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Math problem set Grade 8 Careers & CTE

Willow Draw's Two Sprinkler Packages: Slope as a Water Rate

A grade 8 problem set where students compare an irrigation technician's two sprinkler packages, one given as a table and one as an equation, using unit rate as slope.

Objectives

  1. Students will find the unit rate of a proportional relationship from a table and from an equation.
  2. Students will compare two proportional watering rates presented in different forms.
  3. Students will use each rate to decide which sprinkler package finishes a set within a stated watering window.

Vocabulary

Key words: unit rate, slope, proportional relationship, irrigation set, sprinkler package

Teacher overview

Irrigation system technicians install and tune the sprinklers, drip lines, and controllers that deliver water to orchards and fields, and much of the job is deciding how long a valve stays open. Two sprinkler packages that look similar on a shelf can apply very different amounts of water per minute, which changes both the watering schedule and the water bill. This set has students pull one package's rate from a table of readings and the other's from an equation, then compare the two. The misconception it targets is treating the steeper rate as automatically better; the technician's real question is whether a package finishes the required amount of water inside the time the district allows, and a faster rate can overwhelm the soil's ability to absorb water.

Willow Draw's Two Sprinkler Packages: Slope as a Water Rate — Grade 8 Careers & CTE
Name: ____________________________ Date: ______________

Problems

Suppose the Willow Draw Irrigation District, a fictional district, is helping Sandhurst Orchard choose between two sprinkler packages. Every figure below was invented for this problem set. A technician ran Package A for four timed trials and recorded the gallons applied to one block of trees.

Minutes running Gallons applied
10 340
20 680
30 1,020
45 1,530

Package B comes with a rating card instead of a table. Its card gives the equation g = 26m, where g is gallons applied and m is minutes running.

  1. Find the unit rate for Package A in gallons per minute. Show the division you used.
  2. How many gallons does Package A apply in 20 minutes, according to the table?
  3. What is the unit rate for Package B in gallons per minute?
  4. If both packages were graphed with minutes on the horizontal axis and gallons on the vertical axis, which package's line would be steeper?
  5. How many gallons does Package B apply in 45 minutes? Show your substitution.
  6. At 45 minutes, how many more gallons does Package A apply than Package B?
  7. One irrigation set for this block needs 5,100 gallons. How many minutes does Package A need to deliver it?
  8. How many minutes does Package B need to deliver the same 5,100 gallons? Round to the nearest whole minute.
  9. The district allows Sandhurst Orchard a 3-hour watering window each morning. Which package can finish the set inside that window?
  10. Explain how the slope of each package's line determines the time each one needs to finish the same 5,100-gallon set.

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: 34 gallons per minute.
    Divide gallons by minutes for any row of the table: 340 divided by 10 = 34. Checking a second row, 1,530 divided by 45 = 34, so the relationship is proportional.
  2. Answer: 680 gallons.
    The table pairs 20 minutes with 680 gallons. This is the point (20, 680) on the graph, and 680 divided by 20 also gives the unit rate of 34.
  3. Answer: 26 gallons per minute.
    In g = 26m the coefficient of m is the unit rate, so Package B applies 26 gallons each minute.
  4. Answer: Package A.
    Package A's slope of 34 is greater than Package B's slope of 26, so its line rises faster.
  5. Answer: 1,170 gallons. g = 26(45) = 1,170 gallons.
  6. Answer: 360 gallons.
    1,530 - 1,170 = 360 gallons more from Package A.
  7. Answer: 150 minutes.
    5,100 divided by 34 = 150 minutes.
  8. Answer: 196 minutes.
    5,100 divided by 26 = 196.15 minutes, which rounds to 196 minutes.
  9. Answer: Package A.
    A 3-hour window is 180 minutes. Package A needs 150 minutes, which fits. Package B needs about 196 minutes, which is 16 minutes too long.
  10. Answer: A steeper slope means fewer minutes for the same total.
    Slope here is gallons per minute, so dividing the 5,100-gallon target by the slope gives the minutes required. A larger slope makes that quotient smaller, which is why Package A finishes in 150 minutes and Package B needs about 196.

Differentiation

Support

Give students the first row of the table already divided in this invented example, so the pattern of dividing gallons by minutes is modeled once before problem 1. Provide a number line marked at 60, 120, and 180 minutes for problem 9 so the 3-hour window is visible rather than converted mentally.

Extension

Ask students to find the unit rate a third package would need to finish the same set in exactly 120 minutes, then write that package's equation in the same form as Package B's card.

ELL supports

Pre-teach "unit rate," "slope," and "irrigation set" with the table and the equation side by side, circling the number 34 in the table's division and the number 26 in the equation. Allow problems 4, 9, and 10 to be answered with the package name plus a short phrase.

Extensions

Have students plot both packages on graph paper, draw the horizontal line at 5,100 gallons, and mark where each line crosses it. Ask them to connect the two crossing points to the answers from problems 7 and 8, then discuss why an irrigation technician would still ask about soil type and slope of the ground before recommending the faster package.