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Math problem set Grade 8 Engineering & Design

Ambersyke Meadow Pivot Nozzles: Comparing Two Application Rates

A grade 8 set where students read one nozzle package from a table and one from an equation, compare the unit rates as slopes, and pick the package that fits the pump.

Objectives

  1. Students will find the unit rate of a proportional relationship presented as a table and as an equation.
  2. Students will interpret each unit rate as the slope of a graph of gallons applied against minutes of run time.
  3. Students will use a pump limit and a shift-length criterion to choose between two nozzle packages.

Vocabulary

Key words: unit rate, slope, proportional relationship, center pivot, constraint

Teacher overview

Ambersyke Meadow is a fictional farming community whose irrigation district is replacing the nozzles on one center pivot. Two nozzle packages are under review, and the vendors describe them in different ways: one comes with a table of test results, and the other comes with an equation. Both relationships are proportional, so the whole comparison rests on finding and comparing the two unit rates, then checking each rate against the well pump's limit. The error to watch for is comparing the packages by a single pair of numbers, such as noticing that one table row shows a large total, instead of dividing to find the gallons applied per minute.

Ambersyke Meadow Pivot Nozzles: Comparing Two Application Rates — Grade 8 Engineering & Design
Name: ____________________________ Date: ______________

Problems

The irrigation district in the town of Ambersyke Meadow is choosing between two nozzle packages for one center pivot. Use the information below for every problem.

Package A, the Redwing R-40. The vendor supplied this test table.

Minutes running Gallons applied
10 220
25 550
40 880
60 1,320

Package B, the Larkfield L-2. The vendor supplied this equation, where g is gallons applied and m is minutes running: g = 27.5m.

  1. Use the first row of the Package A table to find how many gallons that package applies in one minute.
  2. Using the rate you found in problem 1, how many gallons does Package A apply in 45 minutes?
  3. What is the unit rate for Package B, in gallons per minute?
  4. How many more gallons per minute does Package B apply than Package A?
  5. One section of the field needs 6,600 gallons. How many minutes must Package A run to apply that amount?
  6. How many minutes must Package B run to apply the same 6,600 gallons?
  7. How many minutes does Package B save on that one section, compared with Package A?
  8. The district's well pump can deliver at most 25 gallons per minute. Which package can run at its full rate on this pump? Name the package and give the two rates that support your choice.
  9. Suppose the district runs Package A for a full 8-hour shift, which is 480 minutes. How many gallons does the pivot apply in that shift?
  10. The district has two rules for this purchase: the package must run at its full rate on the 25-gallon-per-minute pump, and it must finish one 6,600-gallon section inside one 8-hour shift. Which package meets both rules? Explain using your answers from problems 5, 6, and 8.

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: 22 gallons per minute.
    220 gallons ÷ 10 minutes = 22 gallons per minute. Any other row gives the same rate, which is what makes the relationship proportional.
  2. Answer: 990 gallons.
    45 minutes × 22 gallons per minute = 990 gallons.
  3. Answer: 27.5 gallons per minute.
    In g = 27.5m the coefficient of m is the unit rate, so the pivot applies 27.5 gallons for each minute of run time. On a graph of gallons against minutes, 27.5 is the slope.
  4. Answer: 5.5 gallons per minute.
    27.5 - 22 = 5.5 gallons per minute. Package B applies more water in the same time.
  5. Answer: 300 minutes.
    6,600 gallons ÷ 22 gallons per minute = 300 minutes, which is 5 hours.
  6. Answer: 240 minutes.
    6,600 gallons ÷ 27.5 gallons per minute = 240 minutes, which is 4 hours.
  7. Answer: 60 minutes.
    300 minutes - 240 minutes = 60 minutes saved.
  8. Answer: Package A.
    Package A needs 22 gallons per minute and the pump supplies at most 25, so the pump can keep up. Package B needs 27.5 gallons per minute, which is more than the pump can deliver, so that package would run below its rated performance.
  9. Answer: 10,560 gallons.
    480 minutes × 22 gallons per minute = 10,560 gallons.
  10. Answer: Package A.
    Package A satisfies both rules. Its rate of 22 gallons per minute is within the pump's limit of 25, and it finishes the section in 300 minutes, which is well inside the 480-minute shift. Package B finishes faster on paper, in 240 minutes, yet it cannot reach 27.5 gallons per minute on this pump, so its stated time is not achievable here.

Differentiation

Support

Give students the Package A table with a third column already labeled "gallons per minute" and the first row filled in, so the division that produces the unit rate is modeled once before they work the rest. Providing a printed reminder that 8 hours equals 480 minutes removes a conversion step from problems 9 and 10.

Extension

Suppose a third vendor offers a package that applies 24 gallons per minute and costs more to install. Ask students to write its equation, calculate its time for one 6,600-gallon section, and argue whether the time saved over Package A would justify a higher price.

ELL supports

Pre-teach "unit rate," "slope," and "constraint" with a two-column card that pairs each term with a short example from this scenario, such as "unit rate: 22 gallons in 1 minute." Allow students to answer problems 8 and 10 with a labeled comparison table rather than a written paragraph.