Washington Real Estate Fundamentals
Lesson 18:
Real Estate Math
© 2011 Rockwell Publishing
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.
© 2011 Rockwell Publishing
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
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.
© 2011 Rockwell Publishing
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
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
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
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
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
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
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
Area Problems
Formula for calculating
area of square or
rectangular space:
Length
Rectangle formula: A = L × W
Area
Area = Length × Width
A=L×W
Width
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
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
Area Problems
Triangle formula: A = ½ B × H
Formula for calculating
area of triangle:
Area = ½ Base × Height
A=½B×H
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
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.
Triangles
Example, continued
A triangular lot is 140 feet Same result from this:
long and 50 feet wide at its A = 50 ft. × (½ × 140 ft.)
base. What is the area?
A = 50 ft. × 70 ft.
A = (½ × 50 ft.) × 140 ft.
A = 3,500 sq. ft.
A = 25 ft. × 140 ft.
A = 3,500 sq. ft.
And from this:
A = (50 ft. × 140 ft.) ÷ 2
A = 7,000 sq. ft. ÷ 2
A = 3,500 sq. ft.
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
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.
Odd Shapes
Example, continued
Area of rectangle:
A = 60 ft. × 100 ft.
A = 6,000 sq. feet
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.
Odd Shapes
Example, continued
Total area of odd shape:
6,000 sq. ft. (rectangle)
+ 600 sq. ft. (triangle)
6,600 sq. ft.
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.
Odd Shapes
Example
Wrong way to calculate
area of this lot:
25 × 50 = 1,250 sq. ft.
40 × 20 = 800 sq. ft.
1,250 + 800 = 2,050 sq. ft.
This counts middle of
shape (green area) twice.
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.
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.
1,250 + 300 = 1,550 sq. ft.
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
Odd Shapes
Example, continued
Now calculate area of each
rectangle and add them
together:
40 × 20 = 800 sq. ft.
30 × 25 = 750 sq. ft.
800 + 750 = 1,550 sq. ft.
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
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.
Odd Shapes
Example, continued
Break it down into a
rectangle and a triangle
as shown.
Subtract 47 from 319
to find height of triangular
portion:
319 – 47 = 272 ft.
Odd Shapes
Example, continued
Calculate area of rectangle:
426 × 47 = 20,022 sq. ft.
Odd Shapes
Example, continued
Calculate area of triangle:
(½ × 426) × 272 = 57,936 sq. ft.
Odd Shapes
Example, continued
Add together area of
rectangle and area of
triangle to find the lot’s
total square footage:
20,022 + 57,936 = 77,958
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
Volume Problems
Formula: V = L × W × H
To calculate volume, use this formula:
Volume = Length × Width × Height
V=L×W×H
© 2011 Rockwell Publishing
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
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 = 93.33 cubic yards
© 2011 Rockwell Publishing
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
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
Percentage Problems
Formula: P = W × %
Basic formula for solving percentage problems:
Part = Whole × Percentage
P=W×%
© 2011 Rockwell Publishing
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
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
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
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
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
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
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
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
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
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
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
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
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
Interest Problems
Example: Finding interest rate
The interest portion of a loan’s monthly payment is
$517.50. The loan balance is $92,000. What is the
interest rate?
Step 1: Convert monthly to annual figure
$517.50 × 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
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
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
Principal Balance Problems
Example: Finding current balance
The interest portion of a loan’s tenth monthly payment is
$256.67. The interest rate is 7%. What is the loan
balance prior to this payment?
Step 1: Convert monthly to annual figure
$256.67 × 12 = $3,080.04 (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
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
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 $704.41.
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
$704.41 – $640.00 = $64.41
© 2011 Rockwell Publishing
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
Principal Balance Problems
Example: Multiple P&I payments
Continuing with previous example ($96,000 loan @ 8%,
$704.41 P&I pyt, first pyt reduces balance by $64.41):
$96,000 – $64.41 = $95,935.59 (balance after Pyt 1)
$95,935.59 × .08 = $7,674.85 (annual interest)
$7,674.85 ÷ 12 = $639.57 (interest portion of Pyt 2)
$704.41 – $639.57 = $64.84 (principal portion of Pyt 2)
$95,935.59 – $64.84 = $95,870.75 (balance after Pyt 2)
So second payment reduces loan balance to $95,870.75.
For balance after third payment, repeat these steps.
© 2011 Rockwell Publishing
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
Capitalization Problems
Example: Calculating value
A property generates an annual net income of $48,000.
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
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
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
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
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
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
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
Capitalization Problems
Calculating net income
Instead of property’s annual net income, some
problems give annual gross income and:
 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
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
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% vacancy factor
$43,200 × .05 = $2,160
$43,200 – $2,160 = $41,040 (effective gross income)
© 2011 Rockwell Publishing
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
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
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
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
• Vacancy factor
• Operating expenses
• OER: Operating expense ratio
© 2011 Rockwell Publishing
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
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
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 $2,562.50.
© 2011 Rockwell Publishing
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
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
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
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
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
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
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
Seller’s Net Problems
Steps for finding minimum price
1. Add together:
 seller’s desired net
 costs of sale, except for commission
2. Subtract commission rate from 100%.
3. 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
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
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
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
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
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
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
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
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
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
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
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 = $309.66 (seller’s debit, buyer’s credit)
© 2011 Rockwell Publishing
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 = $1,389.47 (buyer’s debit, seller’s credit)
© 2011 Rockwell Publishing
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
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
Insurance Prorations
Example, continued
Step 2: Count number of days
18 (Nov.) + 120 (December to March) = 138 days
Step 3: Multiply per diem rate by number of days
$3.75 × 138 = $517.50 (sellers’ credit)
© 2011 Rockwell Publishing
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
Rent Prorations
Example
The sale of a small 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 = $464.52 per diem
© 2011 Rockwell Publishing
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
$464.52 × 20 = $9,290.40 (seller’s debit, buyer’s credit)
© 2011 Rockwell Publishing
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
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
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
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
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
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
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
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