Course 101 Fundamentals of Real Property Appraisal Solutions Set PDF

COURSE 101 Fundamentals of Real Property Appraisal Solutions Set Solutions Course 101 Fundamentals of Real Property Appraisal Chapter 1 EXERCISE 1-1 Solution How Tax Rates Are Expressed No. 1 Tax Rate $65.00/$1,000 Decimal Equivalent 0.065 2 72 mills 0.072 3 $8.00/$100 0.080 4 $80.00/$1,000 0.080 5...

COURSE 101 Fundamentals of Real Property Appraisal Solutions Set Solutions Course 101 Fundamentals of Real Property Appraisal Chapter 1 EXERCISE 1-1 Solution How Tax Rates Are Expressed No. 1 Tax Rate $65.00/$1,000 Decimal Equivalent 0.065 2 72 mills 0.072 3 $8.00/$100 0.080 4 $80.00/$1,000 0.080 5 115 mills 0.115 6 $4.50/$100 0.045 7 62.5 mills 0.0625 8 $75.25/$1,000 0.07525 9 $12.75/$100 0.1275 10 $6.25/$100 0.0625 EXERCISE 1-2 Solution Expressing Tax Rates Translate the given value into each of the missing values ($1/$1000) $1/$100 Mills Tax Rate $59 $5.90 59 0.059 $85 $8.50 85 0.085 $70 $7.00 70 0.070 $72.5 $7.25 72.5 0.0725 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 1 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-3 Solution Developing a Tax Rate Information Property Tax Revenue (Budget) $200,000,000 Market Value $20,000,000,000 Assessed Value $10,000,000,000 Subject Property Assessed Value $500,000 What is the indicated tax rate? 0.02 What is the assessment ratio? 0.50 What is the amount of property tax for subject? $10,000 Would you want to live in this area? Yes (Note: If your calculator doesn’t handle billions, simply drop the same number of zeroes from the numerator and denominator before you divide.) ( 5 500 500,000 500,000,000 = = = 0.05 = 100 10,000 10,000,000 10,000,000,000 Budget Assessed Value = Tax Rate Assessed Value Market Value Assessed Value X Tax Rate = Taxes (Subject Property) © 2022 IAAO l Revised 06/15 Updated July 2022 200,000,000 0.02 10,000,000,000 (Total) Assessment Ratio ) 10,000,000,000 20,000,000,000 0.50 $500,000 X 0.02 $10,000 Solutions Chapter 1 - 3 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-4 Solution Assessed Value and Market Value You are contemplating buying a home located in the same area where you now rent an apartment. The current owners of the home tell you the real estate taxes this year are $4,284 and the county has just completed its revaluation. At the local assessor’s office, you discover that the tax rate is $60.00 per $1,000 of assessed value and the level of assessment (assessment ratio) is 40 percent. PROCEDURE 1. 2. Divide $60 by $1,000 to convert tax rate to a decimal Divide taxes by the converted tax rate to determine the assessed value Assessed Value X Tax Rate = Taxes SO Assessed Value = taxes tax rate Divide the assessed value by the level of assessment to get market value. AV = AR X MV SO AV MV = AR What is the assessed value of the property? 1. $60 ÷ $1,000 = 0.06 2. $4,284 ÷ 0.06 = $71,400 What is the estimated market value of the property? 3. $71,400 ÷ 0.40 = $178,500 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 5 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-5 Solution Computing a Tax Rate Budget $200,000,000 Total Market Value $10,000,000,000 Assessment Ratio 30% Non-property Tax Revenues $25,000,000 (Note: If your calculator doesn’t handle billions, simply drop the same number of zeroes from the numerator and denominator before you divide.) PROCEDURE 1. Subtract non-property tax revenues from the budget to determine the budgeted revenue to be raised from property taxes. Property Tax Revenue = $200,000,000 − $25,000,000 = $175,000,000 2. Multiply the total market value by the assessment ratio to determine the assessed value. Assessed Value = $10,000,000,000 × 0.30 = $3,000,000,000 3. Divide the the budgeted revenue to be raised from property taxes.by the assessed value to get tax rate. Tax rate = 4. $175,000,000 $3,000,000,000 = 0.0583 Converted to dollars per $100 = 0.0583 × 100 = $5.83 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 7 Solutions Course 101 Fundamentals of Real Property Appraisal Find the jurisdiction’s tax rate expressed in dollars per hundred dollars when the following is known: EXERCISE 1-6 Solution Computing an Effective Tax Rate Find the effective tax rate (ETR) expressed as a percentage when the following is known: Annual Real Estate Taxes $8,000,000 Total Market Value $400,000,000 Assessment Ratio 50% Tax Rate 4% Assessed Value $200,000,000 (Note: If your calculator doesn’t handle billions, simply drop the same number of zeroes from the numerator and denominator before you divide.) PROCEDURE 1. Multiply the assessment ratio by the tax rate to determine the ETR. AR × T = ETR 0.50 × 0.04 = 0.02, or 2% 0.02 ETR AR T 0.50 0.04 X © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 9 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-7 Solution COMPUTING AN EFFECTIVE TAX RATE Find the jurisdiction’s effective tax rate (ETR) when the following is known: Budget $200,000,000 Total Market Value $10,000,000,000 Assessed Value $2,500,000,000 Tax Rate 80 mills (Note: If your calculator doesn’t handle billions, simply drop the same number of zeroes from the numerator and denominator before you divide.) PROCEDURE 1. Determine the assessment ratio (assessed value divided by total market value). Assessment ratio 2. = $2,500,000,000 $10,000,000,000 = 0.25 Multiply the assessment Ratio by the tax rate 0.020, or 2.0% E A © 2022 IAAO l Revised 06/15 Updated July 2022 T Solutions Chapter 1 - 11 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-8 Solution Basic Principles of Value 5 Anticipation 1. The value for a component of property depends on contribution to the whole 7 Balance 2. A property must be valued with a single use for the entire property. 9 Change 3. The amount of goods that producers are willing to sell under various conditions during a given period 10 Competition 4. Net income remaining after the costs of labor, management, and capital have been paid 11 Conformity 5. Present worth of future benefits 2 Consistent Use 6. Market value of a property tends to be set by the cost of acquiring an equally desirable and valuable property. 1 Contribution 7. 12 Increasing/ Decreasing Returns Maximum value is obtained when the four agents of production attain a state of equilibrium. 8. Quantities of various goods that people are willing and able to buy during some period, given the choices available to them. 13 Progression/ 9. Regression The tendency of social and economic forces affecting supply and demand to shift over time. 6 Substitution 10. Availability must be in harmony with demand. If one or the other is in excess, prices will increase or decrease. 4 Surplus Pro- 11. The value of property depends, in part, on its relationship to its surroundductivity ings. 3 Supply 12. After a certain point, the addition of successive increments of one agent of production decreases future incomes or amenities. 8 Demand 13. The value of a lower priced property is increased by its association with better properties of the same type. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 13 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-9 Solution Supply and Demand Discussion Describe what would happen to the price of building materials (i.e., lumber, nails, siding, roofing materials) if a major hurricane hit a community and literally demolished a large percentage of the area? What would buyers and sellers expect? The destruction caused by the hurricane would create a huge demand for building materials, causing a short-term shortage. This shortage would cause the price of building materials to increase. Once the community has completed its rebuilding, the demand for building materials would decrease and the price of the materials would stabilize at the current market price. Describe the supply/demand effect on rental apartments and homes in university towns. Because of the need for temporary housing by college students, a strong demand for this type of housing is created. This demand remains strong as long as the university maintains a small but steady growth rate in enrollment. However, should the cost of tuition or any other economic factor increase to a point that enrollment decreases, demand for rentals would decrease, causing a drop in rental rates. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 15 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-10 Solution Supply and Demand Example The High-Tech Bullet Corporation (High-Tech) has made a decision to relocate its factory to your community. This decision was based on many factors, one of which, the effective tax rate, was the lowest in any community in which the company performed a feasibility analysis. The factory is scheduled to be completed in the coming year and will add approximately 425 families to the community. In addition, HighTech will employ approximately 400 from the surrounding area. 1. What might happen to the price of existing homes in your community? Because of the expected shortage of housing that might occur within the coming year, the price of existing houses will increase. Also, informed buyers will attempt to buy before prices increase. 2. What other change(s) might result from the location of this new factory in your community? Because of the perceived need for additional housing within the coming year, the amount of new construction will increase. Demand for a commodity depends on the expected price of the commodity in a future period. A commodity’s supply depends on its expected future price. The principle of anticipation is a direct result of consumer expectations. Value is affected by the present worth of foreseeable future benefits from the ownership of the commodity. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 17 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 1-11 Solution Highest and Best Use The subject property is a 30-year old, unfurnished, single-family residence, rented at the market rate of $1,050 per month. It is located on commercially zoned land worth $165,000. Several comparable single-family residences indicate a residential value of approximately $195,000. What is your analysis of highest and best use? The property is apparently in a transitional area with properties changing from residential to commercial use. However, the property appears to still have value in the immediate future for residential purposes. In an analysis of the parcel as if vacant, the highest and best use would be for commercial purposes. However, in an analysis of the parcel as if improved the highest and best use would be as residential, because there is still an increment of value in the improvement. In valuing the improvements, the appraiser might investigate their potential for conversion to commercial use to take advantage of the commercial location. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 19 Review Solutions Course 101 Fundamentals of Real Property Appraisal CHAPTER 1 1. The property tax is an ad valorem tax, meaning it is based on value. 2. The assessor is responsible for the discovery, listing, and valuation of all taxable property. 3. The taxable location of personal property is referred to as its situs. 4. An estimate of value, usually in writing, of an adequately described property, as of a given date, is termed an appraisal. 5. A jurisdiction’s tax rate is determined by dividing the budget to be derived from property tax by the total assessed value of the taxing jurisdiction. 6. The type of value usually estimated by the assessor is market value. 7. The effective tax rate reflects the ratio between the current tax bill and the property value. 8. In order for a property to have value, it must have utility, scarcity, desirability, and effective purchasing power. 9. The use that generates the highest net return to a property over a reasonable time period is called its highest and best use. 10. The principle of balance states that maximum value of a neighborhood is attained when the uses of land are perfectly complementary. 11. Which appraisal principle affirms that land cannot be valued on the basis of one use while im provements are valued on the basis of another? Consistent use 12. List the four tests that the appraiser must make in the analysis of the highest and best use. 1. 2. 3. 4. Legally permissible Physically possible Financially feasible Most productive 13. Real estate is a parcel of land and any structures or improvements that are permanently affixed thereto. 14. Property is considered personal if it can be moved without causing any damage or change to either the item of property or the structure to which it is attached. 15. Assessment level times the tax rate equals the effective tax rate. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 21 Review Solutions Course 101 Fundamentals of Real Property Appraisal 16. Equilibrium is the point at which the forces of supply and forces of demand meet. 17. List the two types of highest and best use analysis. 1. As if vacant 2. As if improved 18. The typical life cycle of a neighborhood includes growth, stability, decline, and revitalization. 19. What elements of the marketplace are most likely to contribute to a change in demand? A. Consumer tastes and preferences B. Consumer income C. Price of related commodities D. Consumer expectations E. Price of the commodity 20. Market value is defined as the most probable price of a property in terms of money, assuming A. Buyer and seller are typically motivated B. A reasonable time is allowed for exposure in the open market. C. Payment is made in terms of cash in U.S. dollars or in terms of financial arrangements comparable thereto. D. All of the above. 21. Supply refers to the amount of goods that producers are willing to sell at a given price during a specified time period. 22. Ownership of all legal rights to property is limited by government in four ways: A. Taxation B. Police power C. Eminent domain D. Escheat. 23. The seven steps of the appraisal process are as follows: 1. Definition of the problem 2. Scope of work 3. Preliminary survey and planning 4. Data collection and analysis 5. Highest and best use 6 Application of the data and the three approaches 7. Correlation/reconciliation to final value estimate. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 22 Review Solutions Course 101 Fundamentals of Real Property Appraisal 24. The basis for the adjustments in the sales comparison approach to value is the principle of contribution. 25. Value in use represents the value of a property for a specific use. 26. List any two types of private encumbrances that can be placed on ownership of real property. Rights of co-owners Condo and subdivision restrictions Covenants, conditions and restrictions found in the chain of title to the property Mortgages Easements and rights of way Liens and judgments Leases. 27. The principle of anticipation states that value is created by the expected future benefits to be derived from the property. 28. The six basic rights associated with property are as follows: 1. Sell 2. Lease 3. Use 4. Give away 5. Enter or leave 6. Refuse to do any of the above 29. The ownership of all legal rights to property is known as fee simple (fee simple absolute). 30. Determining the identity of the property to be appraised is a part of the first step in the appraisal process. 31. The three types of boundaries used to delineate neighborhoods are political, man-made, and natural. 32. Under the step in the appraisal process that is termed data collection and analysis, specific data include A. B. C. D. Physical data Sales data Site data Economic data © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 23 Review Solutions 33. Course 101 Fundamentals of Real Property Appraisal The four forces that influence value and must be considered in neighborhood analysis are as follows 1. 2. 3. 4. Physical (environmental) Economic Governmental Social. 34. The typical life cycle of a neighborhood can be extended or shortened by changes in economic factors. 35. An area of complementary land uses in which all properties are similarly influenced by the four forces affecting value is termed a neighborhood. 36 Municipal services, planning and zoning are examples of governmental factors that affect neighborhoods. 37. What are the four basic elements of supply? 1. 2. 3. 4. 38. Cost of production Price of other goods Entrepreneur expectations Number of sellers The underlying principle for the three approaches to appraisal is Substitution. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 1 - 24 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-1 SOLUTION Metes and Bounds Description Use the following legal description to draw the parcel below. Use a scale of 1” = 100’. (Estimate the distances if you don’t have a ruler.) Tract No. 1 A tract of land located in the NW1/4 of the NW1/4 of the NW1/4 of Section 27, Township 41 North, Range 8 East, N.M.P.M. (New Mexico Principal Meridian), containing 1.48 acres, more or less, which tract is more particularly described by metes and bounds as follows, to wit: Beginning at the northwest corner of the tract herein described, which corner is identical with the northwest corner of said Section 27; thence North 89o 50’ East, 209.00 feet along the north line of said NW1/4 of NW1/4 of NW1/4 Section 27 to the northeast corner of the tract herein described; thence South 0o 17.5’ East, 310.00 feet to the southeast corner of the tract herein described; thence South 89o 50’ West, 205.53 feet to the southwest corner of the tract herein described; thence North 0o 56’ West, 310.03 feet along the west line of said NW1/4 of NW1/4 of NW1/4, Section 27 to the place of beginning. 0o N W 90o E 90o S 0o © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 25 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-2 SOLUTION Public Land Survey System Write the Public Land Survey legal description for the shaded areas in the section below. Calculate the acreage for each parcel. (North is up.) Parcel No. Acres 1. 10 2. 60 3. 4. 5. 6. 7. 8. 9. 80 40 20 40 20 20 40 Description NE1/4 NE1/4 NE1/4 NW1/4 NE1/4 & W1/2 NE1/4 NE1/4 E1/2 NW1/4 NW1/4 NW1/4 W1/2 SW1/4 SW1/4 SE1/4 SW1/4 E1/2 NE1/4 SE1/4 N1/2 NW1/4 SE1/4 SW1/4 SE1/4 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 27 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-3 SOLUTION Public Land Survey System - Parcels Locate the following parcels in the section below. Determine the number of acres in each and the total acreage. Parcel Acres Description No. 1. 40 SW ¼ NE ¼ 2. 80 N½ SW ¼ 3. 10 NE¼ NW¼ SE¼ 4. 10 NW¼ NW¼ SE¼ 5. 10 SW¼ NW¼ SE¼ Total 150 N 1 W 4 2 © 2022 IAAO l Revised 06/15 Updated July 2022 3 E 5 S Solutions Chapter 2 - 29 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-4 SOLUTION Adjustments to Comparables The subject parcel is rectangular, containing 9,600 sq ft. The topography is flat, and the view is average. Comparable 1 is rectangular and contains 12,000 sq ft. The topography is flat, and the view is good. Comparable 2 is rectangular and contains 9,600 sq ft. The topography is hilly, and the view is fair. Comparable 3 is rectangular and contains 7,200 sq ft. The topography is flat, and the view is average. Lots in the area sell based on a square-foot basis. Flat lots are worth more than hilly lots. Use the following adjustment grid to determine whether the adjustments should be equal, negative, or positive. Subject Sale 1 Variation Sale price Size Topography View 9,600 sq ft Flat Average Sale 2 Adjustment $xxxx 12,000 sq ft Negative Flat Equal Good Negative © 2022 IAAO l Revised 06/15 Updated July 2022 Variation Sale 3 Adjustment $xxxx 9,600 sq ft Equal Hilly Positive Fair Positive Variation Adjustment $xxxx 7,200 sq ft Positive Flat Equal Average Equal Solutions Chapter 2 - 31 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-5 SOLUTION Calculating the Market Condition Adjustment What is the indicated monthly adjustment for market conditions using these six sales? Sale 1 sold 4 months ago for $40,000. Sale 2 sold 15 months ago for $36,000. The sales are similar in location and physical characteristics. Sale 1, sold 4 months ago $40,000 Sale 2, sold 15 months ago −$36,000 Increase over 11-month period $4,000 Increase = $4,000 ÷ $36,000 = 0.1111, or 11.11% Increase per month = 0.1111 ÷ 11 = 0.0101, or 1% per month Sale 3 sold recently for $40,000. Sale 4 sold 14 months ago for $44,000. Both sales are similar in physical characteristics; however, Sale 3 is located on a major street. This location is less desirable than that of Sale 4, which is considered typical for the area, and requires an adjustment of $10,000. Sale 3, a recent sale $40,000 Adjustment for location +$10,000 Adjusted sale price $50,000 Sale 4 sold 14 months ago $44,000 Increase over 14-month period $6,000 Increase = $6,000 ÷ $44,000 = 0.136, or 13.6% Increase per month = 0.136 ÷ 14 = 0.0097, or 0.97% per month Sale 5 sold 2 months ago for $56,000. Sale 6 sold 17 months ago for $52,000. The sales are similar in location; however, the topography for Sale 6 at the time of sale was considered superior to that for Sale 5, which is considered typical for the area, requiring an adjustment of $4,000. Sale 5, sold 2 months ago $56,000 Sale 6, sold 17 months ago $52,000 Sale 6 adjustment for topography −$4,000 Sale 6 adjusted $48,000 Increase over 15- month period $8,000 Increase = $8,000 ÷ $48,000 = 0.1667, or 16.67% Increase per month = 0.1667 ÷ 15 = 0.0111, or 1.11% per month Sale 1 and Sale 2 market conditions (time adjustment) per month 0.01, or 1% Sale 3 and Sale 4 market conditions (time adjustment) per month 0.0097, or 0.97% Sale 5 and Sale 6 market conditions (time adjustment) per month 0.0111, or 1.1% Indicated monthly adjustment for market condition (time) = 0.01 or 1% per month © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 33 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-6 SOLUTION Adjustment for Location or Physical Characteristics using Paired Sales An analysis of lot sales in two different areas apparently shows a difference in sale prices because of location. Although the lots in both areas vary in size, the following sets of paired sales are similar to each other in all aspects except their location. Also, there is a difference in the time of sale. However, the adjustment for time has been well documented. Sale 1 in Area A sold recently for $80,000. Sale 2 in Area B sold 8 months ago for $66,000. Sale 3 in Area A sold recently for $72,400. Sale 4 in Area B sold recently for $62,400. Sale 5 in Area A sold 6 months ago for $72,000. Sale 6 in Area B sold 12 months ago for $60,000. Lot values in both of these areas have been increasing at a rate of 0.75% (0.0075) per month. The first step in determining the difference between these sale prices is to adjust older sales for time. This requires applying an adjustment of 0.0075 per month to each sale that is not a recent sale. The second step is to determine the difference in sale price between the two areas. Sale 1 (Area A) sold recently Sale 2 (Area B) sold 8 months ago ($66,000 × 1.06) Difference Sale 3 (Area A) sold recently Sale 4 (Area B) sold recently Difference Sale 5 (Area A) sold 6 months ago ($72,000 × 1.045) Sale 6 (Area B) sold 12 months ago ($60,000 × 1.09) Difference $80,000 −$69,960 $10,040 $72,400 −$62,400 $10,000 $75,240 −$65,400 $9,840 The last step is to analyze the results from the paired sales. This analysis indicates an adjustment of $10,000 between Area A and Area B, with Area A being superior to Area B. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 35 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-7 SOLUTION Sales Comparison Approach to Valuing Land Using Dollar Adjustments For an appraisal of vacant site in a suburban area, there are three comparable sales in the general area of the subject. Sale 1 sold for $60,000 and is located in the same area as the subject, but its proximity to a major street subtracts $12,000 from its value. Also, the topography is inferior to that of the subject by $6,000. Sale 2 sold for $72,000 and is located in an area that is less desirable than that of the subject, requiring a $10,000 adjustment. Its other features are similar to those of the subject. Sale 3 sold for $84,000 and is comparable to the subject in everything except topography. The topography for Sale 3 is less desirable than that of the subject, requiring a $6,000 adjustment. Sale 1 sold 2 years ago; Sale 2 sold 1 year ago; and Sale 3 is a current sale. Lot sales have been increasing in value at a rate of 10% per year. Determine the adjusted sale price for each comparable. What is the indicated value of the subject lot? Subject Sale price Market condition Market condition adjusted sale price Location Average Street Interior Topography Average Net adjustments Adjusted sale price Sale 1 Sale 2 Sale 3 $60,000 $72,000 $84,000 +20% +10% None $12,000 $7,200 $72,000 $79,200 $84,000 Equal +$10,000 +$12,000 Equal +$6,000 Equal +$18,000 +$10,000 $90,000 $89,200 Equal Equal +$6,000 +$6,000 $90,000 Sale 3 is most comparable to the subject lot. The indicated value from Sale 3 is well supported by Sales 1 and 2. Thus, the indicated value for the subject lot is $90,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 37 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-8 SOLUTION Sales Comparison Approach to Valuing Land using Percentage Adjustments A 150’ ×290’ lot in Greenville subdivision is being appraised. The lot is flat, with all off-site and on-site improvements. The lot is of average desirability. Lots located in Brownstone subdivision are 10% more desirable while lots located in Hillman subdivision are 5% less desirable than lots in Greenville subdivision. Amenities in Brownstone subdivision are considered 5% better than those in Greenville subdivision. Amenities in Hillman subdivision are equal to the amenities in Greenville subdivision. Hilly lots are 10% less desirable than flat lots. Lots containing more than 1 acre but less than 1 and 1/2 acres sell for 10% more than lots containing approximately 1 acre. Lots have increased in value at a rate of 0.5% per month. Sale 1 Sale 2 Sale 3 Sale price $54,000 $53,400 $39,600 Sale date 10 mo ago Current 12 mo ago Market conditions +5% 0 +6% Market condition adjusted sale price $56,700 $53,400 $41,976 Location −10% 0 +5% Topography 0 0 +10% Amenities −5% 0 0 Size 0 −10% 0 Net percentage adjustments −15% −10% +15% Net dollar adjustments −$8,505 −$5,340 +$6,296 Adjusted sale price $48,195 $48,060 $48,272 1. The value range from the comparable properties = $48,060 − $48,272 2. The estimated value for the subject property, rounded to the nearest $100= $48,100 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 39 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 2-9 SolutIon SaleS CoMparISon appRoaCh to valuIng land uSIng Sale prICe per unIt A tract containing 45 acres of land, located north of the city, is being appraised. The land is flat, making it easy to maintain. Market information for this area is as follows: Land west of the city is 10% less valuable than land north of the city, and land south of the city is 5% less valuable because of access. Flat land sells for 5% more than hilly land. Land values have increased at a rate of 6% per year over the past 3 years. The following land sales are near the city: Sale 1 Sale price Acres Price per acre Sale date Market conditions (time adjust.) Market cond. adjust. price per acre Location adjustment Topography adjustment Net adjustments Net dollar adjustments Indicated value per acre Sale 2 $155,000 50 $3,100 3 yr +0.18 $3,658 +0.05 +0.05 +0.10 $366 $4,024 Sale 3 $209,000 55 $3,800 1 yr +0.06 $4,028 --0 0 $4,028 Sale 4 $136,000 40 $3,400 6 mo +0.03 $3,502 +0.10 +0.05% +0.15 $525 $4,027 $216,000 60 $3,600 1 yr +0.06 $3,816 +0.05 -+0.05 $191 $4,007 The range in sale price per acre = $4,007 − $4,028. The selected sale price per acre = $4,000. The indicated value of the subject parcel $180,000 ($4,000 × 45 acres = $180,000). © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 41 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-10 SOLUTION Allocation Method A lot 75’ x 110’ in an older area where the improvements were destroyed by fire is being appraised. The area is totally developed with no vacant land sales. There is an area comparable in desirability and construction type. Land values have been well documented through vacant land sales. Five sales of improved parcels have occurred in this area. Sale Sale Price Land Value Building Ratio Percent Value $32,000 $128,000 1:4.0 20% $40,000 $160,000 1:4.0 20% $40,000 $156,000 1:3.9 20.4% $40,000 $164,000 1:4.1 19.6% $36,000 $144,000 1:4.0 20% Land Value / Sale Price OR = 32,000/160,000 = 0.20 or 20% 1 $160,000 2 $200,000 3 $196,000 4 $204,000 5 $180,000 Sale Price / Land Value = 160,000/32,000 = 5; land = 1 part, building = 4 parts 1. Sale Determine a land-to-building ratio using the five sales in the comparable area. The indicated land-to-building ratio is 1:4. Therefore, land represents 20% of the value. Sale Price Lot Size Land Value Land Area (sq Land ft) Value (per sq ft) 1 $180,000 75’ x 120’ $36,000 9,000 $4.00 2 $200,000 100’ x 100’ $40,000 10,000 $4.00 3 $160,000 80’ x 100’ $32,000 8,000 $4.00 Sale price X land to building percentage = 180,000 X 20% = 36,000 Lot size in square feet = 75 X 120 = 9,000 Land Value / land area in sq feet= $36,000/9000 = $4.00/sq foot 2. Determine an appropriate unit value based on the three sales in the subject’s neighborhood. The indicated unit value per square foot for the subject lot is $4.00/sq ft 3. Determine the indicated value of the subject lot. The indicated value of the subject parcel is 75’ × 110’ = 8,250 sq ft 8,250 sq ft × $4.00/sq ft = $33,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 43 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-11 SOLUTION Abstraction Method A vacant lot in Cedar Brook Subdivision is being appraised. There are few vacant lots remaining, as this subdivision was developed 10 years ago. There have been no vacant land sales during the past 3 years. However, there are sales of three improved parcels. Each improved parcel is 9 years old and has been well maintained. The three improved parcels have each depreciated 5% since their original construction. What is the indicated value of the subject lot in Cedar Brook Subdivision? The following information is available for each sale: Sale 1 Sale 2 Sale price $258,000 $244,500 Cost new $225,000 $210,000 Sale 3 $273,000 $240,000 What is the indicated value of the subject lot in Cedar Brook Subdivision? Sale price Cost new Depreciation Cost new less depreciation Indicated land value Sale 1 $258,000 $225,000 −$11,250 $213,750 Sale 2 $244,500 $210,000 −$10,500 $199,500 Sale 3 $273,000 $240,000 −$12,000 $228,000 $44,250 $45,000 $45,000 The indicated value for the subject lot in Cedar Brook subdivision = $45,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 45 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 2-12 SOLUTION Anticipated Use or Development Method An 80-acre tract of land on the outskirts of the city is being appraised. The tract is currently raw land in transition to residential use. No sales data are available so this method has been chosen to value the tract. Information on the development of this tract was obtained from several potential developers. Zoning and land use allow a density of four lots per acre. When the tract is fully developed, the lots should sell for $60,000 each. The costs to develop the tract are as follows: Grading $600,000 Site preparation $1,400,000 Streets and street lighting $3,600,000 Curb, gutter, and sidewalks $1,800,000 Electricity $1,200,000 Natural gas $1,400,000 Water $1,000,000 Sewer $1,400,000 Indirect costs $2,800,000 Profit and overhead $960,000 Step 1. Determine gross revenue. Step 2. Subtract all developmental costs (typical market-based) from the gross revenue. The result is the raw land value. Step 3. Divide raw land value by the 80 acres providing the indicated value per acre. Step 1 80 × 4 = 320 lots at $60,000 each Step 2 Grading Site preparation Streets and street lighting Curb, gutter, and sidewalks Electricity Natural gas Water Sewer Indirect costs Profit and overhead Total development costs Indicated value of 80 acre tract $19,200,000 $600,000 $1,400,000 $3,600,000 $1,800,000 $1,200,000 $1,400,000 $1,000,000 $1,400,000 $2,800,000 $960,000 −$16,160,000 $3,040,000 Step 3 ($3,040,000 ÷ 80) = $38,000 per acre © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 47 Review Solutions Course 101 Fundamentals of Real Property Appraisal CHAPTER 2 1. The four factors affecting land values are as follows: 1. Physical (environmental) 2. Economic 3. Governmental 4. Social 2. The Principle of Surplus Productivity states that returns to land are what remain after returns to labor, management, and capital are satisfied. A. B. C. D. 3. supply and demand surplus productivity change anticipation List the six methods of land valuation: 1. 2. 3. 4. 5. 6. Direct sales comparison Allocation Abstraction Anticipated use or development Capitalization of ground rent Land residual technique 4. A complete set of cadastral maps showing boundaries, parcel size, sales information, and other relevant features of all parcels in a jurisdiction is a necessary tool for assessors. 5. The practice of measuring the direction of property lines with compass bearings and the distance with measuring chains or tapes in order to obtain a property description is known as metes and bounds. 6. A township in the Public Land Survey System contains 36 sections, with each section containing 640 acres. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 49 Review Solutions Course 101 Fundamentals of Real Property Appraisal N A S 7. Using the Public Land Survey System, the shaded portion of land labeled “A” above is described as the S 1/2 NE 1/4 NE 1/4. 8. The shaded area in the previous question contains 20 acres. 9. When land is valued by using the direct sales comparison approach, sales must first be stratified into homogeneous groups. 10. Improvements such as streets, sidewalks, and the availability of utilities are considered off-site improvements. 11. Which method of land valuation compares the property being appraised to comparable vacant parcels that have recently sold? direct sales comparison 12. Front foot, square foot, and acre are examples of basic units of comparison. 13. In the direct sales comparison approach to land valuation, adjustments are always made to the comparable property, never to the subject property. 14. If a comparable lot that sold is superior in some aspect to the subject lot, a negative adjustment would be made to the comparable property. 15. A method of valuing land that subtracts the depreciated value of the improvement from the sale price is known as abstraction. 16. The preferred method of valuing land if adequate data is available is the A. Abstraction method B. Allocation method C. Cost of development method D. Direct sales comparison method © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 50 Review Solutions Course 101 Fundamentals of Real Property Appraisal 17. A land-to-building ratio of 1:4 means land represents 20 percent of the value. 18. The appropriate map scale for suburban areas and small towns is 1” = 200’. 19. The appropriate map scale for rural areas is 1” = 400’. 20. The property account number, linking ownership records, tax maps, and assessment records, is known as the parcel identifier. 21. Land that has been developed to the extent that it is ready to be built upon is considered a site. 22. The concept of highest and best use means that the market value of vacant land depends on potential use rather than current use alone. A. B. C. D. 23. Supply and demand Anticipation Change Highest and best use Factors affecting land values such as size, topography, and location are termed physical (environmental) factors. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 2 - 51 Quiz Solutions Course 101 Fundamentals of Real Property Appraisal QUIZ 1 1. D 2. C 3. A 4. B 5. A 6. C 7. A 8. D 9. C 10. B 11. D 12. A 13. C 14. D 15. D 16. B 17. C 18. C © 2022 IAAO l Revised 06/15 Updated July 2022 19. A 20. B 21. C 22. C 23. B 24. C 25. D 26. B 27. D 28. B 29. B 30. D 31. C 32. A 33. B 34. C 35. D Solutions Quiz 1 - 53 4. Quiz Solutions Actual value Assessment level Assessed value Tax rate Taxes Course 101 Fundamentals of Real Property Appraisal $175,000 0.20 $35,000 0.0928 $3,248 8. Market Value Assessment level Assessed value Budget Non-property tax revenue Budget to be paid by property taxes Tax rate: 5,000,000 ÷ 400,000,000 $800,000,000 0.50 $400,000,000 $6,000,000 $1,000,000 $5,000,000 0.0125 18. Parcel Land Improvement Total Land Percent A $50,000 $150,000 $200,000 0.25 B $100,000 $300,000 $400,000 0.25 Subject: 300,000 x 0.25 = 75,000 27. Subject Sale 1 Sale 2 Price $50,000 $64,500 Sale date Current 4 months ago Size 8,400 sq ft 9,600 sq ft 7,200 Topography Level Hilly Level View Average Fair Good Route 66 South North South Adjustments Time Adjusted Sales Price $50,000 +2% (0.005 X 4 months) $65,790 Size -10% +10% Topography +10% View +15% -15% Route 66 +10% Total +25% -5% Adjusted sale price $62,500 $62,500 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Quiz 1 - 54 Quiz Solutions Course 101 Fundamentals of Real Property Appraisal 28. Sale price Sale date Sale 1 $300,000 Current Adjusted sale $300,000 Improvement value $240,000 Land value $60,000 Land percent 0.20 Subject value $280,000 x 0.20 = $56,000 29. Sale price Cost new of improvements Depreciation Depreciated value of improvements Indicated land value 32. Sale 2 $250,000 12 months ago $275,000 $220,000 $55,000 0.20 $220,000 $260,000 $90,000 -170,000 $50,000 242 x 1,800 = 435,600 ÷ 43,560 = 10 x $60,000 = $600,000 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Quiz 1 - 55 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-1 SOLUTION Characteristics of Cost 16’ 6’ 34’ 20’ 30’ Garage 20’ 58’ 32’ Residence 1 50’ 20’ Residence 2 26’ Garage 24’ 78’ 20’ 1. Compute the gross living area of each house and garage. Residence 1 Garage area Area Total area Perimeter Ratio Residence 2 Garage area Area Perimeter Ratio 2. 20’ x 20’ 20’ x 30’ 26’ x 30’ 6’ x 16’ 20’ x 30’ 6’ + 16’ + 32’ + 30’ + 4’ + 20’ + 30’ + 34’ 1,476 ÷ 172 400 square feet 600 + 780 + 96 1,476 square feet 172 linear feet 8.58 20’ x 20’ 26’ x 58’ 58’ + 26’ + 58’ + 26’ 1,508 ÷ 168 400 square feet 1,508 square feet 168 linear feet 8.98 Assuming that the residences are the same quality, design, construction type and story height, which would cost more to build and why? Although Residence 2 is larger, it should cost less to build because it has a more conventional design and fewer angles. Angles generally increase construction costs, especially those of the roof and foundation. *Note: The smaller the ratio, the more costly it is to build* © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 57 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-2 SOLUTION Comparative Unit Method A 1-story residence with 2,376 square feet of living area on a 34,000-square-foot lot is being appraised. There are three recent sales of similar properties in the neighborhood. All the properties were new at the time of sale. Property 1 sold recently for $350,400. It has 2,420 square feet of living area and is located on a 40,000 square foot lot. Property 2 sold recently for $447,000. It has 3,080 square feet of living area and is located on a 60,000 square foot lot. Property 3 sold recently for $291,000. It has 1,980 square feet of living area and is located on a 32,000 square foot lot. Lot values throughout the neighborhood are well established at $2.00 per square foot. What is the indicated cost new for the subject property improvement by using the comparative unit method? Sale price Less lot value at $2.00/sq ft Cost new Gross living area (sq ft) Unit cost Property 1 Property 2 Property 3 $350,400 $447,000 $291,000 − $80,000 − $120,000 − $64,000 $270,400 $327,000 $227,000 ÷ 2,420 ÷ 3,080 ÷ 1,980 $111.74 $106.17 $114.65 Because the subject, with 2,376 square feet, is only 44 square feet smaller than Property 1, the cost new of the subject could probably be estimated at $112 per square foot. The total cost new for the subject improvement is $112 × 2,376 = $266,112. Land value = $68,000. Total value for the property by the cost approach = $334,112. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 59 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-3 SOLUTION Trended Historical Cost or Factored Historical Cost Method Trend factors needed to bring historical costs to current are generally based on an index that surveys costs each year and then publishes an index that reflects the amount of change that has occurred from a base year. As an example, develop a trend factor for a structure that is 20 years old. First, determine the current index and also the index for 20 years ago. Then, divide the current index by the prior index to obtain the trend factor. Assume the cost for the structure 20 years ago was $300,000. The current index is 320, and the index 20 years ago was 200. What is the trend factor and corresponding trended cost of the structure? 320 ÷ 200 = 1.6 The trended cost for the structure = $300,000 × 1.6 = $480,000 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 61 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-4 SOLUTION Trended Historical Cost or Factored Historical Cost Method The reproduction cost new of a residence built 22 years ago is being estimated. The residence contains 1,800 square feet. The cost at the time of construction was $60.00 per square foot. The index 22 years ago was 120. The current index is 180. Also, the owner of the residence added a new detached double-car garage and a patio with a roof this year. The garage cost $15,000, and the patio cost $5,000. What is the reproduction cost new for the residence plus the improvements? The cost factor = 180 ÷ 120 = 1.5 The historical cost trended = $60.00 × 1.5 = $90.00/sq ft The reproduction cost new for the residence would be as follows: 1800 × $90 Garage Patio Reproduction cost new $162,000 + $15,000 + $5,000 $182,000 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 63 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-5 SOLUTION CALCULATING REPLACEMENT COST NEW USING A COST MANUAL The subject property is a one-story, single-family residence built with average quality materials in an extreme climate zone. The concrete block house contains 2,000 square feet with stucco exterior walls. The roof cover is composition shingle. The floor is finished with 1,600 square feet of carpet and 400 square feet of ceramic tile. Heating is a warm air system with short ducts. Plumbing consists of 11 fixtures and 1 rough-in. The dishwasher, range/oven, garbage disposer, and exhaust fan with vent hood are all built-in items in the kitchen. The residence has a full basement with 8” reinforced concrete walls, a staircase with open risers, and 800 square feet with minimal finish. There is a 300-square-foot concrete slab porch with a composite shingle roof on the rear part of the house. The residence also has a 24’ by 25’ two-car attached garage and uses same materials as the house. Yard improvements are estimated to contribute $10,000. Use the cost manual and calculate the indicated replacement cost new for this residence. Item Base Cost- Concrete block w/ stucco Energy Adj-Extreme Climate Foundation Adjustment Roofing- composition shingle floor cover carpet ceramic tile Heating Plumbing fixtures 8 included additional fixtures 1 included rough in Built-in appliances: dishwasher range/oven garbage disposal exhaust fan & hood Basement 8 inch concrete walls unfinished minimal finish stairs with open riser © 2022 IAAO l Revised 06/15 Updated July 2022 Quantity Unit Cost Total Cost Page numbers 2000 $85.69 $171,380.00 AVG-19 2000 $2.21 $4,420.00 AVG-19 2000 $5.21 $10,420.00 AVG-19 2000 $0.00 $0.00 AVG-19 1600 400 3 $1,360.00 $0.00 $5,424.00 $5,404.00 $0.00 $0.00 $4,080.00 $0.00 1 1 1 1 $535.00 $870.00 $175.00 $285.00 $535.00 $870.00 $175.00 $285.00 B-23 B-23 B-23 B-23 17 17 17 17 2000 $18.31 $0.00 $6.48 $780.00 $36,620.00 $0.00 $5,184.00 $780.00 Refinements 8 Refinements 8 800 1 $3.39 $13.51 7 7 7 7 B-10 B-10 AVG-1 13 13 B-22 16 Solutions Chapter 3 - 65 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-5 SOLUTION Cont. CALCULATING REPLACEMENT COST NEW USING A COST MANUAL Item Porch Average concrete slab Composite shingle Roof Garage Attached 24” by 25” Yard Improvements Replacement Cost New © 2022 IAAO l Revised 06/15 Updated July 2022 Quantity 300 300 600 Unit Cost Total Cost Page numbers $6.24 $11.56 $25.78 $1,872.00 $3,468.00 $15,468.00 Refinements 8 Refinements 8 AVG-29 $10,000.00 $10,000.00 $276,385.00 Given Solutions Chapter 3 - 66 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-6 SOLUTION Developing Adjustments for Third-Party Cost Manuals The costs developed from a cost manual are being compared to current costs obtained from new residences that have recently sold. Three residences have just sold, and they are in areas where the land values are well documented. Data on the three residences are as follows: Sale 1 Sale 2 Sale 3 Sale price $243,500 $241,600 $239,500 Land value $40,000 $40,000 $40,000 Area (sq ft) 1,850 1,800 1,750 Cost new from cost $104.85 $106.67 $108.52 manual Determine the cost new per square foot of improvement for each of the three sales and compare the results to the cost new per square foot of improvement from the cost manual. If there is a difference, by what percentage should the cost manual be adjusted? Sale 1 Sale price Land value Sale price of improvements Square foot area Sale price per square foot of improvements Cost new from cost manual Dollar difference between market and cost Percentage difference (dollar difference ÷ cost) Sale 2 $243,500 − $40,000 $203,500 ÷ 1,850 $110 $104.85 − $5.15 4.91% Sale 3 $241,600 − $40,000 $201,600 ÷ 1,800 $112 $106.67 − $5.33 5.00% $239,500 − $40,000 $199,500 ÷ 1,750 $114 $108.52 − $5.48 5.05% A comparison of the cost per square foot for the improvements based on the market against the cost per square foot from the cost manual indicates the cost manual is approximately 5 percent below market. Thus, a factor will need to be applied to the cost manual in order to produce costs that represent the current market. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 67 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-7 SOLUTION Depreciation Using Comparable Sales Step 1 Sale price Indicated land value Depreciated Value of Improvement /Improvement market value Sale 1 $180,000 − $45,000 $135,000 Sale 2 $200,000 − $50,000 $150,000 Sale 3 $190,000 − $47,500 $142,500 $240,000 − $135,000 $105,000 $270,000 − $150,000 $120,000 $250,000 − $142,500 $107,500 $105,000 $240,000 $120,000 $270,000 $107,500 $250,000 0.4375 0.4444 0.4300 RCN Improvement market value $ value of depreciation (from market) Step 4 Depreciation RCN Depreciation converted percentage (depreciation ÷ RCN) Depreciation converted percentage (depreciation ÷ RCN) Effective age (yr) Depreciation percentage per year RCN Depreciation percentage per year (16 yr × 0.025 ) Depreciation amount RCN Depreciation Improvement value Land value Total value © 2022 IAAO l Revised 06/15 Updated July 2022 0.4375 ÷18 2.43% 0.4444 ÷ 17 2.61% 0.4300 ÷ 17 2.53% Subject Property $260,000 X.40 $104,000 $260,000 − $104,000 $156,000 + $39,000 $195,000 Solutions Chapter 3 - 69 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-8 SOLUTION Calculating Economic Age-Life Depreciation What is the percentage depreciation of a 35-year-old improvement that has a total economic life of 60 years and a remaining economic life of 36 years? Step 1. Determine effective age. Total Economic Life (TEL) 60 - Remaining Economic - 36 Life (REL) = = Effective Age (EA) 24 Step 2. Divide the effective age by the total economic life. 24 = 0.40 or 40% 60 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 71 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-9 SOLUTION Curable Physical Deterioration A two-story, frame, single-family residence is being appraised. A physical inspection finds two broken storm doors and four broken window screens. Also, the front landing, which is made of concrete, is cracked and has a corner broken. The RCN for a storm door is $320; replacing the door today costs $400. The RCN for a window screen is $150; a new screen now costs $200. The RCN for the concrete landing is $800; removing the broken concrete and replacing it today costs $1,000. What is the amount of curable physical deterioration for this residence? Item RCNa 2 Storm doors 4 Window screens 1 Front landing Total a RCN is the replacement or reproduction cost new. b Cost to cure is physical deterioration. © 2022 IAAO l Revised 06/15 Updated July 2022 $640 $600 $800 $2,040 Cost to Cureb $800 $800 $1,000 $2,600 Solutions Chapter 3 - 73 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-10 SOLUTION Incurable Physical Deterioration–Short-Lived A one-story single-family residence of masonry construction is being appraised by using the cost approach and the observed condition method of addressing depreciation. The appraiser has identified several items that are not worn out, as of the date of the appraisal, but will need to be replaced before the basic structure reaches the end of its economic life. The appraiser has estimated the age for each of the items and their total useful life. The items are as follows: Item RCN Water heater Floor cover Roof cover Light fixtures Air-conditioning unit $850 $4,500 $6,000 $4,500 $2,000 Actual Age (yr) 5 3 8 10 6 Total Useful Life (yr) 10 15 20 40 20 Use the age-life method to determine the amount of depreciation for the short-lived items. Item RCN Water heater Floor cover Roof cover Light fixtures Air-conditioning unit Total Alternate solution Item Actual Age (yr) $850 $4,500 $6,000 $4,500 $2,000 Total Useful Life (yr) 5 3 8 10 6 10 15 20 40 20 Ratio 5/10 3/15 8/20 10/40 6/20 Percentage Incurable Depreciation Physical Deterioration 50% $425 20% $900 40% $2,400 25% $1,125 30% $600 $17,850 $5,450 Actual Age (yr) Water heater Floor cover Roof cover Light fixtures Air-conditioning unit Total © 2022 IAAO l Revised 06/15 Updated July 2022 5 3 8 10 6 ÷ ÷ ÷ ÷ ÷ Total Useful Life (yr) 10 15 20 40 20 Percentage Depreciation = = = = = 50% 20% 40% 25% 30% RCN X X X X X $850 $4,500 $6,000 $4,500 $2,000 $17,850 Incurable Physical Deterioration $425 $900 $2,400 $1,125 $600 $5,450 Solutions Chapter 3 - 75 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-11 SOLUTION Incurable Physical Deterioration–Long-Lived The reproduction cost new of a single-family residence being appraised is $225,000. The total cost new of curable physical deterioration and incurable physical deterioration–short-lived items included in this amount is $15,000. The actual age of the structure is 12 years, and its total useful life expectancy is 60 years. No other forms of depreciation are evident. Estimate the incurable physical deterioration of the long-lived items. Reproduction cost new (RCN) Less RCN of curable physical and incurable physical (shortlived items) RCN of long-lived items Actual age (yr) Total useful life (yr) Percentage depreciation Total incurable physical deterioration (long-lived) items © 2022 IAAO l Revised 06/15 Updated July 2022 $225,000 − $15,000 $210,000 12 60 12 ÷ 60 x 0.20 $42,000 Solutions Chapter 3 - 77 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-12 SOLUTION Physical Deterioration A one-story residence is being appraised by using the cost approach and applying the observed condition method of depreciation. During the inspection of the property, the four window screens on the west side of the house were found to be damaged and would need to be replaced. RCN for the screens is $500; the cost of replacing all the screens now is $800. Several items were found to be still functional on the date of the appraisal, but will need to be replaced before the total useful life of the residence ends. The items are as follows: Item RCN Water heater Roof cover Floor cover Actual Age (yr) Total Useful Life (yr) 2 10 10 25 4 20 $800 $2,400 $4,500 An RCN for the residence has been calculated at $198,000 using a cost manual. The actual age for the residence is 12 years, with a total useful life of 60 years. Using these data, determine the amount of total physical deterioration for the residence. Curable Physical Deterioration Item Window screens RCN Cost to Cure $500 $800 Incurable Physical Deterioration–Short-Lived Item Water heater Roof cover Floor cover Total RCN RCN $800 $2,400 $4,500 $7,700 © 2022 IAAO l Revised 06/15 Updated July 2022 Actual Age (yr) Total Percentage Incurable Useful Life Depreciation Physical (yr) Deterioration 2 10 20% $160 10 25 40% $960 4 20 20% $900 $2,020 Total depreciation Solutions Chapter 3 - 79 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 3-12 SolutIon Cont. Total RCN Less RCN curable physical deterioration Less RCN incurable physical deterioration–short-lived RCN of basic (bone) structure Actual age of 12 ÷ total useful life of 60 Incurable physical deterioration–long-lived Curable physical deterioration Incurable physical deterioration–short-lived Incurable physical deterioration–long-lived $198,000 − $500 − $7,700 $189,800 x 0.20 $37,960 $800 $2,020 $37,960 $40,780 Total physical deterioration EXERCISE 3-12 Physical Deterioration Alternate Method of Set-up Curable Short-Lived Long Lived A B D TOTAL C RCN Curable Short-Lived Long Lived TOTAL (∑ ABOVE) © 2022 IAAO l Revised 06/15 Updated July 2022 RCN COMPLETE 1ST COMPLETE 3RD COMPLETE 6TH [C – (A+B)] DEPRECIATION COMPLETE 2ND COMPLETE 4TH COMPLETE 7TH D X PERCENTAGE DEPRECIATION COMPLETE 5TH COMPLETE 8TH (∑ ABOVE) DEPRECIATION $500 $7,700 $189,800 ($189,800 X.20) $198,000 $800 $2,020 $37,960 $40,780 Solutions Chapter 3 - 80 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-13 SOLUTION Curable Functional Obsolescence–Normal Deficiency An eight-unit apartment complex is being appraised. It does not have security doors on the two entrances to the building. According to the cost service, the cost to install a security door is $3,200. According to a local contractor, it costs $4,200 per door to install security doors today. Also, today’s market expects security doors. What is the measure of curable functional obsolescence due to this deficiency? Curable functional obsolescence–normal deficiency is measured by how much the cost of the addition exceeds the cost of the item if it were installed new during the construction of the improvement (excess cost to cure). Thus, the amount of functional obsolescence would be $8,400 – $6,400 = $2,000 (2 doors: $4,200 X 2, $3,200 X 2) Cost of existing item* 2. Less any depreciation already charged Added to 3. Cost to cure (all costs) 4. Less cost of security doors if installed new 5. Curable functional obsolescence (excess cost to cure) $0 − $0 +$8,400 − $6,400 $2,000 * Cost of existing item = replacement, reproduction, original, or trended reproduction cost. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 81 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-14 SOLUTION Curable Functional Obsolescence– Modernization Deficiency During the inspection of a residence being appraised, outdated light fixtures are found. The RCN for the existing fixtures is $6,000. An estimate of the amount of deterioration for the existing fixtures is 85 percent. The current cost of removing the old fixtures is $1,000, and the cost of installing modern fixtures is $6,500. There is no salvage value of the existing fixtures. Determine the amount of curable functional obsolescence due to this form of depreciation. 1. Cost of existing light fixtures* $6,000 Less any depreciation already charged (6,000 × 85%) − $5,100 (Step 1 less step 2 equals the depreciated value of the item) 900 Added to 2. Cost to cure (all costs, which include cost of removal + $7,500 less salvage value plus installed new cost) ($1,000 + $6,500) $8,400 3. Less cost if installed new − $6,500 4. Curable functional obsolescence $1,900 * Cost of existing item = replacement, reproduction, original, or trended reproduction cost. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 83 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-15 SOLUTION Developing a GRM A GRM for a single-family residence is being developed. The following are recent sales of comparable houses in the subject neighborhood. What is the GRM for the subject property? Divide sale price by the monthly rent. Sale 1 2 3 4 5 Sale Price Monthly Rent GRM $150,000 $1,550 96.8 $175,000 $1,800 97.2 $210,000 $2,150 97.7 $250,000 $2,550 98.0 $196,000 $2,000 98.0 The indicated GRM is 98. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 85 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-16 SOLUTION Incurable Functional Obsolescence–Normal Deficiency A 10-unit apartment complex is being appraised. There is no fire protection system in the building. Other similar apartment complexes in the area do have fire protection systems in place. A rental study in the area indicates a rent loss of $15 per apartment per month because of the lack of a fire protection system. The following are the sales and gross incomes: Complex Applewood Briarwood Cottonwood Tanglewood Sale Price $525,000 $485,000 $510,000 $495,000 Gross Annual Rent $87,500 $80,800 $85,000 $82,500 Determine the amount of incurable functional obsolescence caused by this deficiency. Step 1. Calculate total annualized rent loss. Multiply monthly rent loss by 12 and then by the number of units. Rent loss = 15 × 12 × 10 = $1,800 annually Step 2. Determine GIM by dividing the sale price by the gross annual rent. Complex Applewood Briarwood Cottonwood Tanglewood Sale Price Gross Annual Rent GIM $525,000 $87,500 $485,000 $80,800 $510,000 $85,000 $495,000 $82,500 6.0 6.0 6.0 6.0 The indicated gross income multiplier is 6.0 Step 3. Calculate incurable functional obsolescence by multiplying the annual rent lost by the GIM. Incurable functional obsolescence = 1,800 × 6 = $10,800 1. Cost of existing item* 2. Less any depreciation already charged Added to 3. Value loss 4. Less cost of fire protection system, if installed new 5. Incurable functional obsolescence–normal deficiency $0 - $0 + $10,800 - $0 $10,800 * Cost of existing item = replacement, reproduction, original, or trended reproduction cost. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 87 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-17 SOLUTION External Obsolescence The house being appraised is located under one of the major flight paths of the city’s airport and rents for $875 per month. Comparable houses in the area, not under the flight path, rent for $950 per month. The land-to-building ratio for this type of property is 1:4. The following are recent sales of comparable houses and their monthly rents. Estimate the external obsolescence for the improvement. Step 1. Calculate GRM for each sale and correlate to a composite GRM. Sale Monthly Sale Price GRM Number Rent 1 $950 $109,300 115.1 2 $900 $103,500 115.0 3 $975 $112,100 115.0 4 $925 $106,400 115.0 The GRM is 115. Step 2. Determine rent loss due to location. Rent in unaffected location Less the rent for the subject Rent loss due to location $950 − $875 $75 Step 3. Multiply rent loss by the GRM. Rent loss due to location GRM Total loss in value (land and building) $75 × 115 $8,625 Step 4. Multiply total loss in value by the building ratio. Total loss in value (land and building) Land-to-building ratio 1:4 (building value = 80%) External obsolescence © 2022 IAAO l Revised 06/15 Updated July 2022 $8,625 × 0.80 $6,900 Solutions Chapter 3 - 89 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-18 SOLUTION External Obsolescence A single-family residence that is being appraised is experiencing significantly more traffic because of a recent zoning change on nearby property. There are two comparable properties that have recently sold, one adjacent to the subject property and one in an area not affected by the zoning change. The property adjacent to the subject property sold for $137,500 with a land value of $27,500. The property outside the area affected by the zoning change sold for $150,000 with a land value of $33,000. Based on paired sales analysis, what is the indicated amount of external obsolescence? Step 1. Isolate building value for each sale:. $137,500 –$ 27,500 $110,000 $150,000 – $ 33,000 $117,000 Step 2. The difference between the two building values represents the obsolescence. $117,000 –$110,000 $ 7,000 External obsolescence measured by analyzing paired sales is $7,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 91 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 3-19 SOLUTION Observed Condition (Breakdown) Method Estimate the value of the subject property using the cost approach. Use the observed condition (breakdown) method to determine depreciation. Vacant land sales have already been analyzed in a site valuation, and the indicated land value for the subject site is $40,000. All necessary information is provided below. Reproduction Cost New $240,000 Curable Physical Deterioration Item RCN Cost to Cure Paint interior $3,900 $4,125 Replace window screens $495 $570 Replace windows $1,845 $2,100 Total $6,240 $6,795 Incurable Physical Deterioration–Short-Lived Item RCN Actual Age (yr) Roof cover $5,775 10 Water heater $945 10 Air-conditioning $6,475 10 unit Total $13,195 Total Useful Life (yr) 20 15 20 Percentage Depreciation 50% 67% 50% Depreciation $2,888 $633 $3,238 $6,759 Incurable Physical Deterioration–Long-Lived The actual age of the structure is 10 years; the total useful life is 50 years. Actual age of 10 years divided by total useful life of 50 years = 20% depreciation RCN curable physical deterioration RCN incurable physical deterioration–short-lived Total RCN of improvement Less RCN already depreciated Remaining RCN to Depreciate © 2022 IAAO l Revised 06/15 Updated July 2022 $6,240 + $13,195 $19,435 $240,000 − $19,435 $220,565 × 0.20 $44,113 Solutions Chapter 3 - 93 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 3-19 SolutIon Cont. oBServed CondItIon (Breakdown) Method Curable Physical Deterioration (From previous page) Cost to Cure $6,795 Incurable Physical Deterioration–Short-Lived $6,759 Incurable Physical Deterioration–Long-Lived $44,113 Curable Functional Obsolescence–Normal Deficiency This home lacks a third bath when the market requires one for this size home. The cost new of installing an additional bath is $22,000. If a third bath had been installed when the home was built, the RCN today would be $15,000. Excess cost to cure is $0 − $0 + $22,000 − $15,000 = $7,000 Curable Functional Obsolescence–Modernization Deficiency Kitchen fixtures need to be replaced. The cost new of existing fixtures is $2,500. Physical deterioration already charged is $1,700. There is no salvage value of old fixtures. Cost to remove the old fixtures is $1,200, and the cost to install new modern fixtures, if installed when built, is $2,000. $2,000 Rent loss of $60 × GRM of 120 = External Obsolescence The rent loss due to the external factor affecting the property’s value is $50; the GRM for the subject property is 120; and the land-tobuilding ratio is 1:4. Rent loss of $50 × GRM of 120 × 0.80 = Total Depreciation by the Observed Condition (Breakdown) Method = Depreciated Improvement Value ($240,000 - $78,667) Land Value Value by the Cost Approach © 2022 IAAO l Revised 06/15 Updated July 2022 $7,200 $4,800 $78,667 (∑ of above) $161,333 + $40,000 $201,333 Solutions Chapter 3 - 94 Review Solutions Course 101 Fundamentals of Real Property Appraisal CHAPTER 3 1. The cost approach is premised primarily on the principle of substitution, which states an informed buyer will not pay more to build a property than the cost of a similar property with equal utility. 2. Costs that occur on the construction site are considered direct costs. 3. The first step in the cost approach is to estimate the land (site) value as if vacant and available for development to its highest and best use. 4. Building A contains 10,000 square feet and is 100’ x 100’. Building B contains 10,000 square feet and is 50’ x 200’. Will it cost more to build Building A or Building B? Building A perimeter = 100 + 100 + 100 + 100 = 400 liner feet. Building B = 50 + 200 + 50 + 200 = 500 linear feet Building B, because as it has more linear feet compared to Building A. 5. Architectural fees, insurance, and title expenses are considered indirect costs and are sometimes called soft costs. 6. Reproduction cost is the cost of producing an exact replica of a building or improvement. 7. Replacement cost is the cost of producing a building or improvement using modern methods and materials with the same utility. 8. Building A was constructed 20 years ago at a cost of $400,000. The index 20 years ago was 200. The current index is 350. 350/200 = 1.75; 1.75 × $400,000 = $700,000. Reproduction cost new for Building A would be $700,000. 9. Cost components such as the floor or roofing are considered horizontal costs, and are expressed as cost per square foot. 10. The easiest, fastest, and most widely used method of estimating cost is the comparative unit (square foot) method. 11. Which of the following is a desirable feature of a cost manual? A. Applicable to most structure types B. Based on actual costs C. Suitable for computerization D. All of the above © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 95 Review Solutions Course 101 Fundamentals of Real Property Appraisal 12. The difference between the market value of an improvement and its cost at the time of appraisal is termed depreciation. 13. The five methods of measuring depreciation are 1. 2. 3. 4. 5. 14. Sales comparison method Capitalization of income method Economic age-life method Modified economic age-life method Observed condition (breakdown) method Property B sold for $180,000 with an estimated land value of $80,000. The RCN for the improvement is $200,000. Depreciation is 50 percent. 15. Curable physical deterioration is measured by cost to cure. 16. Functional obsolescence is a loss in value due to a deficiency, modernization, or superadequacy within the structure. 17. External obsolescence is a loss in value resulting from a decrease in utility and desirability caused by factors outside the property’s boundaries. 18. Incurable physical deterioration is measured by the age-life method. 19. If a depreciation table started at 0 percent and went to 40 percent, a percent good table would start at 100% and go to 60%. 20. Which method of estimating depreciation is the most detailed? Its use is required in the writing of demonstration appraisal reports. Observed condition (breakdown) method 21. An item classified as a short-lived item has a cost new of $5,000. Its actual age is 5 years and its total useful life is 20 years. Actual Age ÷ Total Useful Life = 5 ÷ 20 = 0.25 × Cost New = $1,250 The dollar amount of depreciation would be $1,250. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 96 Review Solutions Course 101 Fundamentals of Real Property Appraisal 22. A two-story residence has only one bath, which is located on the second floor. If a second bath had been included on the first floor when the residence was originally constructed, RCN would be $11,500. The cost to install (cure) the second bath today would be $13,500. Cost to Cure - Cost if installed New = $13,500 - $11,500 = $2,000 The amount of functional obsolescence would be $2,000. 23. A residence is located on a street that has become a very busy thoroughfare. It rents for $800 per month. Residences not located on the busy street rent for $825 per month. The GRM for this area is 150, and the land-to-building ratio is 1:4. Rent loss due to busy street = $825 - $800 = $25/month GRM × Rent loss = 150 × $25 = $3,750. Land to building ratio = 1:4 = 0.80 × $3750 = $3,000 The indicated amount of external obsolescence is $3,000. 24. A residence has an RCN of $200,000. Items classified as curable physical deterioration have an RCN of $10,000 and a cost to cure of $12,000. Incurable physical deterioration (short-lived items) have an RCN of $40,000 and a depreciation amount of $18,000. The actual age for the residence is 6 years, and the total useful life is 60 years. RCN Depreciation (Cost to cure) $12,000 $18,000 Curable Physical Deterioration $10,000 Incurable Physical Deterioration $40,000 Total $50,000 RCN of residence $200,000 Less RCN already depreciated - $50,000 Remaining RCN to depreciate $150,000 × 0.10 Actual age ÷ Total Useful Life = 6 ÷ 60 = 10 % depreciation Total Curable + Incurable + Remaining RCN The total amount of physical deterioration for the residence would be $45,000. 25. $30,000 $15,000 $45,000 If the cost to cure a functional deficiency within a residence exceeds the increase in value added to the property, it is called incurable functional obsolescence. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 3 - 97 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 4-1 SolutIon SelECTIng unItS of CoMparISon Subject Sale Price Area (sq. ft.) Sale price per square foot Number of Units Sale price per unit Number of Rooms Sale price per room Number of Bedrooms Sale price per Bedroom Potential gross income (annual) GIM Sale 1 4,180 Sale 3 Sale 4 Percentage Spread $904,000 4,000 $226.00 $984,000 4,800 $205.00 $800,000 3,600 $222.22 $925,000 4,200 $220.24 8 8 8 8 $113,000 $123,000 $100,000 $115,625 36 32 24 36 $25,111 $30,750 $33,333 $25,694 16 16 8 16 $56,500 $61,500 $100,000 $57,813 $115,200 $124,800 $100,800 $115,200 7.85 7.88 7.94 8.03 8 32 16 $115,200 Sale 2 10.24% 23.00% 32.74% 76.99% 2.29% The Gross Income Multiplier (GIM) appears to be the preferred unit of comparison. If Sale 3 were not included in the calculation for number of bedrooms, the percentage spread would be 8.85%, which would be much more acceptable. However, the GIM would still be the preferred unit of comparison, and a GIM of 8.00 appears to be appropriate for the subject property. Applying the GIM to the subject parcel would provide an indicated value for that parcel of: $115,200 × 8.00 = $921,600 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 99 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-2 SOLUTION Market Condition Analysis using Resales Property 1 sold 2 years ago for $150,000 and sold again 6 months ago for $163,500. Property 2 sold 1 year ago for $160,000 and sold again last week for $169,600. Property 3 sold 30 months ago for $140,000 and sold again 6 months ago for $156,800. Use the three sales to estimate the monthly adjustment for change in market conditions over time. Step 1. Calculate number of months between sales. First sale = 2 yr × 12 = 24 mo. ago Property 1 Second sale = 6 mo. ago First sale = 1 yr × 12 = 12 mo. ago Property 2 Second sale = 0 mo. ago Property 3 First sale = 30 mo. ago Second sale = 6 mo. ago 24 – 6 = 18 12 – 0 = 12 30 – 6 = 24 Step 2. Subtract first sale amount from second sale amount. Step 3. Divide result of step 2 by first sale amount. Step 4. Divide result of step 3 by the number of months (found in step 1). Step 2 (second sale – first sale) Property 1 $163,500 – $150,000 = $13,500 Property 2 $169,600 – $160,000 = $9,600 Property 3 $156,800 – $140,000 = $16,800 Step 3 (result of step 2 ÷ first sale) $13,500 ÷ $150,000= 0.09 $9,600 ÷ $160,000 = 0.06 $16,800 ÷ $140,000 = 0.12 Step 4 (result of step 3 ÷ mo.) 0.09 ÷ 18 = 0.005 0.06 ÷ 12 = 0.005 0.12 ÷ 24 = 0.005 The monthly adjustment would be 0.005, or 0.5% per month. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 101 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-3 SOLUTION Market Condition Analysis using Paired Sales Formula: Step 1. The current sale minus the prior sale equals the overall dollar amount of change. Step 2. The overall dollar amount of change divided by the prior sale equals the percentage of change. Step 3. The percentage of change divided by the number of months between sales provides the percentage of change per month. Current sale = $275,000; prior sale 10 mo. ago = $250,000 Step 1. $275,000 – $250,000 = $25,000 Step 2. $25,000 ÷ $250,000 = 0.10, or a 10% increase Step 3. 0.10 ÷ 10 mo. = 0.01, or 1% per month Current sale = $231,000; prior sale 5 mo. ago = $220,000 Step 1. $231,000 – $220,000 = $11,000 Step 2. $11,000 ÷ $220,000 = 0.05, or a 5% increase Step 3. 0.05 ÷ 5 mo. = 0.01, or 1% per month Current sale = $218,000; prior sale 9 mo. ago = $200,000 Step 1. $218,000 – $200,000 = $18,000 Step 2. $18,000 ÷ $200,000 = 0.09, or a 9% increase Step 3. 0.09 ÷ 9 mo. = 0.01, or 1% per month The indicated monthly adjustment based on the three groups of paired sales is 0.01, or 1% per month. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 103 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-4 Adjustment for Location Using Paired Sales A percentage adjustment for location must be determined using paired sales analysis. Paired sales have already been used to estimate an appropriate adjustment for market conditions (time adjustment) with a result of 0.5 percent per month. The following sales are very similar in all aspects except for location. The adjustment for location must be estimated because not all the comparables are in the same subdivision. Lake View is considered the subject area. Use the following sales and determine the indicated percentage adjustment for location: Sale 1 2 3 Location Lake View Spring Creek Spring Creek Sale Date Current 6 mo. ago 12 mo. ago Sale Price $180,000 $200,000 $196,000 Step 1. Apply the market conditions adjustments to sale price 2 and 3. Step 2. Subtract the adjusted sale prices 2 and 3 from adjusted sale price 1. Step 3. Divide the results from Step 1 by the adjusted sale prices of sales 2 and 3. Based on these data, the indicated percentage adjustment for location of Spring Creek is -13%. Since the subject property is in Lake View, the location adjustment would be to adjust Spring Creek to Lake View. Spring Creek is more valuable than Lake View since the sale prices of both properties in Spring Creek are greater than Lake View. Thus, the adjustment would be negative from Spring Creek to Lake View. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 105 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 4-5 SolutIon SaleS CoMparISon approaCh The appraisal assignment is a one-story brick residence containing three bedrooms, one bathroom, a fireplace, central heat and air conditioning, and a single-car garage. The following sales are comparables: Sale 1 is a one-story brick residence containing four bedrooms, two bathrooms, a fireplace, central heat and air conditioning, and a two-car garage. It sold 6 months ago for $243,000. Sale 2 is a one-story brick residence containing three bedrooms, two bathrooms, a fireplace, central heat and air conditioning, and a single-car garage. It sold 2 months ago for $234,000. Sale 3 is a one-story frame residence containing three bedrooms, one bathroom, central heat and air conditioning, and a single-car garage. It is a current sale for $215,000. Market analysis provided the following adjustments: Fourth bedroom $12,000 Second bathroom $11,000 Single-car garage $12,000 Two-car garage $18,000 Fireplace $3,000 Frame construction (brick is superior to frame) $10,000 Market conditions (time of sale) 1% per month Prepare a lump-sum adjustment grid and estimate the value of the subject property. Adjustment Grid Subject Sale price Market conditions adjustment Adjusted sale price Bedrooms Bathrooms Fireplace Garage Construction Net adjustments Adjusted sale price Sale 1 3 1 1 1 Brick Sale 2 Sale 3 $243,000 6% $234,000 2% $215,000 $257,580 –$12,000 – $11,000 $238,680 $215,000 – $11,000 + $3,000 – $6,000 – $29,000 $228,580 –$11,000 $227,680 + $10,000 + $13,000 $228,000 The indicated value of the subject property is $228,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 107 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-6 SOLUTION Sales Comparison Approach The appraisal assignment is a single-family residence with three bedrooms, one bathroom, forced-air heat, no fireplace, and an attached two-car garage. There are four recent sales of single-family residences in the neighborhood that are similar in age, construction, and amenities to the subject: Sale 1 is a four-bedroom, two-bathroom residence with a large fireplace, hot-water heat, and an attached two-car garage. This property sold 2 years ago for $185,000. Sale 2 is a three-bedroom, one-bathroom residence with forced air heat and an attached single-car garage. This sale is a current sale and sold for $170,000. Sale 3 is a three-bedroom, two-bathroom residence with an average fireplace, hot-water heat, and an attached two-car garage. This property sold 2 years ago for $175,000. Sale 4 is a four-bedroom, two-bathroom residence with a large fireplace, hot-water heat, and an attached two-car garage. This property sold 1 year ago for $210,000 and included personal property worth $10,000. Adjustment Grid Subject Sale price Sale conditions Adjusted sale price Market condition adjustment Adjusted sale price Heat Bathrooms Bedrooms Fireplace Garage Sale 1 Sale 2 Sale 3 Sale 4 $185,000 $170,000 $175,000 $210,000 – $10,000 $185,000 $170,000 $175,000 $200,000 10% $203,500 Forced air 1 3 None Attached 2-car Net adjustments Adjusted sale price $170,000 – $2,200 – $11,000 –$12,000 – $5,000 10% 5% $192,500 $210,000 – $2,200 – $11,000 – $4,000 – $2,200 – $11,000 – $12,000 – $5,000 – $17,200 $175,300 – $30,200 $179,800 + $5,500 – $30,200 $173,300 + $5,500 $175,500 The indicated value of the subject property is $175,500. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 109 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-7 SOLUTION Sales Comparison Approach The appraisal assignment is a single-family residence of average quality. The subject has three bedrooms, two bathrooms, one fireplace, and a two-car attached garage. It is in average condition and is located in the southwest suburbs. The following are comparable sales: Sale 1 is an average-quality residence containing four bedrooms, two bathrooms, one fireplace, and a two-car attached garage. Its condition is good compared to the subject. It is also located in the southwest suburbs. It sold 1 year ago for $250,000. Sale 2 is a good-quality residence containing four bedrooms, two bathrooms, one fireplace, and a three-car attached garage. It is in average condition and is located in the northeast suburbs. It sold 6 months ago for $280,000. Sale 3 is an average-quality residence containing three bedrooms, two bathrooms, one fireplace, and a two-car attached garage. It is in good condition and is located in the southwest suburbs. It is a current sale for $260,000. However, financing requires an adjustment because of seller financing. Adjustment Grid Subject Sale price Financing Adjusted sale price Market condition adjustment Adjusted sale price Bedrooms Quality Condition Garage Location Net adjustments Adjusted sale price Sale 1 3 Average Average 2-car Southwest Sale 2 Sale 3 $250,000 n/a $250,000 +10% $280,000 n/a $280,000 +5% $260,000 – 5% $247,000 0% $275,000 – 10% Equal – 5% Equal Equal – 15% $233,750 $294,000 – 10% – 15% Equal – 5% + 10% – 20% $235,200 $247,000 Equal Equal – 5% Equal Equal – 5% $234,650 The indicated value for the subject is $235,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 111 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-8 SOLUTION Sales Comparison Approach The appraisal assignment is an average-quality, frame, two-story single-family residence, located on a 1-acre tract in Benton subdivision. The residence has three bedrooms, two bathrooms, one fireplace, and a two-car attached garage. It is in average condition. The following are comparable sales: Sale 1 is an average-quality, frame, two-story single-family residence, located on a 2-acre tract in Benton subdivision. The residence has three bedrooms, two and one-half bath rooms, one fireplace, and a three-car attached garage. Its condition is good compared to the subject. It sold 1 year ago for $300,000. Sale 2 is a good-quality, frame, two-story single-family residence, located on a 1-acre tract in Aspen subdivision. The residence has four bedrooms, two bathrooms, one fireplace, and a three-car attached garage and is in average condition. It sold 6 months ago for $275,000. Sale 3 is an average-quality, brick, two-story single-family residence, located on a 1-acre tract in Benton subdivision. The residence has three bedrooms, two and one-half bathrooms, one fireplace, and a two-car attached garage. It is in average condition and is a current sale for $279,000. Adjustment Grid Subject Sale price Market conditions adjustment Adjusted sale price Location Lot size Bedrooms Half-bathrooms Garage Fireplace Quality Condition Construction Net adjustments Adjusted sale price Sale 1 Benton 1 acre 3 None 2-car 1 Average Average Frame Sale 2 Sale 3 $300,000 + 6% $275,000 + 3% $279,000 0% $318,000 Equal – $25,000 Equal – $4,000 – $8,000 Equal Equal - 15,900 Equal – $52,900 $265,100 $283,250 + $28,325 Equal – $12,000 Equal – $8,000 Equal – $28,325 Equal Equal – $20,000 $263,250 $279,000 Equal Equal Equal – $4,000 Equal Equal Equal Equal – $10,000 – $14,000 $265,000 The indicated value for the subject is $265,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 113 Solutions Course 101 Fundamentals of Real Property Appraisal EXERCISE 4-9 SOLUTION Sales Comparison Approach The appraisal assignment is an eight-unit apartment. It is brick construction containing 6,400 square feet, of average quality, and located in Fremont subdivision. Its condition is average, and each unit has its own forced air heat with air conditioning. There are two entrances to the building, and each entrance has a security door. There is a fire protection system throughout the building. Also, there is covered parking, and each tenant has one space. Market analysis indicates the sale price per square foot is the most appropriate unit to use in the analysis. The following are comparable sales: Sale 1 is an eight-unit apartment of brick construction containing 6,100 square feet. It is a fair-quality building in Sherwood subdivision. Its condition is average, and each unit has its own forced-air heat with air conditioning. The two entrances have security doors, and there is a fire protection unit throughout the building. However, there is no covered parking; there is a limited parking area on a first-come first-served basis. This property sold 6 months ago for $480,000. Sale 2 is an eight-unit apartment of brick construction containing 7,200 square feet. It is an average-quality building in Fremont subdivision. Its condition is average. It has hot-water heat with window air-conditioning units. The two entrances have security doors, but there is no fire protection unit in the building. Also, there is no covered parking; there is a limited parking area on a first-come first-served basis. This property sold recently for $570,000. Sale 3 is an eight-unit apartment of frame construction containing 5,800 square feet. It is a good-quality building in Fremont subdivision. Its condition is fair, and each unit has its own forced-air heat with air-conditioning. The two entrances do not have security doors, but there is a fire protection unit throughout the building. Also, there is no covered parking; there is a limited parking area on a first-come first-served basis. This property sold 1 year ago for $485,000. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 115 Solutions Course 101 Fundamentals of Real Property Appraisal eXerCISe 4-9 SolutIon Cont. SaleS CoMparISon approaCh Adjustment Grid Subject Sale price Market conditions adjustment Market-adjusted sale price Area (square feet) Sale price per square foot Heat and cooling Quality Condition Security doors Fire protection Location Construction Parking Net adjustments Adjusted sale price per square foot Sale 1 Sale 2 Sale 3 $480,000 $570,000 $485,000 1.5% 0 3% $487,200 $570,000 $499,550 6,400 6,100 $79.87 7,200 $79.17 5,800 $86.13 Yes Average Average Yes Yes Fremont Brick Covered Equal + 10% Equal Equal Equal + 5% Equal + 5% + 20% $95.84 + 10% Equal Equal Equal + 5% Equal Equal + 5% + 20% $95.00 Equal – 10% + 5% + 5% Equal Equal + 5% + 5% + 10% $94.74 The indicated square foot value for the subject property is $95.00 The indicated value for the subject property is 6,400 (square foot area) × $95.00 (adjusted sale price per square foot) = $608,000 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 116 Review Solutions Course 101 Fundamentals of Real Property Appraisal CHAPTER 4 1. The Principle of Substitution holds that the value of a property tends to be set by the price that would be paid to acquire a property of similar utility and desirability within a reasonable amount of time. 2. The Principle of Contribution is the underlying principle in the adjustment process in the sales comparison approach. 3. A major principle in the sales comparison approach holds that the relationship between a property and its environment must be in balance for a property to reflect its optimum market value. 4. A major advantage to the sales comparison approach is that it reflects the actions of buyers and sellers in the market. 5. A unit of comparison that would be applicable to single-family residential property is the price per square foot of gross living area or total property price. 6. A unit of comparison that would be applicable to agricultural farmland or larger rural residential tracts would be the price per acre. 7. In the selection of comparable properties in the sales comparison approach, list four comparability items that should be considered. 1. 2. 3. 4. 5. 6. 7. 8. 8. List the order of adjustments in the sales comparison approach. 1. 2. 3. 9. Financing Current market conditions (date of sale) Similar in location Similar in size, age, condition Similar in design type Similar in construction type Affected by the same four forces that influence value Similar physical attributes Financing Market conditions (time of sale) Location and other physical characteristics Which method of determining adjustments in the sales comparison approach cannot be used to determine market conditions (changes in value over time)? Depreciated cost © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 117 Review Solutions 10. Course 101 Fundamentals of Real Property Appraisal The first step in the sales comparison approach is to define the appraisal problem. List the five items that need to be included in defining the appraisal problem. 1. 2. 3. 4. 5. Identify the property to be appraised Determine the property rights to be appraised Define the purpose and function of the appraisal Specify the date of appraisal Define the type of value 11. A cardinal rule in the sales comparison approach holds that all adjustments are made to the comparable property, and never to the subject. 12. The Principle of Change states that market value is never constant because physical (environmental), governmental, social, and economic forces are at work, affecting the property. 13. When the four forces—physical (environmental), governmental, social, and economic—are in balance, the market achieves a temporary state of equilibrium. 14. Paired sales analysis, which is used primarily in single-property appraisal, requires the sale properties to be identical in all attributes except the one being measured (or adjustments have already been made for any differences). 15. The sales comparison approach usually cannot be used to value special-purpose or one-of-a-kind properties. 16. In the sales comparison approach, step two is data collection and verification. This requires the appraiser to collect and verify general, specific, and comparative data. 17. The unit of comparison that is generally used in valuing office properties is the price per square foot of net rentable area. 18. Price per animal unit might be the appropriate unit to use in the valuation of agricultural pastureland. © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Chapter 4 - 118 Quiz Solutions Course 101 Fundamentals of Real Property Appraisal QUIZ 2 1. C 2. C 3. B 4. A 5. D 6. B 7. A 8. D 9. B 10. C 11. B 12. D 13. D 14. B 15. C 16. A 17. B 18. B © 2022 IAAO l Revised 06/15 Updated July 2022 19. C 20. C 21. D 22. D 23. A 24. D 25. B 26. A 27. B 28. C 29. B 30. D Solutions Quiz 2 - 119 4. Quiz Solutions Course 101 Fundamentals of Real Property Appraisal You are analyzing resales to determine a time adjustment. You have found a sale that sold 18 months ago for $200,000 and sold again 8 months ago for $220,000. Based on this sale, the monthly adjustment for time would be: Older sale New sale Difference Months 18 8 10 Price $200,000 $220,000 $20,000 Difference per month $20,000 ÷ 10 = $2,000 Percentage per month $2,000 ÷ $200,000 = 0.01 9. $80.00 × 1.20 = $96.00 × 1,500 = $144,000 10. Improvement cost new Less depreciation: (0.005 × 10)=0.05 × 144,000 Replacement cost new less depreciation Land value Indicated value by the cost approach $144,000 -$7,200 $136,800 $50,000 $186,800 The following data are for questions 15, 16, and 17. RCN Actual age Total useful life RCN: physical curable items Total depreciation: physical curable items RCN: physical incurable short-lived items Total depreciation: physical incurable short-lived items Land value 15. Total RCN RCN physical curable items RCN physical incurable short-lived items Remaing RCN to be depreciated Actual age Total useful life Depreciation © 2022 IAAO l Revised 06/15 Updated July 2022 $368,000 20 years 80 years $8,800 $10,000 $60,000 $17,200 $75,000 $368,000 $8,800 $60,000 $68,800 $299,200 20 ÷80 0.25 $74,800 Solutions Quiz 2 - 120 Quiz Solutions Course 101 Fundamentals of Real Property Appraisal 16. Total depreciation: physical curable items l Total depreciation: physical incurable short-lived items Total depreciation: physical incurable long-lived items Total physical depreciation 17. Total RCN Total depreciation Cost new less depreciation Land value Indicated value by the cost approach $10,000 17,200 74,800 $102,000 $368,000 102,000 266,000 75,000 $341,000 18. Item Subject Sale 1 Sale 2 Sale 3 Sale 4 Sale price $360,000 $277,500 $330,000 $310,000 Sale date 10 mo. ago 12 mo. ago 10 mo. ago Current Bedrooms 3 4 3 3 3 Bathrooms 2 and 1/2 2 and 1/2 1 and 1/2 2 and 1/2 1 and 1/2 Den 1 1 None 1 1 Garage stalls 2 2 1 2 1 Market conditions +$18,000 +$16,650 +$16,500 Adj. Sales Price $378,000 $294,150 $346,500 $310,000 Bedroom -$22,000 Bathroom +$12,000 +$12,000 Den +$15,000 Garage +$8,500 +8,500 Total adjustments -$22,000 +$35,500 +$20,500 Adjusted sale price $356,000 $329,650 $346,500 $330,500 © 2022 IAAO l Revised 06/15 Updated July 2022 Solutions Quiz 2 - 121 Quiz Solutions 27. 1. Cost of existing outlets 2. Less depreciation @ 50% 3. Cost to cure (all costs) Depreciation Cost to remove old and install new 4. Less cost if installed new 5. Curable functional obsolescence Course 101 Fundamentals of Real Property Appraisal $3,600 − 1,800 +$4,600 29. Rent off the busy street Rent on the busy street Rent loss Gross rent multiplier Total value loss Loss to building at 1:4 land to building ratio Indicated depreciation © 2022 IAAO l Revised 06/15 Updated July 2022 − $3,800 $2,600 $900 850 $50 × 150 $7,500 × 0.80 $6,000 Solutions Quiz 2 - 122