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Worksheet Grade 7 Math

Aldermoor Plant Turbidity: A Grade 7 Percent-Removal Worksheet

A grade 7 worksheet where students apply the removal-efficiency formula to a fictional plant's turbidity readings, then check finished water against the NTU limit.

Objectives

  1. Students will calculate the percent of turbidity a treatment stage removes with the removal-efficiency formula.
  2. Students will compare two treatment stages by percent removed rather than by the raw change in NTU.
  3. Students will find percent increase and percent error for turbidity readings.

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.

Also: NTU, removal efficiency, percent error

Teacher overview

Turbidity is a measure of how cloudy water is, reported in NTU units, and a treatment plant's job is to lower it before the water reaches homes. This worksheet uses a fictional plant with three readings, one before treatment and two after, so students practice the same percent-removal math operators use on a daily log sheet. A common student error is to compare stages by the raw drop in NTU instead of the percent removed; item 4 and item 10 are built to surface that mistake.

Aldermoor Plant Turbidity: A Grade 7 Percent-Removal Worksheet — Grade 7 Math
Name: ____________________________ Date: ______________

Student Worksheet

The Aldermoor Water Treatment Plant is a fictional plant used for this worksheet. It cleans river water in two steps. First the water rests in a basin so dirt settles out. Then it passes through sand filters. The plant checks how cloudy the water is in NTU units. A lower NTU means clearer water.

Use these three readings for every question:

  • Raw river water: 18 NTU
  • After the settling basin: 4.5 NTU
  • After the sand filters: 0.24 NTU

The EPA rule says finished water must stay at or below 0.3 NTU in 95 percent of monthly samples, and it must never go above 1 NTU. The removal formula is: percent removed = (start NTU − end NTU) ÷ start NTU × 100.

  1. Find the percent of cloudiness removed across the whole plant, from 18 NTU down to 0.24 NTU. Round to the nearest tenth of a percent.
  2. Find the percent removed by the settling basin alone, from 18 NTU to 4.5 NTU.
  3. Find the percent removed by the sand filters, from 4.5 NTU to 0.24 NTU. Round to the nearest tenth of a percent.
  4. The basin removed more NTU than the filters did. But which step removed the greater percent? Use your answers to items 2 and 3, and explain in one sentence.
  5. Does the finished water at 0.24 NTU meet the 0.3 NTU limit? If it does, by how many NTU is it under the limit?
  6. The plant's target for finished water is 0.20 NTU, but the sample read 0.24 NTU. Find the percent error. Use percent error = (measured − target) ÷ target × 100.
  7. A storm makes the raw water jump from 18 NTU to 27 NTU. Find the percent increase in the raw reading.
  8. The finished water is always the same fraction of the raw water. Item 1 used 0.24 out of 18. Find the finished NTU when the raw water is 27 NTU. Round to the nearest hundredth.
  9. Does that finished water meet the 0.3 NTU limit? If not, by how many NTU does it miss?
  10. Explain why percent removed is a fairer way to compare the two steps than the plain drop in NTU.

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: 98.7%.
    (18 − 0.24) ÷ 18 × 100 = 17.76 ÷ 18 × 100 = 98.6667, which rounds to 98.7 percent.
  2. Answer: 75%.
    (18 − 4.5) ÷ 18 × 100 = 13.5 ÷ 18 × 100 = 75 percent exactly.
  3. Answer: 94.7%.
    (4.5 − 0.24) ÷ 4.5 × 100 = 4.26 ÷ 4.5 × 100 = 94.6667, which rounds to 94.7 percent.
  4. Answer: The sand filters.
    The filters removed 94.7 percent while the basin removed 75 percent, so the filters removed the greater percent even though the basin removed more raw NTU (13.5 NTU versus 4.26 NTU).
  5. Answer: 0.06 NTU under.
    Yes, it meets the limit. 0.3 − 0.24 = 0.06 NTU, so the finished water sits 0.06 NTU below the limit.
  6. Answer: 20%.
    (0.24 − 0.20) ÷ 0.20 × 100 = 0.04 ÷ 0.20 × 100 = 20 percent.
  7. Answer: 50%.
    (27 − 18) ÷ 18 × 100 = 9 ÷ 18 × 100 = 50 percent.
  8. Answer: 0.36 NTU.
    27 × (0.24 ÷ 18) = 27 × 0.013333 = 0.36 NTU. The finished reading stays the same fraction of the raw reading.
  9. Answer: 0.06 NTU over.
    No, it does not meet the limit. 0.36 − 0.30 = 0.06, so the finished water is 0.06 NTU above the 0.3 NTU limit.
  10. Answer: Percent removed compares each step to the water it started with.
    A step that starts with cloudier water can drop more NTU yet still clean a smaller share, so percent removed shows how well each step does its job.

Differentiation

Support

Give students the removal formula worked once on a separate example (from 40 NTU down to 10 NTU) before they start item 1, and let them keep a calculator for every division step. Provide a number line marked at 0.3 NTU so item 5 and item 9 become a visual comparison.

Extension

Ask students to find the raw turbidity that would make the finished water land exactly at the 0.3 NTU limit, using the same 0.24-out-of-18 fraction from item 8, and to explain what that means for the plant during a large storm.

ELL supports

Pre-teach "turbidity," "settling basin," and "sand filter" with a simple three-box diagram showing the water getting clearer left to right. Provide a sentence frame for items 4 and 10: "The ___ removed the greater percent because ___."