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Quiz & test questions Grade 10 Math

Quarrywood Plant Operations: A Grade 10 Conversions Unit Test

A grade 10 math unit test on a fictional plant: multiple-choice and computation items on the pounds formula, basin volume, detention time, and conversions.

Objectives

  1. Students will apply the pounds formula and volume and detention-time formulas to a treatment-plant scenario.
  2. Students will convert a flow rate through a chain of standard water-industry units.
  3. Students will compare a chlorine residual to the federal maximum residual disinfectant level.

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, filter loading rate, residual, unit conversion

Teacher overview

This 45-minute unit test (26 points total) checks whether students can run the multi-step, unit-driven math that treatment-plant operations depend on. It uses the fictional Quarrywood Water Treatment Plant as its running scenario and mixes two multiple-choice recall items with computation items covering the pounds formula, basin volume, detention time, filter loading, and chained flow conversions. The item mix moves from recall to application, so students both state standard conversions and carry them through calculations.

Quarrywood Plant Operations: A Grade 10 Conversions Unit Test — Grade 10 Math
Name: ____________________________ Date: ______________

Assessment

Every item below refers to the fictional Quarrywood Water Treatment Plant.

  1. (2 points) One million gallons per day (1 MGD) is equal to about how many gallons per minute?

    • A) 448.8 gpm
    • B) 694 gpm
    • C) 1,547 gpm
    • D) 325,851 gpm

  2. (4 points) Quarrywood treats a flow of 1.8 MGD and doses alum at 12 mg/L. Using the pounds formula, Loading (lbs/day) = Concentration (mg/L) × Flow (MGD) × 8.34, calculate the pounds of alum fed per day. Round to the nearest tenth of a pound.

  3. (4 points) Quarrywood's rectangular sedimentation basin is 70 feet long, 20 feet wide, and holds water 10 feet deep. Calculate the basin's volume in gallons, using 1 cubic foot = 7.48 gallons.

  4. (4 points) The basin from item 3 receives the plant's flow of 1.8 MGD. Using the detention-time formula, detention time = Volume ÷ Flow, calculate the basin's detention time in hours. Round to the nearest tenth of an hour.

  5. (3 points) Quarrywood runs 1,250 gpm through 500 square feet of filter surface. Calculate the filter loading rate in gpm per square foot, and state whether it falls within the standard 2-to-6 gpm per square foot range.

  6. (3 points) Convert Quarrywood's flow of 1.8 MGD to gallons per minute (using 1 MGD = 694 gpm), then to cubic feet per second (using 1 MGD = 1.547 cfs). Round the gpm to the nearest whole number and the cfs to the nearest hundredth.

  7. (4 points) A pressure-zone sample shows a chlorine residual of 0.9 mg/L. The federal maximum residual disinfectant level (MRDL) for chlorine is 4.0 mg/L. How far below the MRDL is this residual, and in one or two sentences, explain why the plant keeps a residual in the distribution system at all.

  8. (2 points) Conventional filtration plants must keep finished-water turbidity at or below which value in 95 percent of their monthly samples?

    • A) 0.3 NTU
    • B) 1 NTU
    • C) 5 NTU
    • D) 15 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) 694 gpm.
    By the standard conversion, 1 MGD equals 694 gpm. Option A is the cfs-to-gpm figure, option C is close to the cfs relationship, and option D is the gallons in an acre-foot, so all three are plausible mix-ups.

  2. Answer: 180.1 lbs/day.
    Loading = 12 mg/L × 1.8 MGD × 8.34 = 21.6 × 8.34 = 180.144, which rounds to 180.1 pounds per day.

  3. Answer: 104,720 gallons.
    Volume = 70 ft × 20 ft × 10 ft = 14,000 cubic feet. Converting to gallons: 14,000 × 7.48 = 104,720 gallons.

  4. Answer: 1.4 hours.
    Flow of 1.8 MGD is 1,800,000 gallons per day. Detention time = 104,720 ÷ 1,800,000 = 0.05818 days. Converting to hours: 0.05818 × 24 = 1.396, which rounds to 1.4 hours.

  5. Answer: 2.5 gpm per square foot.
    Loading rate = 1,250 gpm ÷ 500 sq ft = 2.5 gpm per square foot, which is inside the standard 2-to-6 range, so the filters are not overloaded.

  6. Answer: 1,249 gpm (about 2.78 cfs).
    1.8 MGD × 694 gpm/MGD = 1,249.2, which rounds to 1,249 gpm. 1.8 MGD × 1.547 cfs/MGD = 2.7846, which rounds to 2.78 cfs.

  7. Answer: 3.1 mg/L below the MRDL.
    4.0 mg/L − 0.9 mg/L = 3.1 mg/L below the limit. Accept any explanation noting that keeping a chlorine residual in the pipes continues to protect the water from bacteria as it travels to homes, guarding against contamination that could enter through a leak or a pressure drop.

  8. Answer: A) 0.3 NTU.
    Conventional plants must keep finished-water turbidity at or below 0.3 NTU in 95 percent of monthly samples. The higher values in options B, C, and D are above that limit and would signal a treatment problem.

Differentiation

Support (below grade level)

Provide the pounds formula, basin-volume conversion, and detention-time formula on a labeled reference card, and break items 3 and 4 into separate steps with the cubic-feet result carried forward for the student, so they complete only the gallon and time conversions.

Extension (above grade level)

Ask students to recalculate item 4's detention time if Quarrywood's flow dropped to a fictional 1.2 MGD, then explain in one or two sentences how a lower flow changes the time water spends settling in the basin and why that matters for clarity.

ELL supports

Pre-teach detention time, loading rate, and residual with a labeled treatment-plant diagram showing where each is measured. Allow item 7's explanation to be written with a sentence frame: "The plant keeps a residual because ___," and let students show items 2 through 6 as numbered calculation steps.