1. Washington Real Estate Fundamentals
Lesson 18:
Real Estate Math
© 2011 Rockwell Publishing

2. Solving Math Problems Four basic steps
1. Read the question.
What answer are you being asked to find?
2. Write down the formula.
3. Substitute.
Replace elements in formula with relevant numbers from problem.
4. Calculate the answer.
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3. Solving Math Problems Using formulas
Each of these choices expresses the same formula, but in a way that lets you solve it for A, B, or C:
A = B × C
B = A ÷ C
C = A ÷ B
Choose the one that isolates the unknown.
© 2011 Rockwell Publishing

4. Solving Math Problems Isolating the unknown
The unknown is the element that you’re trying to determine.
The unknown should always sit alone on one side of the equals sign.
All the information that you already know should be on the other side.
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5. Solving Math Problems Example
If a lot has an area of 7,000 square feet and is 100 feet long, what is its width?
Formula needed: Area = Length × Width
Substitute numbers from problem: 7,000 sq. ft. = 100 ft. × Width
Width is the unknown, so isolate Width by switching formula to Width = Area ÷ Length.
Calculate: 7,000 sq. ft. ÷ 100 ft. = 70 ft.
Width = 70 feet, so the lot is 70 feet wide.
© 2011 Rockwell Publishing

6. Decimal Numbers Converting fraction to decimal
Calculators use only decimals, not fractions.
If problem contains a fraction, convert it to a decimal:
Divide top number (numerator) by bottom number (denominator).
1/4 = 1 ÷ 4 = 0.25
1/3 = 1 ÷ 3 = 0.333
5/8 = 5 ÷ 8 = 0.625
© 2011 Rockwell Publishing

7. Decimal Numbers Converting percentage to decimal
If problem contains a percentage, may be necessary to:
convert percentage to decimal
do calculations to find answer
then convert decimal back to percentage
(Calculator’s percent key converts percentage to decimal automatically.)
© 2011 Rockwell Publishing

8. Decimal Numbers Converting percentage to decimal
To convert a percentage to a decimal:
Remove percent sign.
Move decimal point two places to the left. (May be necessary to add a zero.)
2% becomes .02
17.5% becomes .175
80% becomes .80
123% becomes 1.23
© 2011 Rockwell Publishing

9. Decimal Numbers Converting decimal to percentage
To convert a decimal to a percentage, reverse the process:
Move decimal point two places to the right.
Add percent sign.
.02 = 2%
.175 = 17.5%
.80 = 80%
1.23 = 123%
© 2011 Rockwell Publishing

10. Summary Solving Math Problems
Four steps: read problem, write formula, substitute, calculate
Using formulas
Isolating the unknown
Converting fraction to decimal number
Converting percentage to decimal number
Converting decimal number to percentage
© 2011 Rockwell Publishing

11. Area Problems
Real estate agent may need to calculate area of lot, building, or room.
Area usually stated in square feet or square yards.
Formula needed depends on shape.
Area problem may also involve calculating other elements, such as cost per square foot or rental rate.
© 2011 Rockwell Publishing

12. Area Problems Rectangle formula: A = L × W
Formula for calculating area of square or rectangular space:
Area = Length × Width
A = L × W
Length
Area
Width

13. Rectangles Example
An office is 27 feet wide by 40 feet long. It rents for $2 per square foot per month. How much is the monthly rent?
Step 1: Calculate area
A = 27 ft. × 40 ft.
A = 1,080 sq. ft.
Step 2: Calculate rent
Rent = 1,080 sq. ft. × $2 per sq. ft.
Rent = $2,160 per month
© 2011 Rockwell Publishing

14. Rectangles Square yards
Some problems express area in square yards rather than square feet.
Remember: 1 square yard = 9 square feet
1 yard = 3 feet
1 square yard is 3 feet on each side
3 feet × 3 feet = 9 square feet
Divide square footage by 9 to find number of square yards.
© 2011 Rockwell Publishing

15. Area Problems Triangle formula: A = ½ B × H
Formula for calculating area of triangle:
Area = ½ Base × Height
A = ½ B × H

16. Triangles Order of operations
Note that in applying triangle formula, order of operations doesn’t matter.
These will all reach same result:
A = ½ B × H
A = B × ½ H
A = (B × H) ÷ 2
© 2011 Rockwell Publishing

17. Triangles Example
A triangular lot is 140 feet long and 50 feet wide at its base. What is the area?
A = (½ × 50 ft.) × 140 ft.
A = 25 ft. × 140 ft.
A = 3,500 sq. ft.

18. Triangles Example, continued
A triangular lot is 140 feet long and 50 feet wide at its base. What is the area?
A = (½ × 50 ft.) × 140 ft.
A = 25 ft. × 140 ft.
A = 3,500 sq. ft.
Same result from this:
A = 50 ft. × (½ × 140 ft.)
A = 50 ft. × 70 ft.
A = 3,500 sq. ft.
And from this:
A = (50 ft. × 140 ft.) ÷ 2
A = 7,000 sq. ft. ÷ 2

19. Area Problems Odd shapes
To find area of an irregular shape:
Divide figure up into squares, rectangles, and triangles.
Find area of each of the shapes that make up the figure.
Add the areas together.
© 2011 Rockwell Publishing

20. Odd Shapes Example
The lot’s western side is 60 feet long. Its northern side is 100 feet long, but its southern side is 120 feet long.
To find the area of this lot, break it into a rectangle and a triangle.

21. Odd Shapes Example, continued
Area of rectangle:
A = 60 ft. × 100 ft.
A = 6,000 sq. feet

22. Odd Shapes Example, continued
To find length of triangle’s base, subtract length of northern boundary from length of southern boundary.
120 ft. – 100 ft. = 20 ft.
Area of triangle:
A = (½ × 20 ft.) × 60 ft.
A = 600 sq. ft.

23. Odd Shapes Example, continued
Total area of odd shape:
6,000 sq. ft. (rectangle)
+ 600 sq. ft. (triangle)
6,600 sq. ft.

24. Odd Shapes Avoid counting same section twice
Common mistake when working with odd shapes:
Calculating area of part of the figure twice.
Can happen with a figure like this one.

25. Odd Shapes Example Wrong way to calculate area of this lot:
25 × 50 = 1,250 sq. ft.
40 × 20 = 800 sq. ft.
1, = 2,050 sq. ft.
This counts middle of shape (green area) twice.

26. Odd Shapes Example, continued
Avoid the mistake by breaking the shape down like this instead.
Find height of smaller rectangle by subtracting height of top rectangle (25 ft.) from height of whole shape (40 ft.).
40 – 25 = 15 ft.

27. Odd Shapes Example, continued
Now calculate area of each rectangle and add them together:
25 × 50 = 1,250 sq. ft.
20 × 15 = 300 sq. ft.
1, = 1,550 sq. ft.

28. Odd Shapes Example, continued
Another way to break the odd shape down into rectangles correctly:
To find width of rectangle on right, subtract width of left rectangle from width of whole shape:
50 – 20 = 30 feet

29. Odd Shapes Example, continued
Now calculate area of each rectangle and add them together:
40 × 20 = 800 sq. ft.
30 × 25 = 750 sq. ft.
= 1,550 sq. ft.

30. Odd Shapes Narrative problems
Some odd shape problems are expressed only in narrative form, without a diagram.
Useful to draw the shape yourself, then break it down into rectangles and triangles.
© 2011 Rockwell Publishing

31. Odd Shapes Example
A lot’s boundary begins at a certain point and runs due south for 319 feet, then east for 426 feet, then north for 47 feet, and then back to the point of beginning.
To solve this problem, first draw the shape.

32. Odd Shapes Example, continued
Break it down into a rectangle and a triangle as shown.
Subtract 47 from to find height of triangular portion:
319 – 47 = 272 ft.

33. Odd Shapes Example, continued
Calculate area of rectangle:
426 × 47 = 20,022 sq. ft.

34. Odd Shapes Example, continued
Calculate area of triangle:
(½ × 426) × 272 = 57,936 sq. ft.

35. Odd Shapes Example, continued
Add together area of rectangle and area of triangle to find the lot’s total square footage:
20, ,936 = 77,958

36. Volume Problems Area: Measurement of a two-dimensional space.
Volume: Measurement of a three-dimensional space.
Width, length, and height.
Cubic feet or cubic yards instead of square feet or square yards.
© 2011 Rockwell Publishing

37. Volume Problems Formula: V = L × W × H
To calculate volume, use this formula:
Volume = Length × Width × Height
V = L × W × H
© 2011 Rockwell Publishing

38. Volume Problems Cubic yards
If problem asks for cubic yards:
1 cubic yard = 27 cubic feet
1 cubic yard measures 3 feet on each dimension
3 feet × 3 feet × 3 feet = 27 cubic feet
Divide cubic footage by 27 to find number of cubic yards.
© 2011 Rockwell Publishing

39. Volume Problems Example
A trailer is 40 feet long, 9 feet wide, and 7 feet high. How many cubic yards does it contain?
40 × 9 × 7 = 2,520 cubic feet
2,520 ÷ 27 = cubic yards
© 2011 Rockwell Publishing

40. Summary Area and Volume Problems
Area of square or rectangle: A = L × W
1 square yard = 9 square feet
Area of triangle: A = ½ B × H
Divide odd shapes into squares, rectangles, and triangles.
Volume: V = L × W × H
1 cubic yard = 27 cubic feet
© 2011 Rockwell Publishing

41. Percentage Problems
Many math problems ask you to find a certain percentage of another number.
Example: What is 85% of $150,000?
This means you need to multiply that other number by the percentage:
$150,000 × .85 = $127,500
(Note conversion of percentage to decimal, explained earlier.)
© 2011 Rockwell Publishing

42. Percentage Problems Formula: P = W × %
Basic formula for solving percentage problems:
Part = Whole × Percentage
P = W × %
© 2011 Rockwell Publishing

43. Percentage Problems Formula: P = W × %
“Whole” is larger figure.
Example: property’s sales price
“Part” is smaller figure.
Example: amount of commission owed
A percentage of the whole equals the part.
© 2011 Rockwell Publishing

44. Percentage Problems Isolating the unknown
Use basic formula (P = W × %) if part is the unknown.
If whole is the unknown: W = P ÷ %
If percentage is the unknown: % = P ÷ W
© 2011 Rockwell Publishing

45. Percentage Problems Multiply or divide?
Knowing whether to divide or multiply can be hardest part of solving percentage problems.
If missing element is the part (smaller number), it’s a multiplication problem.
If missing element is either the whole (larger number) or the percentage, it’s a division problem.
© 2011 Rockwell Publishing

46. Percentage Problems Types of problems
Many real estate math problems are basically percentage problems:
commission problems
loan problems (interest and principal)
profit or loss problems
capitalization problems
Percentage sometimes referred to as “rate.”
Examples: 7% commission rate, 5% interest rate, 10% rate of return
© 2011 Rockwell Publishing

47. Commission Problems For commission problems, use P = W × %.
In commission problems:
Percentage (%) is commission rate.
Whole (W) is amount commission based on.
Usually sales price of house.
Part (P) is amount of commission.
© 2011 Rockwell Publishing

48. Commission Problems Example
A home sells for $300,000. Listing firm is paid a 6% commission on the sales price. Listing broker is entitled to 60% of the commission. How much is the broker’s share?
Step 1: Calculate total commission (6%)
P = W × %
P = $300,000 × .06
P = $18,000
Step 2: Calculate broker’s share (60%)
P = $18,000 × .60
P = $10,800
© 2011 Rockwell Publishing

49. Commission Problems Two-tiered commission
Some problems involve two-tiered commission:
one rate on sales price up to certain amount
different rate on amount by which price exceeds that amount
© 2011 Rockwell Publishing

50. Two-tiered Commission Example
Listing firm’s commission is 7% on first $300,000 of home’s sales price, 5% on any amount over $300,000. If commission is $23,500, how much was the sales price?
Step 1: Calculate commission on first $300,000
P = W × %
P = $300,000 × .07
P = $21,000
Step 2: Subtract to find commission on rest of price
$23,500 – $21,000 = $2,500
© 2011 Rockwell Publishing

51. Two-tiered Commission Example, continued
Step 3: Calculate amount of price over $300,000
P = W × %
$2,500 = W × .05
Isolate unknown: W = $2,500 ÷ .05
W = $50,000
Step 4: Add to find total sales price
$300,000 + $50,000 = $350,000
© 2011 Rockwell Publishing

52. Summary Percentage Problems
Percentage formula: P = W × %
W = P ÷ %
% = P ÷ W
Types of percentage problems: commissions, loans, profit or loss, capitalization
Commission problems:
% is commission rate
W is usually sales price of home
P is commission amount
© 2011 Rockwell Publishing

53. Loan Problems Loan problems include: interest problems
principal balance problems
For loan problems, again use P = W × %.
Now P is amount of interest.
W is loan amount or principal balance.
% is interest rate, expressed as an annual rate.
© 2011 Rockwell Publishing

54. Loan Problems Interest problems
Some problems state interest amount (the part, P) in semiannual, quarterly, or monthly installments.
Semiannual: Twice a year (two payments per year).
Quarterly: Four times a year (four payments per year).
First step in problems like these: convert interest payments into annual figure.
© 2011 Rockwell Publishing

55. Interest Problems Example: Finding annual interest
A lender makes an interest-only loan of $140,000. The interest rate is 6.5%. How much does the annual interest come to?
P = W × %
P = $140,000 × .065
P = $9,100
© 2011 Rockwell Publishing

56. Interest Problems Example: Finding interest rate
The interest portion of a loan’s monthly payment is $ The loan balance is $92,000. What is the interest rate?
Step 1: Convert monthly to annual figure
$ × 12 = $6,210 annual interest
Step 2: Calculate interest rate
P = W × %
Substitute: $6,210 = $92,000 × %
Isolate the unknown: % = $6,210 ÷ $92,000
% = .0675, or 6.75%
© 2011 Rockwell Publishing

57. Loan Problems Principal balance problems
Some problems give interest amount and interest rate and ask you to find loan amount or principal balance.
Once again, use basic percentage formula, P = W × %
© 2011 Rockwell Publishing

58. Principal Balance Problems Example: Finding loan amount
A real estate loan calls for semiannual interest-only payments of $3,250. The interest rate is 9%. What is the loan amount?
Step 1: Convert semiannual to annual figure
$3,250 × 2 = $6,500 annual interest
Step 2: Calculate loan amount
P = W × %
Substitute: $6,500 = W × .09
Isolate the unknown: W = $6,500 ÷ .09
W = $72,222.22
© 2011 Rockwell Publishing

59. Principal Balance Problems Example: Finding current balance
The interest portion of a loan’s tenth monthly payment is $ The interest rate is 7%. What is the loan balance prior to this payment?
Step 1: Convert monthly to annual figure
$ × 12 = $3, (annual interest)
Step 2: Calculate principal balance
P = W × %
Substitute: $3,080 = W × .07
Isolate the unknown: W = $3,080 ÷ .07
W = $44,000
© 2011 Rockwell Publishing

60. Principal Balance Problems Principal and interest payments
Some problems give the monthly principal and interest payment (instead of just the interest portion of monthly payment) and ask how much payment reduces balance.
These require an additional step: subtracting interest portion from payment amount.
© 2011 Rockwell Publishing

61. Principal Balance Problems Example: Finding principal reduction
A loan’s balance is $96,000. The interest rate is 8%. The monthly principal and interest payment is $ How much will this payment reduce the loan balance?
Step 1: Calculate annual interest
P = W × %
$96,000 × .08 = $7,680 annual interest
Step 2: Divide to find interest portion of payment
$7,680 ÷ 12 = $640
Step 3: Subtract to find principal portion
$ – $ = $64.41
© 2011 Rockwell Publishing

62. Principal Balance Problems Multiple P&I payments
If question asks how much loan balance will be after second, third, or sixth principal and interest payment:
calculate first payment’s effect on principal balance
repeat steps again as many times as necessary, using new balance each time
© 2011 Rockwell Publishing

63. Principal Balance Problems Example: Multiple P&I payments
Continuing with previous example ($96,000 8%, $ P&I pyt, first pyt reduces balance by $64.41):
$96,000 – $64.41 = $95, (balance after Pyt 1)
$95, × .08 = $7, (annual interest)
$7, ÷ 12 = $ (interest portion of Pyt 2)
$ – $ = $64.84 (principal portion of Pyt 2)
$95, – $64.84 = $95, (balance after Pyt 2)
So second payment reduces loan balance to $95,
For balance after third payment, repeat these steps.
© 2011 Rockwell Publishing

64. Summary Loan Problems Use percentage formula: P = W × %
P is interest amount
% is interest rate
W is loan amount or principal balance
Convert semiannual, quarterly, or monthly interest into annual interest.
© 2011 Rockwell Publishing

65. Profit or Loss Problems
Another common type of percentage problem: calculating property owner’s profit or loss.
Use this variation on percentage formula:
Now = Then × %
“Then” is property’s value at earlier point.
“Now” is property’s value at later point.
Percentage refers to rate of profit (appreciation) or loss (depreciation).
© 2011 Rockwell Publishing

66. Profit or Loss Problems Then and Now formula
Basic “Then and Now” formula:
Now = Then × %
To isolate the unknown, change it to:
Then = Now ÷ % or
% = Now ÷ Then
© 2011 Rockwell Publishing

67. Profit or Loss Problems Percentage to use in formula
Key to solving these problems: finding correct percentage to use in formula.
Percentage is 100% plus the percentage of profit, or minus the percentage of loss.
If no profit or loss, Now value is 100% of Then value.
If profit, Now value is greater than 100% of Then value.
If loss, Now value less than 100% of Then.
© 2011 Rockwell Publishing

68. Percentage to Use in Formula Property sold at a loss
For loss, subtract percentage of loss from 100% to find percentage to use in formula.
Example: 30% loss
House didn’t sell for 30% of original value.
It sold for 30% less than original value.
100% – 30% = 70%
So Now value is 70% of Then value.
Use 70% (.70) in formula.
© 2011 Rockwell Publishing

69. Percentage to Use in Formula Example: Property sold at a loss
A seller sells her house for $420,000, which represents a 30% loss. How much did she originally pay for it?
Step 1: Determine percentage to use in formula
100% – 30% = 70%, or .70
Step 2: Calculate Then value
Now = Then × %
Substitute: $420,000 = Then × .70
Isolate the unknown: Then = $420,000 ÷ .70
Then = $600,000
© 2011 Rockwell Publishing

70. Percentage to Use in Formula Property sold for a profit
If property sold for a profit, add percentage of profit to 100% to find percentage to use in formula.
Example: 30% profit
Means house sold for 130% of what seller originally paid.
100% + 30% = 130%
So Now value is 130% of Then value.
Use 130% (1.30) in formula.
© 2011 Rockwell Publishing

71. Percentage to Use in Formula Property sold for a profit
Note that if seller sells house for 130% of what she paid for it, that doesn’t means she has made a 130% profit.
She received 100% of what she paid, plus 30%.
She made a 30% profit.
© 2011 Rockwell Publishing

72. Percentage to Use in Formula Example: Property sold for a profit
A seller sells her house for $420,000, which represents a 30% profit. How much did she originally pay for it?
Step 1: Determine percentage to use in formula
100% + 30% = 130%, or 1.30
Step 2: Calculate Then value
Now = Then × %
Substitute: $420,000 = Then × 1.30
Isolate the unknown: Then = $420,000 ÷ 1.30
Then = $323,077
© 2011 Rockwell Publishing

73. Profit or Loss Problems Calculating selling price
Examples so far gave selling price and profit or loss percentage, asked how much seller originally paid for property (the Then value).
Other problems will:
give original price and percentage of profit or loss
ask how much property is being sold for (the Now value)
© 2011 Rockwell Publishing

74. Profit or Loss Problems Example: Calculating selling price
A seller bought his house four years ago for $200,000 and sold it this year for 5% more than he paid for it. How much did he sell it for?
Step 1: Determine percentage to use in formula
100% + 5% = 105%, or 1.05
Step 2: Calculate Now value
Now = Then × %
Substitute: Now = $200,000 × 1.05
Now = $210,000
© 2011 Rockwell Publishing

75. Profit or Loss Problems Calculating profit or loss percentage
Other problems give original price and selling price, ask for percentage of profit or loss.
© 2011 Rockwell Publishing

76. Profit or Loss Problems Example: Calculating profit
A seller bought his house four years ago for $200,000 and sold it this year for $210,000. What was his percentage of profit on the sale?
Now = Then × %
Substitute: $210,000 = $200,000 × %
Isolate the unknown: % = $210,000 ÷ $200,000
% = 1.05, or 105%
105% – 100% = 5%
The seller made a 5% profit on the sale.
© 2011 Rockwell Publishing

77. Profit or Loss Problems Example: Calculating loss
Calculating a loss works the same way:
A seller bought his house seven years ago for $184,000 and sold it this year for $160,000. What was his percentage of loss on the sale?
Now = Then × %
Substitute: $160,000 = $184,000 × %
Isolate the unknown: % = $160,000 ÷ $184,000
% = .869, or 87%
100% – 87% = 13%
The seller took a 13% loss on the sale.
© 2011 Rockwell Publishing

78. Profit or Loss Problems Appreciation or depreciation
Profit or loss problem may also be expressed in terms of appreciation or depreciation.
Problem solved same way as ordinary profit or loss problem.
If problem asks how much property appreciated or depreciated per year over several years, repeat same calculation for each year.
© 2011 Rockwell Publishing

79. Profit or Loss Problems Example: Compound depreciation
A property is currently worth $420,000. It has depreciated four and a half percent per year for the past five years. What was the property worth five years ago?
Step 1: Find percentage to use in formula
100% – 4.5% = 95.5%, or .955
Step 2: Calculate value one year ago
Now = Then × %
Substitute: $420,000 = Then × .955
Isolate the unknown: Then = $420,000 ÷ .955
Then = $439,791 (value 1 year ago)
© 2011 Rockwell Publishing

80. Profit or Loss Problems Example, continued
Now that you know how much the house was worth one year ago, calculate the Then value four more times, to find how much the house was worth five years ago:
$439,791 ÷ .955 = $460,514 (value 2 years ago)
$460,514 ÷ .955 = $482,214 (value 3 years ago)
$482,214 ÷ .955 = $504,936 (value 4 years ago)
$504,936 ÷ .955 = $528,729 (value 5 years ago)
The house was worth around $528,729 five years ago.
© 2011 Rockwell Publishing

81. Profit or Loss Problems Compound appreciation
If you’re told that a property gained value at a particular rate over several years, you’ll use the same process.
The difference is that you’ll need to add the rate of change to 100%, instead of subtracting it from 100%.
© 2011 Rockwell Publishing

82. Profit or Loss Problems Example: Compound appreciation
A property is currently worth $380,000. It has appreciated in value 4% per year for the last four years. What was it worth four years ago?
Step 1: Find percentage to use in formula
100% + 4% = 104%, or 1.04
Step 2: Calculate Then value for four years
$380,000 ÷ 1.04 = $365,385 (value 1 year ago)
$365,385 ÷ 1.04 = $351,332 (value 2 years ago)
$351,332 ÷ 1.04 = $337,819 (value 3 years ago)
$337,819 ÷ 1.04 = $324,826 (value 4 years ago)
© 2011 Rockwell Publishing

83. Summary Profit or Loss Problems
Now = Then × Percentage
To find percentage to use in formula:
If loss or depreciation, subtract rate of change from 100%.
If profit or appreciation, add rate of change to 100%.
Compound appreciation and depreciation: repeat profit or loss calculation as needed.
© 2011 Rockwell Publishing

84. Capitalization Problems
Capitalization: In appraisal of income-producing property, process used to convert property’s income into estimate of its value.
Value is price an investor would be willing to pay for the property.
Property’s annual net income is investor’s return on the investment (ROI).
© 2011 Rockwell Publishing

85. Capitalization Problems Formula: I = V × %
Capitalization problems are another type of percentage problem.
Formula is another variation on P = W × %
Part is property’s income, Whole is property’s value, % is capitalization rate.
Income = Value × Capitalization Rate
Value = Income ÷ Rate
Rate = Income ÷ Value
© 2011 Rockwell Publishing

86. Capitalization Problems Capitalization rate
Capitalization rate represents rate of return an investor would be likely to want on this investment.
Investor who wants higher rate of return won’t pay as much for the property as investor who’s willing to accept lower rate of return.
Higher cap rate means lower value.
© 2011 Rockwell Publishing

87. Capitalization Problems Example: Calculating value
A property generates an annual net income of $48, An investor wants a 12% rate of return on his investment. How much could he pay for the property and realize his desired rate of return?
Income = Value × Rate
Substitute: $48,000 = Value × .12
Isolate the unknown: Value = $48,000 ÷ .12
Value = $400,000
The investor could pay $400,000 for the property and realize his desired 12% return. Thus, the property is worth $400,000 to this investor.
© 2011 Rockwell Publishing

88. Capitalization Problems Example: Finding the cap rate
An investment property is valued at $425,000 and its net income is $40,375. What is the capitalization rate?
Income = Value × Rate
Substitute: $40,375 = $425,000 × Rate
Isolate the unknown: Rate = $40,375 ÷ $425,000
Rate = .095, or 9.5%
© 2011 Rockwell Publishing

89. Capitalization Problems Changing the cap rate
Capitalization rate is up to the investor, and depends on many factors.
One investor might be satisfied with 9% return.
Another might want 10.5% return from same property.
Some problems ask how property’s value will change if different cap rate is applied.
© 2011 Rockwell Publishing

90. Capitalization Problems Example: Changing the cap rate
Using a 10% cap rate, a property is valued at $1,500,000. What would its value be using an 11.5% cap rate?
Step 1: Calculate property’s annual net income
Income = Value × Rate
Substitute: Income = $1,500,000 × .10
Income = $150,000
© 2011 Rockwell Publishing

91. Capitalization Problems Example, continued
Step 2: Calculate value at higher cap rate
Income = Value × Rate
Substitute: $150,000 = Value × .115
Isolate the unknown: Value = $150,000 ÷ .115
Value = $1,304,348 ($195,652 less)
© 2011 Rockwell Publishing

92. Capitalization Problems Example: Changing the cap rate
A property with net income of $16,625 is valued at $190,000. If the cap rate is increased by 1%, what is the property’s value?
Step 1: Find current capitalization rate
Income = Value × Capitalization Rate
Substitute: $16,625 = $190,000 × Rate
Isolate the unknown: Rate = $16,625 ÷ $190,000
Rate = .0875
© 2011 Rockwell Publishing

93. Capitalization Problems Example, continued
Step 2: Increase cap rate by 1%
8.75% + 1% = 9.75%, or .0975
Step 3: Calculate new value
Income = Value × Capitalization Rate
Substitute: $16,625 = Value × .0975
Isolate the unknown: Value = $16,625 ÷ .0975
Value = $170,513
© 2011 Rockwell Publishing

94. Capitalization Problems Calculating net income
Instead of property’s annual net income, some problems give annual gross income and:
bad debt/vacancy factor
list of operating expenses
These must be subtracted from gross income to arrive at net income before capitalization formula can be applied.
© 2011 Rockwell Publishing

95. Capitalization Problems Example: Calculating net income
A six-unit apartment building rents three units for $650 a month and three units for $550 a month.
Allow 5% for vacancies and uncollected rent.
The annual operating expenses are $4,800 for utilities, $8,200 for property taxes, $1,710 for insurance, $5,360 for maintenance, and $2,600 for management fees.
If the cap rate is 8¾%, what is the property’s value?
© 2011 Rockwell Publishing

96. Capitalization Problems Example, continued
Step 1: Calculate gross income
$550 × 3 × 12 = $19,800 (three units at $550/month)
$650 × 3 × 12 = $23,400 (three units at $650/month)
$19,800 + $23,400 = $43,200 (gross income)
Step 2: Subtract 5% bad debt/vacancy factor
$43,200 × .05 = $2,160
$43,200 – $2,160 = $41,040 (effective gross income)
© 2011 Rockwell Publishing

97. Capitalization Problems Example, continued
Step 3: Subtract operating expenses
List of operating expenses adds up to $22,670
$41,040 – $22,670 = $18,370 (net income)
Step 4: Calculate value with 8¾% cap rate
Income = Value × Rate
Substitute: $18,370 = Value × .0875
Isolate the unknown: Value = $18,370 ÷ .0875
Value = $209,943
© 2011 Rockwell Publishing

98. Capitalization Problems Calculating net income with OER
Some problems give property’s operating expense ratio (OER) instead of list of operating expenses.
OER is percentage of gross income that goes to pay operating expenses.
Multiply gross income by OER to determine annual operating expenses.
Then subtract expenses from gross income to determine net income.
© 2011 Rockwell Publishing

99. Capitalization Problems Example: OER
A store grosses $758,000 annually. It has an operating expense ratio of 87%. With a capitalization rate of 9¼%, what is its value?
Step 1: Multiply gross income by OER
$758,000 × .87 = $659,460 (operating expenses)
Step 2: Subtract expenses from gross income
$758,000 – $659,460 = $98,540 (net income)
Step 3: Apply capitalization formula (I = V × %)
Value = $98,540 ÷ .0925
Value = $1,065,297
© 2011 Rockwell Publishing

100. Summary Capitalization Problems
Income = Value × Rate (I = V × %)
Income is annual net income
Rate is capitalization rate
Higher the cap rate, the lower the value.
Gross income
Bad debt/vacancy factor
Operating expenses
OER: Operating expense ratio
© 2011 Rockwell Publishing

101. Tax Assessment Problems
Some tax assessment problems can be solved using this variation on basic percentage formula (Part = Whole × %):
Tax = Assessed Value × Tax Rate
© 2011 Rockwell Publishing

102. Tax Assessment Problems Assessment ratio
Problem may state property’s assessed value.
Or it may give market value and assessment ratio, and assessed value must be calculated.
Multiply market value by assessment ratio to find assessed value.
Example: Property’s market value is $500,000 and assessment ratio is 80%.
$500,000 × .80 = $400,000
© 2011 Rockwell Publishing

103. Tax Assessment Problems Example: Assessment ratio
The property’s market value is $410,000. It’s subject to a 25% assessment ratio and an annual tax rate of 2.5%. How much is the annual tax the owner must pay?
Step 1: Calculate assessed value
$410,000 × .25 = $102,500
Step 2: Calculate tax
Tax = Assessed Value × Tax Rate
Substitute: Tax = $102,500 × .025
Tax = $2,562.50
The property owner must pay $2,
© 2011 Rockwell Publishing

104. Tax Assessment Problems Tax rate per $100 or $1,000
In some problems, tax rate not expressed as a percentage.
Instead, stated as a dollar amount per hundred dollars or per thousand dollars of assessed value.
Divide value by 100 or 1,000 to find number of $100 or $1,000 increments.
Then multiply that number by tax rate.
© 2011 Rockwell Publishing

105. Tax Assessment Problems Example: Tax rate per $100
A property is assessed at $125,000. The tax rate is $2.10 per hundred dollars of assessed value. What is the annual tax?
Step 1: Find number of increments in assessed value
$125,000 ÷ 100 = 1,250 ($100 increments)
Step 2: Multiply number of increments by tax rate
1,250 × $2.10 = $2,625 (annual tax)
© 2011 Rockwell Publishing

106. Tax Assessment Problems Example: Tax rate per $1,000
A property is assessed at $396,000. The tax rate is $14.25 per thousand dollars of assessed value. What is the annual tax?
Step 1: Find number of increments in assessed value
$396,000 ÷ 1,000 = 396 ($1,000 increments)
Step 2: Multiply number of increments by tax rate
396 × $14.25 = $5,643 (annual tax)
© 2011 Rockwell Publishing

107. Tax Assessment Problems Tax rate in mills
One other way in which a tax rate may be expressed: mills per dollar of assessed value.
One mill is one-tenth of one cent (.001), or one-thousandth of a dollar.
To convert mills to a percentage rate in decimal form, divide by 1,000.
Example: 30 mills ÷ 1,000 = .03 (3%)
© 2011 Rockwell Publishing

108. Tax Assessment Problems Example: Tax rate in mills
A property is assessed at $290,000 and the tax rate is 23 mills per dollar of assessed value. How much is the annual tax?
Step 1: Convert mills to a percentage rate
23 mills/dollar ÷ 1,000 = .023 (2.3%)
Step 2: Multiply assessed value by tax rate
Tax = Assessed Value × Tax Rate
Substitute: Tax = $290,000 × .023
Tax = $6,670
© 2011 Rockwell Publishing

109. Summary Tax Assessment Problems
Tax = Assessed Value × Tax Rate
Assessed value
Assessment ratio
Dollar amount per $100 or $1,000 of value
Mills
© 2011 Rockwell Publishing

110. Seller’s Net Problems
These ask how much seller needs to sell property for in order to get specified net amount from sale.
© 2011 Rockwell Publishing

111. Seller’s Net Problems Steps for finding minimum price
Add together:
seller’s desired net
costs of sale, except for commission
Subtract commission rate from 100%.
Divide result of Step 1 by result of Step 2.
If problem states mortgage balance to be paid off, treat that as one of the costs of sale.
© 2011 Rockwell Publishing

112. Seller’s Net Problems Example
A seller wants to net $120,000 from the sale of his property. He’ll pay $1,650 in attorney’s fees, $700 for the escrow fee, $550 for repairs, and a 6% commission. What’s the minimum he can sell the property for?
Step 1: Add desired net and costs of sale
$120,000 + $1,650 + $700 + $550 = $122,900
Step 2: Subtract commission rate from 100%
100% - 6% = 94%, or .94
Step 3: Calculate minimum sales price
$122,900 ÷ .94 = $130,744.68
© 2011 Rockwell Publishing

113. Seller’s Net Problems Example
A seller wants to net $24,000 from selling her home. She will have to pay $3,500 in closing costs, $1,600 in discount points, and a 7% commission. She’ll also have to pay off the mortgage balance, which is $86,050. What is the minimum amount she’ll have to sell the home for?
Step 1: Add desired net, mortgage, and sale costs
$24,000 + $86,050 + $3,500 + $1,600 = $115,150
© 2011 Rockwell Publishing

114. Seller’s Net Problems Example, continued
Step 2: Subtract commission rate from 100%
100% – 7% = 93%, or .93
Step 3: Calculate minimum sales price
$115,150 ÷ .93 = $123,817.20
© 2011 Rockwell Publishing

115. Summary Seller’s Net Problems
Desired Net + Costs of Sale + Loan Payoff
Subtract commission rate from 100%
Divide Step 1 total by Step 2 rate.
Result is how much property must sell for.
© 2011 Rockwell Publishing

116. Proration Problems
Proration: Dividing an expense proportionally, when one party is responsible for only part of the expense.
Items often prorated in real estate transactions include:
property taxes
insurance premiums
rent
mortgage interest
© 2011 Rockwell Publishing

117. Proration Problems Closing date is proration date
In most cases, property expenses prorated as of closing date.
Seller’s responsibility for expense usually ends on closing date.
Buyer’s responsibility for expense usually begins on closing date.
© 2011 Rockwell Publishing

118. Proration Problems 3 Steps
1. Calculate per diem (daily) rate of expense.
2. Determine number of days one party is responsible for expense.
3. Multiply per diem rate by number of days.
Share = Rate × Days
© 2011 Rockwell Publishing

119. Proration Problems 365 days or 360 days
Problem will say whether to use:
365-day year (calendar year), or
360-day year (banker’s year).
With 365-day year, use exact number of days in each month.
In 360-day year, each month has 30 days.
© 2011 Rockwell Publishing

120. Proration Problems Property taxes
In Washington:
General real estate taxes levied annually.
Tax year: January 1 – December 31
First installment due April 30 (covers January through June)
Second installment due October 31 (covers July through December)
© 2011 Rockwell Publishing

121. Property Tax Prorations Paid in advance or in arrears
Seller may already have paid tax installment or full year’s taxes.
Buyer owes seller prorated share at closing.
Debit for buyer, credit for seller.
Or seller may not yet have paid taxes.
Seller owes buyer prorated share at closing.
Debit for seller, credit for buyer.
Buyer’s responsibility starts on closing date.
© 2011 Rockwell Publishing

122. Property Tax Prorations Example: Taxes in arrears
The annual property taxes are $2,860. The seller has paid the first installment, but not the second. The sale is closing on August 10. How much will the seller have to pay in taxes at closing? Use a 360-day year.
Step 1: Calculate per diem rate
$2,860 ÷ 360 = $7.94
Step 2: Count number of days (starting July 1)
30 (July) + 9 (August) = 39 days
Step 3: Multiply rate by number of days
$7.94 × 39 = $ (seller’s debit, buyer’s credit)
© 2011 Rockwell Publishing

123. Property Tax Prorations Example: Taxes paid in advance
A home sale is due to close on September 20. The annual property taxes, $4,924, have already been paid through the end of the year. How much will the buyer owe at closing for property taxes? Use a 365-day year.
Step 1: Calculate per diem rate
$4,924 ÷ 365 = $13.49
Step 2: Count number of days
11 (Sept) + 31 (Oct) + 30 (Nov) + 31 (Dec) = 103 days
Step 3: Multiply per diem rate by number of days
$13.49 × 103 = $1, (buyer’s debit, seller’s credit)
© 2011 Rockwell Publishing

124. Proration Problems Hazard insurance
Property insurance is paid for in advance.
Seller entitled to refund of prorated share of insurance premium for coverage extending beyond closing date.
Refund is credit for seller.
Not a debit for buyer, unless buyer is assuming policy.
© 2011 Rockwell Publishing

125. Insurance Prorations Example
The annual premium for the sellers’ hazard insurance policy is $1,350. It’s been paid for through next March, but the sale is closing November 12. The sellers are responsible for insuring the property until the day after closing. How much is their refund? Use a 360-day year.
Step 1: Calculate per diem rate
$1,350 ÷ 360 = $3.75
© 2011 Rockwell Publishing

126. Insurance Prorations Example, continued
Step 2: Count number of days
18 (Nov.) (December to March) = 138 days
Step 3: Multiply per diem rate by number of days
$3.75 × 138 = $ (sellers’ credit)
© 2011 Rockwell Publishing

127. Proration Problems Rent
When rental property sold, seller owes buyer prorated share of any rent paid in advance.
Tenants generally expected to pay rent at beginning of month.
© 2011 Rockwell Publishing

128. Rent Prorations Example
The sale of an apartment house is closing January 12. The seller collected January rent from all tenants; it totals $14,400. How much will the seller owe the buyer for rent at closing? Use the exact number of days in the month.
Step 1: Calculate per diem rate
$14,400 ÷ 31 days = $ per diem
© 2011 Rockwell Publishing

129. Rent Prorations Example, continued
Step 2: Count number of days
January 12 through January 31 = 20 days
Step 3: Multiply per diem rate by number of days
$ × 20 = $9, (seller’s debit, buyer’s credit)
© 2011 Rockwell Publishing

130. Proration Problems Mortgage interest
For interest prorations, remember that mortgage interest is almost always paid:
on a monthly basis
in arrears (at end of month in which it accrues)
If problem doesn’t give amount of annual interest, first use loan amount and interest rate to calculate it.
Then do the other proration steps.
© 2011 Rockwell Publishing

131. Mortgage Interest Prorations Two types of prorated interest
Two types of mortgage interest usually have to be prorated as of closing date and paid at closing:
seller’s final interest payment
buyer’s prepaid interest (interim interest)
For both types, lender charges interest for day of closing.
© 2011 Rockwell Publishing

132. Mortgage Interest Prorations Example: Seller’s final payment
Ann is selling her home for $275,000, closing May 14. Her mortgage at 7% interest has a balance of $212,500. How much is her final interest payment? (Use 365 days.)
Step 1: Calculate annual interest
$212,500 × .07 = $14,875
Step 2: Calculate per diem rate
$14,875 ÷ 365 = $40.75
© 2011 Rockwell Publishing

133. Mortgage Interest Prorations Example, continued
Step 3: Count number of days
May 1 through May 14 = 14 days
Step 4: Multiply per diem by number of days
$40.75 × 14 = $570.50
© 2011 Rockwell Publishing

134. Mortgage Interest Prorations Buyer’s prepaid interest
Prepaid interest: At closing, buyer is charged interest for closing date through end of month in which closing occurs.
Example: Sale closing on April 8
Buyer’s first loan payment due June 1.
June 1 payment includes May interest, but not April interest.
At closing, buyer must pay interest for April 8 through April 30.
© 2011 Rockwell Publishing

135. Mortgage Interest Prorations Example: Buyer’s prepaid interest
A home buyer has a $350,000 loan at 5.5% interest. The closing date is January 17. How much prepaid interest will the buyer have to pay? (Use a 360-day year.)
Step 1: Calculate annual interest
$350,000 × .055 = $19,250
Step 2: Calculate per diem rate
$19,250 ÷ 360 = $53.47
© 2011 Rockwell Publishing

136. Mortgage Interest Prorations Example, continued
Step 3: Count number of days
January 17 through January 30 = 14 days
Step 4: Multiply per diem rate by number of days
$53.47 × 14 = $748.58
© 2011 Rockwell Publishing

137. Summary Proration Problems
Closing date
365-day year vs. 360-day year
Payment in advance or in arrears
Steps:
Calculate per diem rate
Count number of days
Multiply per diem rate by number of days
© 2011 Rockwell Publishing