Set up the problem, based on what's given and what's asked for, multiply by the necessary identities, and convert the units of measurement by cancelling. Do not do any arithmetic. You will not be graded on the answer, only on the setup and conversion.

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1. Students in a turfgrass class wish to apply the fertilizer urea (45-0-0) to 3 foot X 6 foot plots at a rate of 1 pound nitrogen (N) per thousand square feet. The fertilizer will be weighed with a gram balance. How many grams of urea are required per plot?

Set up the problem: What's asked for?

Amount = area x rate = (3 ft x 6 ft) x (1 pound N / 1000 ft²)

Use identities to solve conversions

= (3 ft x 6 ft) x (1 pound N / 1000 ft²) x (454 g / pound) x (1 unit urea / 0.45 unit N)

Cancel units of measurement, top and bottom

= (3 ft x 6 ft) x (1 pound N / 1000 ft²) x (454 g / pound) x (1 unit urea / 0.45 unit N)

Simplify

= (3 x 6 x 454 g urea) / (1000 x 0.45)

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2. A turf manager wants to apply a 10-5-5 fertilizer at the rate of 1 pound of N per thousand square feet. How many pounds of fertilizer should be applied to a 2000 square foot lawn?

Set up the problem: What's asked for?

Amount = area x rate = 2000 ft² x (1 pound N / 1000 ft²)

Use identities to solve conversions

= 2000 ft² x (1 pound N / 1000 ft²) x (1 unit fertilizer / 0.10 unit N)

Cancel units of measurement, top and bottom

= 2000 ft² x (1 pound N / 1000 ft²) x (1 unit fertilizer / 0.10 unit N)

Simplify

= (2000 pound fertilizer) / (1000 x 0.10)

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3. A liquid fertilizer solution is to be applied at the rate of 100 gallons solution per acre. This is how many milliliters solution per square meter?

Set up the problem: What's asked for?

rate = rate = 100 gallons solution / acre (pure conversions here, but lots of them)

Use identities to solve conversions

= (100 gal. solution / acre) x (3785 ml / gal.) x (1 acre / 43560 ft²) x (1 ft / 12 inch) x (1 ft / 12 inch) x (1 inch / 2.54 cm)² x (100 cm / 1 m)²

Cancel units of measurement, top and bottom

= (100 gal. solution / acre) x (3785 ml / gal.) x (1 acre / 43560 ft²) x (1 ft / 12 inch) x (1 ft / 12 inch) x (1 inch / 2.54 cm)² x (100 cm / 1 m)²

Simplify

= (100 x 3785 x 100 x 100 ml solution) / (43560 x 12 x 12 x 2.54 x 2.54 m²)

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1. The fertilizer Milorganite (6-2-0) is to be applied at the rate of 1 pound N per thousand square feet. How large a lawn (in square feet) can be treated with one bag (50 pounds)?
   
   Set up the problem: What's asked for?
   
   area = amount / rate = 50 pounds Milorganite / (1 pound N / 1000 ft²)
   
   = (50 pounds Milorganite x 1000 ft²) / (1 pound N) rewritten to simplify double denominators
   
   Use identities to solve conversions
   
   = [(50 pounds Milorganite x 1000 ft²) / (1 pound N)] x (0.06 unit N / 1 unit Milorganite)
   
   Cancel units of measurement, top and bottom
   
   = [(50 pounds Milorganite x 1000 ft²) / (1 pound N)] x (0.06 unit N / 1 unit Milorganite)
   
   
   Simplify
   
   = 50 x 1000 x 0.06 ft²
   

^{18 September 1997}