Worksheet Grade 9 Math

Dunmere Clarifier Testing: Removal Efficiency and Oxygen

A grade 9 water-testing worksheet at the fictional Dunmere plant, calculating turbidity removal efficiency and checking dissolved oxygen against a fish limit.

Objectives

  1. Students will calculate removal efficiency as a percent from influent and effluent readings.
  2. Students will compare dissolved-oxygen readings against a fish-health benchmark.
  3. Students will report a measured value at a precision that matches the instrument.

Vocabulary

Turbidity
Turbidity is a measure of how cloudy or murky water looks because of tiny particles floating in it. Clear water has low turbidity, and muddy water has high turbidity.
Dissolved oxygen
Dissolved oxygen is the oxygen gas mixed into water that fish and other underwater animals breathe. Cold, moving water holds more oxygen than warm, still water.

Also: removal efficiency, NTU, precision

Teacher overview

The fictional Dunmere treatment plant tests its water at every step and reports the results to the state. Two habits from a real lab drive this worksheet: measuring how much of a pollutant a step removes, stated as a percent, and matching the number of reported digits to what the meter can actually read. Students calculate removal efficiency, check dissolved oxygen against a fish-health limit, and clean up a reading that claims false precision.

Answer Key

  1. Answer: 80 %. (45 minus 9) divided by 45, times 100 = 36 divided by 45, times 100 = 80 percent.
  2. Answer: Yes. 0.2 NTU is at or below the 0.3 NTU goal.
  3. Answer: 80 %. (60 minus 12) divided by 60, times 100 = 48 divided by 60, times 100 = 80 percent.
  4. Answer: No. 80 percent is below the 85 percent target.
  5. Answer: 9 NTU. Solve 85 = (60 minus x) divided by 60, times 100. Then 0.85 = (60 minus x) divided by 60, so 60 minus x = 51 and x = 9 NTU.
  6. Answer: Yes. 6.4 mg/L is above the 5 mg/L fish limit.
  7. Answer: No, short by 0.8 mg/L. 4.2 is below 5; 5 minus 4.2 = 0.8 mg/L short.
  8. Answer: 6.07 mg/L. (6.4 plus 5.8 plus 6.0) divided by 3 = 18.2 divided by 3 = 6.0667, about 6.07 mg/L.
  9. Answer: 6.1 mg/L. The meter reads to the nearest 0.1 mg/L, so report only one decimal place; extra digits claim precision the meter does not have.
  10. Answer: A percent shows how much of the incoming load was removed, not just what is left. Two plants can leave the same settled number yet remove very different shares if their influent differs.

Differentiation

Support

Provide the removal-efficiency formula on a reference card and one fully worked example using numbers not on this worksheet. For items 1, 3, and 5, print the subtraction step already set up so students focus on the division and the percent.

Extension

Ask students to find the settled turbidity needed to reach 90 percent removal from a fictional 75 NTU influent, then write a general equation for the settled turbidity that meets any target percent from a 75 NTU start.

ELL supports

Pre-teach "influent," "effluent," "removal," and "precision" with two labeled jars, one cloudy and one clear, and a meter face showing one decimal place. Provide a sentence frame for item 10: "A percent tells us how much was ___, not just how much is ___."