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Quiz & test questions Grade 9 Math

Glenmara Treatment Plant: A Process Math Quiz

A grade 9 unit quiz on the fictional Glenmara plant, with computation and multiple-choice items on detention time, filter loading, removal efficiency, and dosing.

Objectives

  1. Students will calculate detention time, loading rate, removal efficiency, and chlorine feed.
  2. Students will convert a plant flow between MGD and gallons per minute.
  3. Students will identify treatment steps and a finished-water turbidity limit.

Vocabulary

Sedimentation
Sedimentation is the step where water sits almost still in a big basin so the heavy clumps of dirt can sink to the bottom. The cleaner water on top then moves on to the filters.
Detention time
Detention time is how long water stays inside a tank or basin before flowing out. Water must stay long enough for each treatment step, like settling or disinfecting, to finish its job.
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: loading rate, removal efficiency, pounds formula

Teacher overview

This 40-minute unit quiz, worth 30 points total, checks grade 9 skills on the fictional Glenmara treatment plant. It mixes multiple-choice, short-answer, and computation items. Students move through the treatment train: sedimentation detention time, filter loading rate, removal efficiency, and chlorine dosing, then convert a plant flow and recall a finished-water turbidity limit.

Glenmara Treatment Plant: A Process Math Quiz — Grade 9 Math
Name: ____________________________ Date: ______________

Assessment

Glenmara (a fictional plant) treats river water. Use these facts: detention time = volume divided by flow; loading rate = flow in gpm divided by area in square feet; removal efficiency, in percent, = (influent minus effluent) divided by influent, times 100; pounds per day = mg/L, times MGD, times 8.34; one cubic foot holds 7.48 gallons; one MGD equals 694 gpm.

  1. (2 points) Which step lets particles settle out of slow-moving water?
    • A) Coagulation
    • B) Sedimentation
    • C) Disinfection
    • D) Distribution
  2. (2 points) Which expression gives chemical feed in pounds per day?
    • A) mg/L, times MGD, times 8.34
    • B) mg/L divided by MGD
    • C) gpm, times 2.31
    • D) NTU, times 0.433
  3. (4 points) A sedimentation basin holds 80,000 gallons and passes 480,000 gallons per day. Find the detention time in hours.
  4. (2 points) A typical sedimentation basin is designed for how many hours of detention?
    • A) 2 to 4 hours
    • B) 20 to 40 hours
    • C) 1 to 2 minutes
    • D) 10 to 12 hours
  5. (4 points) A filter passes 900 gpm across 180 square feet of surface. Find the loading rate in gpm per square foot.
  6. (3 points) Conventional filters run 2 to 6 gpm per square foot. Is the item-5 loading rate inside that range? Explain in one sentence.
  7. (4 points) The influent turbidity is 40 NTU and the settled water is 10 NTU. Find the removal efficiency as a percent.
  8. (4 points) The plant doses chlorine at 2.5 mg/L into a 2 MGD flow. Find the chlorine feed in pounds per day.
  9. (3 points) Convert the plant's 2 MGD flow to gallons per minute.
  10. (2 points) Finished drinking water must stay at or below what turbidity in 95 percent of samples?
    • A) 0.3 NTU
    • B) 5 NTU
    • C) 15 NTU
    • D) 40 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: B) Sedimentation. In sedimentation, slow-moving water lets heavy particles fall to the bottom before filtration.
  2. Answer: A) mg/L, times MGD, times 8.34. This is the pounds formula for daily chemical feed.
  3. Answer: 4 hours. Detention time = 80,000 divided by 480,000 = 0.1667 day; 0.1667 times 24 = 4 hours.
  4. Answer: A) 2 to 4 hours. Sedimentation basins are commonly designed for 2 to 4 hours of detention.
  5. Answer: 5 gpm/sq ft. Loading rate = 900 gpm divided by 180 square feet = 5 gpm per square foot.
  6. Answer: Yes. 5 gpm per square foot falls between 2 and 6 gpm per square foot, so it is inside the range.
  7. Answer: 75 %. (40 minus 10) divided by 40, times 100 = 30 divided by 40, times 100 = 75 percent.
  8. Answer: 41.7 lb/day. 2.5 mg/L times 2 MGD times 8.34 = 41.7 pounds per day.
  9. Answer: 1,388 gpm. 2 MGD times 694 gpm per MGD = 1,388 gpm.
  10. Answer: A) 0.3 NTU. Finished water must stay at or below 0.3 NTU in 95 percent of samples.

Differentiation

Support

Provide a formula strip listing detention time, loading rate, removal efficiency, and the pounds formula, each with one worked example that is not on the quiz. Allow a calculator on all computation items and read the multiple-choice stems aloud.

Extension

Ask students to find the largest flow, in gpm, the item-5 filter at the fictional plant can take while staying at or below 6 gpm per square foot, and then to convert that flow to MGD.

ELL supports

Provide the formula strip with a small labeled picture for each step: settling tank, filter bed, and a chlorine feed pump. Let students answer item 6 with a sentence frame: "The loading rate is ___ the range because 5 is between ___ and ___."