Cost Approach Flashcards

1
Q

A house cost $200,000 to build new. It is 10 years old and has sustained a total of 15% depreciation. It sits on a lot worth $40,000. What is its value by the cost approach?

A

$200,000
X 85%
= $170,000
$40,000 plus
= $210,000

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2
Q

A house contains 2,826 square feet. It was constructed in 1995 at a cost of $54.18 per SF. The cost index at that time was 193.4.
The cost index today is 286.3.
What is the estimated cost to build the building today?

A

Index Method

2,826 SF X $54.18 = $153,113
286.3 I 193.4 = 1.480
$153,113 X 1.480 = $226,607

Therefore, costs have gone up 48% since 1995, and it would cost $226,607 to build the same house today.

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3
Q

You find a comparable sale of a new house similar to the one you are appraising. It just sold for $320,000. You estimate the value of the site at $75,000 and the cost of the site improvements at $15,000.

The house has 2,550 square feet (SF) of gross living area (GLA).
What is the unit cost of the structure?

If your subject property has 2,475 SF, what should it cost to build?

Okay, let’s do a quick cost approach for our subject property. Your research shows the land is worth $65,000 and you project the site improvements will cost $12,000. When completed, it will be brand new with no depreciation.

A

Comparative Unit Method / Market Derivation
$320,000 - $75,000 -$15,000 = $230,000.

All you have to do is divide $230,000 by 2,550 and the answer is $90.20. That is a market- derived comparative unit cost.

2,475 x $90.20 = $223,245. That works well when the houses are about the same size. We will see later that the costs vary according to the size of the structure.

Cost new - Depreciation + Land Value+ Site Improvements
= Value by Cost Approach
$223,245 - 0 + $65,000 + $12,000
= $300,245, rounded to $300,000.

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4
Q

If a total economic life is estimated to be 60 years and the current effective age is estimated to be 20 years, then the remaining economic life is x

A

40

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5
Q

A 20 year old building that would cost $220,000 to build new today. It is in good condition, and you estimate the effective age to be 15 years. Your estimate of total economic life is 60 years, based on analysis of similar properties. How much is the estimated depreciation calculated with the age-life method?

A

15
/60
x $220,000 = $55,000 depreciation

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6
Q

Formula Age Life Method:

A

Effective age
/ Total economic life
X total cost
= depreciation

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7
Q

The cost new of the improvements is $245,200, the site value is $60,000, and the effective age is 17. The total economic life expectancy judged by comparable sales is 55 years. What is the value using the cost approach?

A

17 / 55 = .309 or 30.9%
$245,200 X .309 = $ 75,767
Total cost of the improvements $245,200
Less total depreciation - 75,767
Depreciated cost $169,433
Plus site value + 60,000
Indicated value by the cost approach $229,433
The indicated value should be appropriately rounded. In this case, we would probably round to $230,000.

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8
Q

The cost new of the subject property is $235,000, the site value is $70,000 and the effective age is 25. The total economic life is estimated to be 60 years.
The subject has deferred maintenance items that total $4,500. After these repairs, the effective age will be reduced to 20 years. What is the value using the cost approach?

A

Modified Age Life Method:

$235,000 - $4,500 = $230,500 Remaining Improvement Cost

20 (effective age)
/ 60 (total economic life)
= .333 (percent of depreciation)

Cost Approach
Remaining Improvement Cost $230,500
Minus Depreciation ($230,500 x .33) - 76,065
Depreciated Cost of Improvements $154,435
Plus Site Value $70,000
Value by Cost Approach $224,435
Rounded to $225,000

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9
Q

A roof on a house has an expected life of 20 years. The economic life of the house is 60 years.
We are appraising a house when it was 15 years old. We might estimate the cost to replace the roof at $4,000. How much is the depreciation for the roof?

A

Depreciation of short lived items:
Fifteen out of 20 years means it has lost .75 or 75% of its value. However, it still retains 25% of its original value.
$4,000 x .75 = $3,000 depreciation.

The roof still has a remaining value of $1,000, but the short-lived depreciation charged to the roof is $3,000.

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10
Q

The subject property is a house adjoining a busy interstate highway. A vacant lot next door recently sold for $60,000 and a similar lot with the same zoning, in a quiet location, recently sold for $65,000.

A house along the interstate in the next block recently sold for $340,000, and a similar house away from the interstate sold for $360,000. What is the external obsolescence for the building?

A

Market Data Analysis using the breakdown method:

There is a premium of $5,000 for less traffic noise for land sales.

Thus, the property located along the interstate suffered a $20,000 loss for the location. The external obsolescence attributable to the whole property is $20,000 ($360,000 - $340,000).
Pairing the two land sales indicates a loss of $5,000 for the land portion. Therefore, the external obsolescence attributable to the building is $15,000 ($20,000 - $5,000).

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11
Q

A building has a cost new of $400,000. Deferred maintenance items have been identified at a cost of $9,000. Short-­lived depreciation has been identified at $15,000; these short­-lived items have a total cost of $45,000. The building has an effective age of 20 years and a total economic life of 50 years. What is the amount of depreciation attributable to the long-­lived items?

A

$400,000 - $9,000 - $45,000 = $346,000 (cost of long-lived items). 20 / 50 = 40% depreciation. $346,000 x 40% = $138,400.

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12
Q

A very simplistic example would be as follows. A house would cost $200,000 to build new. It is 10 years old and has sustained a total of 15% depreciation. It sits on a site worth $40,000. What is its value by the cost approach?

A

Cost new
Accrued Depreciation ($200,000 x .15) Depreciated value of improvements site value
Indicated value by cost approach
$200,000 - $ 30,000
$170,000 + $ 40,000
=$210,000

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13
Q

The contract cost for constructing a house in January 1996 was $324,500.

The index for January 1996 was 212.6 and the current index is 306.4.

What is the current cost estimate?

A

Cost Index Method:
The contract cost for constructing a house in January 1996 was $324,500.

The index for January 1996 was 212.6 and the current index is 306.4.

306.4 divided by 212.6 = 1.441. This means the costs have increased 44.1%. For the sake of consistency, let’s round to two decimal places, and use 1.44.

$324,500 X 1.44 = $467,280

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14
Q

You find a comparable sale of a new house similar to the one you are appraising.
It just sold for $320,000. You estimate the value of the site at $75,000 and the cost of the site improvements at $15,000.
The house has 2,550 square feet of gross living area.
Your research shows the subject land is worth $65,000 and you project the site improvements will cost $12,000. 2,475 SF. When completed it will be brand new with no depreciation.

What is the value according to the cost approach?

A

Comparative Unit Method
$320,000 - $75,000 -$15,000 = $230,000 improvement cost new
$230,000
/2,550 SF
= $90.20/SF

2,475 SF
x $90.20
= $223,245

$223,245 building new
+$65,000 site
+ $12,000 site improvements
= $300,245
Rounded $300,000

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15
Q

Marshall & Swift may state that a 2,400 square foot home costs $90.00 per square foot, and a 2,500 square foot costs $85.00 per square foot. (Larger homes typically cost less per square foot to build.)
But what if your subject home is 2,425 square feet?

A

Interpol:

It would cost less than a 2,400 square foot home, but more than a 2,500 square foot home. We can interpolate between these known costs.

If a 2,400 square foot home costs $90.00 per SF, and a 2,500 SF home costs $85.00 per SF, we can add the costs together ($90 + $85) and divide by 2 to get the midpoint between the two costs ($87.50). This would be the interpolated cost per SF of a 2,450 SF home, because 2,450 is the midpoint between 2,400 and 2,500.
Now, we can interpolate between the costs of a 2,400 SF home ($90.00) and a 2,450 SF home ($87.50) and get the cost of a 2,425 SF home ($90.00 + $87.50 = $177.50 / 2 = $88.75). So $88.75 would be the interpolated cost new of a 2,425 SF home, based on the above cost figures.

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16
Q

If a home costs $102.48 per square foot based on the Marshall & Swift manual, and the current cost multiplier is 1.02, and the local multiplier is .96, if the subject home has 2,152 square feet, cost new would be $x.

A

$102.48 x 1.02 x 0.96 and this would indicate a cost per square foot of $100.35.
215,953

(Site valuation & cost approach ch 5, page 56)

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17
Q

Formula depreciation with age life method:

A

Effective age

/Total economic life
X total cost = depreciation

OR
Total Economic life = 100%
Effective Age = x

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18
Q

Assume a 20-year-old building that would cost $220,000 to build new today. It is in good condition and you estimate the effective age to be 15 years. Your estimate of total economic life is 60 years, based on the experience of similar neighborhoods. How much is the depreciation in % and $?

A

Age-Life-Method:

15
/60
x $220,000
= $55,000 depreciation

60 = 100%
15 = x
25% depreciation

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19
Q

The cost new of the improvements is $245,200, the land value is $75,000 and the effective age is 17. The total economic life expectancy as indicated by comparable sales is 55 years.
What is the indicated value by the cost approach?

A

Age-Life Method
17 / 55 = .309 or 30.9%
$245,200 x .309 = $ 75,767
$245,200 Total cost of the improvements
- 75,767 Less total depreciation
= $169,433 Depreciated cost

 + 75,000 Plus land value
= $244,433 Indicated value by the cost approach ($244,000 rounded)

20
Q

The cost new of the subject property is $235,000, the land value is $70,000 and the effective age is 25. The total economic life is estimated to be 60 years.
The subject has deferred maintenance items that total $4,500. After these repairs, the effective age will be reduced to 20 years. What is the value using the cost approach?

A

Modified Age Life Method:
$235,000 - $4,500 = $230,500 remaining improvement cost (after deferred maintenance)
20 (new effective age) / 60 (total economic life) = .333 (percent of depreciation)

$235,000 Total cost new

- 4,500 Minus deferred maintenance
= $230,500 Remaining improvement cost

- 76,065 Minus depreciation ($230,500 x .33)
= $154,435 Depreciated cost of Improvements
+ $70,000 Plus land value

= $224,435 Indicated value by cost approach
Rounded to $225,000

21
Q

What is the total amount of short-lived depreciation?

A

$6,124

22
Q

Assume a replacement cost of $200,000. Let’s say the short-lived items cost $40,000 and there is no deferred maintenance. If the actual age of the building is 15 years and we estimate the total economic life to be 60 years, how much is depreciation for long-lived items?

A

We subtract $40,000 from $200,000 and get $160,000. This $160,000 figure represents, by default, the cost of the long-lived items.
15/60 or 25%.
.25 X $160,000, or $40,000.

23
Q

The house you are appraising has a total cost of $182,844. It is 22 years old (which is its effective age) and it has an estimated remaining economic life of 28 years.
The cost to cure deferred maintenance items is $1,290.
Short-lived items of depreciation have been identified totaling $3,530. The cost new of these short-lived items is $9,500.
What is the amount of long-lived depreciation attributable to the building?
What is the total physical depreciation attributable to the building?

A

$182,844 Total cost new

- 1,290 Minus deferred maintenance
- 9,500 Minus short-lived items cost
= $172,054 Cost of long-lived items
Total life = 22 + 28 = 50 years
Depreciation = 22 / 50 = .44 or 44%

Long-lived depreciation = $172,054 x .44 = $75,704 (rounded to nearest $1)

$75,704 Long-lived depreciation

+ 1,290 Deferred maintenance cost to cure
+ 3,530 Short-lived depreciation
= $80,524 Total physical depreciation

24
Q

The subject is a building with three apartments without air conditioning, in a market where air conditioning is standard and expected by purchasers. If it had been installed when the building was built, it would have cost $4,000 but the cost of retrofitting now is $4,900.
Installing air conditioning would allow the owner to raise rents and increase his gross monthly rent by a total of $60. The current gross rent multiplier (GRM) is 95. Therefore the increase in property value would be $60 x 95 or $5,700.
The functional obsolescence is curable as the value increase ($5,700) is greater than the cost to cure ($4,900).
How much is the functional obsolescence?

A

Functional Obsolescence Calculation:
* Cost of existing item 0 

* Less depreciation previously charged - 0 

* Plus cost to cure (all costs) OR 

* value of the loss +5,700 

* Less cost if installed new - 4,000 

* Equals depreciation for functional obsolescence $1,700 
Steps 1 and 2 are zero because the item does not currently exist in the property. 
In Step 3, the loss in value is $5,700. The property would be worth that much more if it had air conditioning.

25
Q

Assume that if the building has air conditioning but it is inadequate and breaks down frequently. The reproduction cost now is $3,000 and it is 30% depreciated. 
It will cost $2,000 to remove the old system but there is a salvage value of $1,000. It will still cost $4,900 to retrofit the new system and it would have cost $4,000 when the building was new. 
It will still add $5,700 to the value of the property. However, the depreciation is now incurable as it costs more to cure the problem ($5,900) than the value it will add.

A
  • Cost of existing item $ 3,000 

  • Less depreciation previously charged - 900 

  • Plus cost to cure (all costs) OR 

  • value of the loss + 5,900 

  • Less cost if installed new - 4,000 

  • Equals depreciation for functional obsolescence $ 4,000
    
In Step 2, the depreciation was calculated as .30 x $3,000, or $900. We already took 30% physical depreciation on this item, so we have to make sure that we don’t double-depreciate. 

    In Step 3, the cost to cure will be $5,900 total. This includes $4,900 to retrofit the installation plus $2,000 to remove the old system, minus $1,000 that can be reclaimed as salvage value of the old equipment.
26
Q

Your subject property is a house adjoining a busy interstate highway. A vacant lot next door recently sold for $60,000 and a similar lot with the same zoning, in a quiet location, recently sold for $65,000.
An improved property along the interstate in the next block recently sold for $340,000 and a similar improved property away from the interstate sold for $360,000.
What is the external obsolescence attributable to the building?

A

There is a premium of $5,000 for less traffic noise.
Here we can see the total property enjoyed a bonus of $20,000 for the location away from traffic noise.
The external obsolescence attributable to the whole property is $20,000 ($360,000 - $340,000).
Pairing the two land sales indicates a loss of $5,000 for the land portion. Therefore, the external obsolescence attributable to the building is $15,000 ($20,000 - $5,000).

27
Q

We have a property that sold for $320,000, of which $60,000 was the land value.
Cost new of the building $340,000
Contributory value of the building $260,000
Physical deterioration $30,700
Functional Obsolescence $$30,000
How much is the External Obsolescence?

A

Allocation of Market-Based Depreciation:
$340,000
Minus $260,000
= $80,000 Total Depreciation
Minus $30,700
Minus $30,000
= $19,300 External Obsolescence

28
Q

Assume a house that would rent for $1,200 per month in a normal market. In this market that has experienced overbuilding, it can only bring $1,150 per month. The appropriate GRM is 120. How much is the External Obsolescence?

A

$50 (rent loss) x 120 = $6,000 (external obsolescence)

29
Q

Calculate the absorption rate for the market area for the previous 12 month period. There have been 18 sales in the past year in this neighborhood. There are currently 6 houses offered for sale in the market area. What is the absorption period required to sell these houses?

A

18 sales = 12 months
6 sales = x
4 months

30
Q

If there were 60 sales during a 6 month period, the absorption rate is x.
Calculate the months of housing supply if there are 240 active listings.

A

10 sales per month (60/6).

There is a 24-month supply of homes on the market (240 active sales/10 sales per month).

31
Q

Number of months in the calculating period (12) 
Number of closed sales during the calculating period (10) 
Number of offerings that did not sell or listings that expired during the calculation period (4)
What is expiration ratio?

A

14 total = 100%
4 expired listings = x

29%

32
Q

Using researched historical data for the past 3 years, the appraiser found 36 comparable closed property sales. During the past three years there were a total of 14 expired offerings that did not sell in the neighborhood. Calculate the expiration ratio in the neighborhood for the past three years.

Looking at data for the past 12 months, the appraiser found a total of 18 comparable closed sales during this most recent time period. There were 6 expired offerings noted during the same time period. Does the expiration ratio for the most recent 12 month period appear to be normal compared to the three year analysis?

A

50 = 100%
14 = x
28% expiration ratio

24 = 100%
6 = 25% expiration ratio

The 12 month expiration rate appears to be normal when compared to the 36 month expiration rate for this market area.

33
Q

Example to Supporting Effective Age of the Subject property
Use several comparable sales from the same neighborhood. Assumptions are that all sales and the subject property have had similar upkeep, with no major upgrading, remodeling, rehabilitation, or additions.
Sale Price of the Comparable Sale: $100,000.
Land Value: $ 20,000.
Cost New of the Improvements: $130,000.
Assume that the comparable sale is 24 years old.
Let’s further assume that two other comparable sales indicate average annual depreciation rates of 1.4% and 1.5%.
The subject property is 26 years old.
If the subject has a Total Economic Life of 70 years, what is its Effective Age?

A

Sale Price of the Comparable Sale: $100,000.
Less: Land Value: $ 20,000.
Present Value of the Improvements: $ 80,000.
Cost New of the Improvements: $130,000.
Less: Present Value of the Improvements: $80,000.
Total accrued Depreciation of Improvements: $50,000.
Total Depreciation of Improvements / Divided by Cost New of Improvements:
$50,000 / $130,000 = .3846
This is the total percentage of loss from cost new: .3846 or 38.5%
Assume that the comparable sale is 24 years old. That means it has depreciated 38.5% over 24 years – or an average of 1.6% per year (.385 / 24 = 0.016).
Let’s further assume that two other comparable sales indicate average annual depreciation rates of 1.4% and 1.5%.
If we reconcile the three indications to be 1.5% depreciation a year; then we can use that to estimate the Effective Age of the subject property.
The subject property is 26 years old. 26 X 0.015 = .39 or 39% depreciated.
If the subject has a Total Economic Life of 70 years, then its Effective Age could be calculated as 70 X .39 = 27.3; rounded to 27 years.

34
Q

How is Total Economic Life of the Subject Property calculated?

A

As with effective age, use several comparable sales from the same neighborhood.
Sale Price of the Comparable
Less: Land Value
Equals: Contributing Value of the Improvement
Gross Living Area of the Comparable is multiplied by the appropriate cost per square foot for the type of construction to obtain:
Replacement Cost New of the Improvement.
Replacement Cost New minus the Contributing Value of the Improvement
Equals: Total Accrued Depreciation from Cost New.
Total Accrued Depreciation divided by the Replacement Cost New equals:
Percent (%) of total Accrued Depreciation.
Percent of Total Accrued Depreciation (%) divided by the Chronological Age of the Comparable Sale equals:
Rate of Depreciation per year.
1.0(this is 100 percent depreciation) divided by the Rate of Depreciation per year equals:
Total Economic Life
The results of these calculations can support the appraiser’s conclusion for the subject property’s Total Economic Life.

35
Q

22,740 Number of people employed in the community
12,633 Number of core industries employees
For each new core employee hired, how many additional jobs will be created?

A

22,740 divided by 12,633 will indicate a base core multiplier of 1.8. This economic base multiplier indicates that for each new core employee hired, there will be an additional .8 jobs created in the community.

12,633 = 100%
22,740 = x
x = 180%

36
Q

The cost to install the air conditioning at the time of construction would have been $4,000. The cost to install the air conditioning now is $7,000. How much is the functional obsolescence? Is it curable if the value will increase by $8,000?

A

$7,000
Minus $4,000
= $3,000 functional obsolescence

The value needs to increase by more than $7,000 to be curable. In this case, it is curable.

37
Q

The blueprints of a building indicate the measurement of one wall to be 43’9” and the intersection wall to be 29’3”. What is the indicated square footage of the building?

A

1,280 SF

need to convert inches to the tenth of a foot
1’ = 12”
x = 9”
x = .7’
.25’ = 3”

43.75’ x 29.25’

38
Q

The dining room measures 9’ x 12’. The living room measures 9’ x 15’. Carpet costs $25/sq yd. installed. What is the cost of installing carpet assuming no waste and that both rooms are carpeted?

A

$675

108 SF + 135 SF =243 SF

1 sq yard = 9 SF
x = 243 SF
x = 27 sq yards

$25 = 1 sq yard
x = 27 sq yard
x = $675

39
Q

Jenny is performing an appraisal of a single‐family residence for mortgage lending. The subject residence is 50’ x 30’ and consists of 2 stories over a full, unfinished basement. The replacement cost new of the residence (including basement) is $175 per square foot. There is an attached 20’ x 22’ garage and a concrete patio. The actual age of the structure is 20 years. The effective age is 15 years. The remaining economic life is 60 years. The patio has a depreciated cost of $500. The garage has a depreciated cost of $7,040. The land to building value ratio is 1:3. What is the total value of the subject property (rounded to the nearest thousand).

A

Exam Crammer
A:
$570,000

50 x 30 x 2 = 3,000 SF
X $175/SF
= RCN $525,000

75 years total economic life = 100%
15 years effective age = x
20% depreciation

$525,000 x 20% = $105,000 depreciation

$525,000 RCN
Minus $105,000 depreciation
= $420,000 depreciated cost of improvements

$420,000
Plus $7,040
Plus $500
$427,540 = 75% (building value)
x = 100 % total value
x = $570,053

40
Q

A 15‐year‐old office property recently sold for $1,400,000. The cost to replace the building is $1,625,000. Land in that office complex sells for $30,000 per acre and this sale had 10 acres. While the paving and landscaping cost $80,000, they contributed $50,000 to the sale. What is the annual depreciation rate of the building?

A

Exam Crammer
2.3%

$1,400,000
Minus $300,000
Minus $50,000
= $1,050,000

CN $1,625,000 = 100%
Minus $1,050,000
= $575,000 = x

35.38%
/15
= 2.3

41
Q

For the purpose of estimating depreciation via the extraction method, you have utilized Comparable Sale #1, which is a 2,400 sq. ft. home with a two‐car, 440 sq. ft. attached garage, having a replacement cost of $85/sq. ft. for gross living area and $18/sq. ft. for the garage area. This property sold for $240,000 and has a chronological age of 10 years. Through your research, the indicated land value is $50,000 and the site improvements have a replacement cost of $25,000. What is the percentage total amount of accrued depreciation indicated by this sale?

A

Exam Crammer:
A: 22.14% depreciation

$204,000 building (2,400 x 85)
$7,920 garage (440 x 18)
$211,920 RCN

$240,000 Sale
Minus 25,000 site improvements
Minus $50,000 land value
= $165,000 depreciated cost of the improvements

$211,920 RCN = 100%
$165,000 = 77.68%
22.14% depreciation

42
Q

The subject property is a five-year-old house that contains 5,200 sq. ft. of livable area, located on 1.8 acres, with a two-car garage with 650 sq. ft., and a granite swimming pool that measures 85 perimeter feet. The costs are estimated to be as follows: House, $85 per sq. ft.; garage, $25 per sq. ft.; indicated land value is $0.35 per sq. ft.; and the swimming pool is estimated to cost $275 per perimeter foot. The cost multiplier for the area is 0.9678. The chronological age is 5 years, and the effective age is 2 years. The remaining economic life is 63 years. If the total cost of improvements is $466,117, what is the indicated value via the cost approach?

A

$479,576

The effective age is 2 years and the remaining economic life is 63 years for a total life of 65 years. The total cost new of the improvements equals $466,117. Depreciation is 3%, calculated as (2 / 65 = 3%). Multiplying cost new of $466,117 x 3% (0.03) = $13,984 which is total depreciation. $466,117 - $13,984 = $452,133 depreciated cost of improvements. $452,133 + $27,443 (land value) = $479,576.

43
Q

You have been asked by ABC Mortgage to do an appraisal of a single-family residence that has a substantial amount of deferred maintenance. The value of the property upon completion of the items of deferred maintenance is $62,000, of which $10,000 is land value. The cost of curing the deferred maintenance is estimated to be $25,000. Rounded to the nearest thousand, what is the highest dollar amount that can be paid if the market warrants a 10% return for completing the work?

A

$31,000

$62,000 divided by 1.10 (100 percent plus 10 percent) equals $56,363. This, less the $25,000 in repairs, equals $31,363 (rounded to $31,000).

$62,000 = 110% value with repairs
X = 100% value without repairs
$56,363 value without repairs
Minus $25,000 repair cost
= $31,363

44
Q

Calculate the remaining economic life of a building using the following information: Total market value: $160,000; Site value: $30,000; Actual Age: 20 years; Effective age: 10 years; Replacement cost new: $170,000.

A

75

First, figure the market value of just the building: $160,000 - $30,000 (site) = $130,000. Next, subtract the market value of the building from the replacement cost new: $170,000 - $130,000 = $40,000. Then, calculate the depreciation per year: $40,000 / 20 years (actual age) = $2,000 depreciation per year. $170,000 / $2,000 = 85 years (economic life). 85 years (economic life) - 10 years (effective age) = 75 years remaining economic life.

$170,000 = TEL
$2,000 = 1 year

45
Q

You have been asked by ABC Mortgage to do an appraisal of a single-family residence that has a substantial amount of deferred maintenance. The value of the property upon completion of the items of deferred maintenance is $62,000, of which $10,000 is land value. The cost of curing the deferred maintenance is estimated to be $25,000. Rounded to the nearest thousand, what is the highest dollar amount that can be paid if the market warrants a 10% return for completing the work?

A

$31,000

$62,000 divided by 1.10 (100 percent plus 10 percent) equals $56,363. This, less the $25,000 in repairs, equals $31,363 (rounded to $31,000).

$62,000 = 110%
$x. = 100%
$56,362
Minus $25,000 already paid/invested
= $31,363 remaining amount that can be max paid