Math problem set Grade 7 Math

Dunmore Reclamation Plant: A BOD Percent-Removal Problem Set

A grade 7 problem set at the fictional Dunmore plant, where students find the percent of waste removed across treatment stages and check it against the standard.

Grade: Grade 7
Subject: Math
Material: Math problem set
Time: 45 min
Prep: No prep
Materials: Calculator (optional)

Standards

7.RP.A.3 Common Core Math

Use proportional relationships to solve multistep ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error.

How this material aligns: Problems 1, 2, 4, and 5 each compute a percent removed from before-and-after concentrations, and problems 6 through 8 apply a percent increase and a percent removal, the multistep percent work this standard targets.

All materials for 7.RP.A.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 compute percent removed from an incoming and outgoing concentration.
Students will apply a percent increase and a percent removal to predict a treated value.

Vocabulary

Key words: percent removed, BOD, treatment stage

Teacher overview

A wastewater plant is judged by how much waste it takes out, and that removal is reported as a percent. This set uses the fictional Dunmore Water Reclamation Facility, following its BOD (a measure of how much waste is in the water) from the raw sewage coming in to the treated water leaving. Students compute the percent removed at each stage, check the final water against the 30 mg/L secondary standard, and then see what a storm does to the numbers. The misconception this set is built to correct is adding stage percents together: students often add a primary stage's percent and a secondary stage's percent to get the total removal, but each stage acts only on the waste still left, so the total must be computed from the original value.

Dunmore Reclamation Plant: A BOD Percent-Removal Problem Set — Grade 7 Math

Name: ____________________________ Date: ______________

Problems

Dunmore Water Reclamation Facility is a fictional plant. BOD (biochemical oxygen demand) measures how much oxygen microbes use to break down the waste in the water, in mg/L; more waste gives a higher BOD. The raw sewage coming in has a BOD of 240 mg/L. The secondary standard says the treated water leaving must be 30 mg/L or less. Percent removed is the change divided by the value coming in.

Primary settling lowers the BOD from 240 mg/L to 168 mg/L. Find the percent removed in this stage.
After all treatment, the BOD leaving is 24 mg/L. Find the total percent removed from the 240 mg/L coming in.
The standard is 30 mg/L or less. Is the final 24 mg/L within the standard, and by how many mg/L?
For solids (TSS), the value coming in is 220 mg/L and the value leaving is 22 mg/L. Find the percent removed.
The secondary stage lowers BOD from 168 mg/L to 24 mg/L. Find the percent removed in that stage alone. Round to the nearest tenth of a percent.
A storm raises the incoming BOD by 25% above the usual 240 mg/L. What is the new incoming BOD?
If the plant still removes 90% of that storm load, what is the BOD leaving? Is it within the 30 mg/L standard?
To stay safely under the standard, the operator wants 92% removal of the storm load. What BOD would leave the plant then?

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: 30%.
The change is 240 − 168 = 72 mg/L. 72 ÷ 240 = 0.30, which is 30%.
Answer: 90%.
The change is 240 − 24 = 216 mg/L. 216 ÷ 240 = 0.90, which is 90%.
Answer: 6 mg/L below.
24 mg/L is less than 30 mg/L, so the water meets the standard, with 30 − 24 = 6 mg/L to spare.
Answer: 90%.
The change is 220 − 22 = 198 mg/L. 198 ÷ 220 = 0.90, which is 90%.
Answer: 85.7%.
The change is 168 − 24 = 144 mg/L. 144 ÷ 168 = 0.857, which rounds to 85.7%.
Answer: 300 mg/L.
A 25% increase adds 0.25 × 240 = 60 mg/L, so 240 + 60 = 300 mg/L.
Answer: 30 mg/L; yes, it just meets the 30 mg/L-or-less standard, with no margin.
Removing 90% leaves 10%, and 0.10 × 300 = 30 mg/L, which is exactly at the 30 mg/L standard, leaving no margin.
Answer: 24 mg/L.
Removing 92% leaves 8%, and 0.08 × 300 = 24 mg/L, which is safely under the standard.

Differentiation

Support

Give students the percent-removed formula on a printed strip with problem 1 worked as a model: change first, then divide by the value coming in. Let students use a calculator so they can focus on choosing the right two numbers rather than on long division.

Extension

Ask students to show that removing 30% and then removing another amount cannot simply be added to reach 90% total. Have them compute the secondary-stage percent needed after a 30% primary removal to reach 90% overall, and explain why it is larger than 60%.

ELL supports

Provide a labeled diagram with "coming in" on the left, "leaving" on the right, and an arrow marked "percent removed" between them. Pair each problem with the frame "from ___ to ___ is ___ percent removed" so students can state the two values they are comparing.

Extensions

Have students find a local wastewater plant's annual report, which utilities post publicly, and read the percent removal it reports for BOD or solids, using only the figures the plant itself publishes. Comparing a real plant's reported removal with Dunmore's 90 percent shows how consistently secondary treatment performs.