Quiz & test questions Grade 11 Math

Coalport Plant Process Math: A Grade 11 Treatment Unit Test

A 45-minute, 30-point grade 11 unit test mixing multiple-choice, short-answer, and computation items on treatment process math for a fictional plant.

Objectives

  1. Students will apply the pounds formula, volume, and detention-time formulas to treatment scenarios.
  2. Students will compute removal efficiency and filter loading rate and judge them against ranges.
  3. Students will solve the pounds formula for an unknown dose.

Vocabulary

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.

Also: pounds formula, removal efficiency, filter loading

Teacher overview

This 45-minute unit test (30 points total) checks whether grade 11 students can carry out the core process math a treatment-plant operator uses every shift. It is built around the Coalport Water Treatment Plant, a fictional plant, and mixes two multiple-choice items, two short-answer items, and five computation items covering the pounds formula, tank volume, detention time, removal efficiency, and filter loading. The items move from recall of a formula to solving that formula for an unknown, so the test samples more than one level of understanding.

Answer Key

  1. Answer: A.
    The pounds formula multiplies concentration in mg/L by flow in MGD by 8.34 pounds per gallon; the other options misplace a term.
  2. Answer: 50.0 lb/day.
    2.0 mg/L x 3.0 MGD x 8.34 = 50.04, which rounds to 50.0 pounds per day.
  3. Answer: 215,424 gal.
    60 ft x 30 ft x 16 ft = 28,800 cubic feet; 28,800 x 7.48 = 215,424 gallons.
  4. Answer: 1.7 hr.
    215,424 gallons divided by 3,000,000 gallons per day = 0.0718 days; 0.0718 x 24 = 1.72 hours.
  5. Answer: C.
    Removal efficiency = (12 minus 3) divided by 12 x 100 = 75%, so C is correct; 300% would require the effluent to be negative.
  6. Answer: 3.0 gpm per sq ft.
    900 gpm divided by 300 square feet = 3.0 gpm per square foot, which falls inside the typical 2-to-6 range.
  7. Answer: 1.8 mg/L.
    45 = dose x 3.0 x 8.34, so dose = 45 divided by 25.02 = 1.80 mg/L.
  8. Answer: yes, it complies.
    The residual left after chlorine demand cannot exceed the 1.8 mg/L applied, so it stays below the 4.0 mg/L maximum residual disinfectant level, which limits the residual in distribution rather than the dose. Operators keep the residual well below the limit to hold down disinfection byproducts and taste-and-odor complaints while still disinfecting.
  9. Answer: 2,082 gpm.
    3.0 MGD x 694 gpm per MGD = 2,082 gallons per minute.

Differentiation

Support

Provide a formula reference strip listing the pounds formula, the rectangular-volume formula, and the detention-time formula, along with the conversions 1 cubic foot = 7.48 gallons and 1 MGD = 694 gpm. Allow partial credit for correct setup even when the final arithmetic slips.

Extension

Ask students to write and solve one additional item of their own: choose a flow and a target pounds-per-day feed, then solve for the dose, and explain why the residual that reaches customers, not the dose, is what must stay below the 4.0 mg/L maximum residual disinfectant level.

ELL supports

Provide a labeled diagram of the treatment train (basin, filter, clearwell) with each item's key term marked, and allow short-answer items 6 and 8 to be answered with a labeled calculation plus a short phrase rather than full sentences.