Math problem set Grade 6 Math

Kilmory Farm's Irrigation Timer: Inches per Hour on Four Zones

A grade 6 problem set on a farm's sprinkler timer: students work with inches per hour, convert depth of water to gallons, and adjust the schedule after a rainy week.

Objectives

  1. Students will use a rate in inches per hour to find the run time a crop's weekly need requires.
  2. Students will convert a depth of water on a known area into gallons using a stated conversion.
  3. Students will adjust an irrigation schedule after rainfall and express the saving as a percent.

Vocabulary

Key words: irrigation zone, unit rate, depth of water, timer

Teacher overview

Irrigation is measured the same way rain is measured, as a depth of water spread over an area, so a sprinkler's output is written in inches per hour rather than in gallons. The fictional Kilmory Farm waters four equal zones from one timer and needs 1.5 inches of water on its crop each week. The misconception this set corrects is that inches of water describe how deep the water sits on the soil. They describe the depth the water would reach if none soaked in, which is why a quarter inch per hour still adds up to thousands of gallons on a large field.

Answer Key

  1. Answer: 6 hours.
    Divide the weekly need by the hourly rate. 1.5 divided by 0.25 is 6, so each zone runs 6 hours a week.
  2. Answer: 3 days.
    6 hours divided by 2 hours per run is 3 runs, so the zone runs on 3 days.
  3. Answer: 0.5 inch.
    2 hours times 0.25 inch per hour is 0.5 inch.
  4. Answer: 2,492 gallons.
    Multiply area, depth, and the conversion: 8,000 times 0.5 is 4,000, and 4,000 times 0.623 is 2,492 gallons.
  5. Answer: 7,476 gallons.
    8,000 times 1.5 is 12,000, and 12,000 times 0.623 is 7,476 gallons. This is three times the answer to problem 4, which matches 1.5 inches being three times 0.5 inch.
  6. Answer: 29,904 gallons.
    4 zones times 7,476 gallons each is 29,904 gallons.
  7. Answer: 186.9 minutes.
    7,476 gallons divided by 40 gallons per minute is 186.9 minutes, which is a little over 3 hours.
  8. Answer: 0.9 inch.
    1.5 inches needed minus 0.6 inch of rain leaves 0.9 inch for the timer.
  9. Answer: 3.6 hours.
    0.9 divided by 0.25 is 3.6 hours, which is 2.4 hours less than the usual 6 hours.
  10. Answer: 40 percent.
    0.6 divided by 1.5 is 0.4, which is 40 percent.

Differentiation

Support

Give these students a conversion strip for the fictional Kilmory Farm that reads "hours times 0.25 equals inches" and "area times inches times 0.623 equals gallons." Assign problems 1, 2, 3, 8, and 9, which stay in inches and hours, then add problem 4 with the multiplication already set up.

Extension

Ask these students to design a new schedule for the fictional Kilmory Farm using sprinklers that apply 0.4 inch each hour, and to state how many hours a week each zone would run and how many gallons the farm would use. They should explain whether the faster sprinklers change the total gallons, and why.

ELL supports

Pre-teach zone, timer, and depth of water using a sketch of four rectangles labeled with the same area and a ruler standing upright to show a quarter inch. Provide this sentence frame for problem 1: "The crop needs ___ inches, the sprinkler gives ___ inch each hour, so the zone runs ___ hours."

Extensions

Set out a line of straight-sided cans under a sprinkler or a hose on the school grounds, run it for a measured time, and have students read the depth caught in each can with a ruler to work out the inches applied per hour. This is the catch-can test an irrigation technician runs on a real system, and comparing the measured rate with the 0.25 inch figure used for the fictional Kilmory Farm shows the class why a watering schedule depends on an accurate measured rate.