Question

This question requires solving for the answers requested in 5a thru 5e, below, and neatly showing your work with appropriate labels in 5f.

Subject Property

You are appraising 640 acres located near Collegeville. The property has 440 acres of Class I soil and 200 acres of Class II soil. Building improvements contribute $275,000 to total property value.

Market Data

Sale 1: 660 acres located near Collegeville recently sold for $8,680,000. This property has 450 acres of Class I soil and 210 acres of Class III soil. Improvements contributed $205,000 to the sale price.

Sale 2: 510 acres located near Collegeville recently sold for $5,075,000. This property has 510 acres of Class III soil. Improvements contributed $230,000 to the sale price.

Sale 3: 710 acres located near Collegeville recently sold for $8,498,500. This property has 240 acres of Class I soil, 175 acres of Class II soil, and 295 acres of Class III soil. Improvements contributed $350,000 to the sale price.

Show your work for problems 5a through 5e on the next page (problem 5f.). For problems 5a, 5b, and 5c round each answer to nearest $1 per acre.

5a. (4 Points): What is the indicated $/acre value of land with Class I soil?

      $                 

5b. (4 Points): What is the indicated $/acre value of land with Class II soil?

      $                 

5c. (4 Points): What is the indicated $/acre value of land with Class III soil?

      $                 

5d. (4 Points): What is indicated as the subject property’s total land value? Round to nearest $1,000.

      $                    

5e. (4 Points): What is the total indicated value of the subject property? Round to nearest $10,000.

      $                    

           

5f. (6 Points): Neatly show and label all of your work showing how you arrived at your conclusions for questions 5a through 5e. Add additional worksheet(s), if necessary.

         

Answer 1

5a) The market data shows :

Sale 1

Class I Soil = 450 acres out of 660 acres. The estimated market price is 450/660 x $8,680,000 = $5,918,182

Sale 2

Class I soil = 0 acres out of 510 acres.

Sale 3

Class I Soil = 240 acres out of 710 acres. The estimated market price is 240/710 x $8,948,500 = $3,024,845

So the total indicated value of class I soil = $5,918,182 + $3,024,845 = $8,943,027

5b) The market data shows :

Sale 1

Class II Soil = 200 acres out of 640 acres . The estimated market price is 200/640 x $8,680,000 = $2,712,500

Sale 2

Class II Soil = 0 acres out of 510 acres

Sale 3

Class II Soil = 175 acres out of 710 acres. The estimated market price is 175/710 x $8,948,500 = $2,205,616

So the total indicated value of class II soil = $2,712,500 + $2,205,616 = $4,918,116

5c) The market data shows :

Case I

Class III Soil = 0 acres out of 660 acres

Case II

Class III Soil = 510 acres out of 510 acres. The market price is $5,075,000.

Case III

Class III Soil = 295 acres out of 710 acres. The estimated market price is 295/710 x $8,948,500 = $3,718,039

So the total indicated value of Class III soil = $5,075,000 + $3,718,039 = $8,793,039

5d) The subject properties indicated total land value :

Class I land = 440 acres and Class II land = 200 acres

The total indicated value of Class I soil is $8,943,027. Per acre price is $8,943,027/690 = $12961

So the the total value of class I soil  in subject property is = $12961 x 440 acres = $5,702,840

The total indicated value of class II soil  is $4,918,116. Per acre price is $4,918,116/375 = $13,115

So the total value of class II soil in subject property is = $13,115 x 200 acres = $2,623,000

so the indicated total land value of subject property is = $5,702,840 + $2,623,000 + $275,000 = $8,600,840

5e) I have shown systematically, how the total value of subject property is evaluated. The total value of subject property is evaluated from the market related information.