salt remain in the permeate?
  8. The plant wants permeate salt below 250 mg/L. Does the permeate from problem 7 meet that goal, and by how many mg/L?

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

Answer Key
  1. Answer: 300,000 gallons.
    0.60 × 500,000 = 300,000 gallons of permeate.
  2. Answer: 200,000 gallons; 40%.
    Reject = 500,000 − 300,000 = 200,000 gallons. 200,000 ÷ 500,000 = 0.40, or 40 percent of the feed.
  3. Answer: 8.3%.
    The increase is 325,000 − 300,000 = 25,000 gallons. 25,000 ÷ 300,000 = 0.0833, which rounds to 8.3 percent.
  4. Answer: 1,080 kWh.
    300,000 ÷ 1,000 = 300 thousand-gallon units. 300 × 3.6 = 1,080 kWh.
  5. Answer: $151.20.
    1,080 kWh × $0.14 = $151.20.
  6. Answer: $128.52.
    New price = 0.14 × (1 − 0.15) = 0.14 × 0.85 = $0.119 per kWh. New cost = 1,080 × 0.119 = $128.52.
  7. Answer: 240 mg/L.
    The membranes remove 98 percent, so 2 percent of the salt remains. 0.02 × 12,000 = 240 mg/L.
  8. Answer: 10 mg/L under.
    Yes, it meets the goal. 250 − 240 = 10, so the permeate is 10 mg/L below the 250 mg/L goal.

Differentiation

Support

Provide a labeled diagram of the plant showing feed splitting into permeate and reject, with the 60 percent and 40 percent shares marked, before problem 1. Supply the energy setup for problem 4 as "300 groups of 1,000 gallons" so students see why they divide first.

Extension

Suppose the plant runs at the higher 65 percent recovery from problem 3. Ask students to find the daily energy cost, using the same 3.6 kWh per 1,000 gallons and the rebate price from problem 6, then to explain whether higher recovery raises or lowers the cost per gallon of permeate.

ELL supports

Pre-teach "feed," "permeate," "reject," and "recovery" with a three-arrow picture. Provide a sentence frame for problems 2 and 8: "The ___ is ___ because ___." Allow the salt percents in problem 7 to be shown as 98 percent and 2 percent on a bar before calculating.