Washington Real Estate Law © Copyright 2010 Rockwell Publishing, Inc. Lesson 14: Condominiums.
Washington Real Estate Fundamentals Lesson 18: Real Estate Math © 2011 Rockwell Publishing.
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Transcript of Washington Real Estate Fundamentals Lesson 18: Real Estate Math © 2011 Rockwell Publishing.
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Washington Real Estate Fundamentals
Lesson 18:
Real Estate Math
© 2011 Rockwell Publishing
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Solving Math ProblemsFour 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
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Solving Math ProblemsUsing 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
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Solving Math ProblemsIsolating 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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Solving Math ProblemsExample
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
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Decimal NumbersConverting 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 = 0.251/3 = 1 ÷ 3 = 0.3335/8 = 5 ÷ 8 = 0.625
© 2011 Rockwell Publishing
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Decimal NumbersConverting percentage to decimal
If problem contains a percentage, may be necessary to:
convert percentage to decimaldo calculations to find answerthen convert decimal back to percentage
(Calculator’s percent key converts percentage to decimal automatically.)
© 2011 Rockwell Publishing
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Decimal NumbersConverting 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 .0217.5% becomes .17580% becomes .80123% becomes 1.23
© 2011 Rockwell Publishing
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Decimal NumbersConverting 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
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SummarySolving 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
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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
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Area ProblemsRectangle formula: A = L × W
Formula for calculating area of square or rectangular space:
Area = Length × Width
A = L × W
Area
Width
Le
ng
th
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RectanglesExample
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
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RectanglesSquare yards
Some problems express area in square yards rather than square feet. Remember: 1 square yard = 9 square feet
1 yard = 3 feet1 square yard is 3 feet on each side3 feet × 3 feet = 9 square feet
Divide square footage by 9 to find number of square yards.
© 2011 Rockwell Publishing
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Area ProblemsTriangle formula: A = ½ B × H
Formula for calculating area of triangle:
Area = ½ Base × Height
A = ½ B × H
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TrianglesOrder of operations
Note that in applying triangle formula, order of operations doesn’t matter.
These will all reach same result:A = ½ B × HA = B × ½ HA = (B × H) ÷ 2
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TrianglesExample
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.
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TrianglesExample, 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
A = 3,500 sq. ft.
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Area ProblemsOdd 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.
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Odd ShapesExample
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.
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Odd ShapesExample, continued
Area of rectangle:
A = 60 ft. × 100 ft.
A = 6,000 sq. feet
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Odd ShapesExample, 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.
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Odd ShapesExample, continued
Total area of odd shape:
6,000 sq. ft. (rectangle)+ 600 sq. ft. (triangle) 6,600 sq. ft.
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Odd ShapesAvoid 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.
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Odd ShapesExample
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.
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Odd ShapesExample, 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.
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Odd ShapesExample, 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.
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Odd ShapesExample, 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
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Odd ShapesExample, 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.
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Odd ShapesNarrative 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.
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Odd ShapesExample
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.
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Odd ShapesExample, 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.
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Odd ShapesExample, continued
Calculate area of rectangle:
426 × 47 = 20,022 sq. ft.
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Odd ShapesExample, continued
Calculate area of triangle:
(½ × 426) × 272 = 57,936 sq. ft.
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Odd ShapesExample, continued
Add together area of rectangle and area of triangle to find the lot’s total square footage:
20,022 + 57,936 = 77,958
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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.
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Volume ProblemsFormula: V = L × W × H
To calculate volume, use this formula:
Volume = Length × Width × Height
V = L × W × H
© 2011 Rockwell Publishing
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Volume ProblemsCubic 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
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Volume ProblemsExample
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
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SummaryArea 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
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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
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Percentage ProblemsFormula: P = W × %
Basic formula for solving percentage problems:
Part = Whole × Percentage
P = W × %
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Percentage ProblemsFormula: 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.
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Percentage ProblemsIsolating 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
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Percentage ProblemsMultiply 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.
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Percentage ProblemsTypes of problems
Many real estate math problems are basically percentage problems:
commission problemsloan problems (interest and principal)profit or loss problemscapitalization problems
Percentage sometimes referred to as “rate.”Examples: 7% commission rate,
5% interest rate, 10% rate of return © 2011 Rockwell Publishing
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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.
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© 2011 Rockwell Publishing
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 × .06P = $18,000
Step 2: Calculate broker’s share (60%)P = $18,000 × .60P = $10,800
Commission ProblemsExample
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Commission ProblemsTwo-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
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© 2011 Rockwell Publishing
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
Two-tiered CommissionExample
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© 2011 Rockwell Publishing
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
Two-tiered CommissionExample, continued
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SummaryPercentage 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
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Loan Problems
Loan problems include:interest problemsprincipal 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
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Loan ProblemsInterest 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
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© 2011 Rockwell Publishing
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
Interest ProblemsExample: Finding annual interest
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© 2011 Rockwell Publishing
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%
Interest ProblemsExample: Finding interest rate
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Loan ProblemsPrincipal 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 × %
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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
Principal Balance ProblemsExample: Finding loan amount
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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
Principal Balance ProblemsExample: Finding current balance
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Principal Balance ProblemsPrincipal 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.
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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 interestP = 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
Principal Balance ProblemsExample: Finding principal reduction
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Principal Balance ProblemsMultiple 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
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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.
Principal Balance ProblemsExample: Multiple P&I payments
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SummaryLoan 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.
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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
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Profit or Loss ProblemsThen and Now formula
Basic “Then and Now” formula:
Now = Then × %
To isolate the unknown, change it to:
Then = Now ÷ %
or
% = Now ÷ Then
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Profit or Loss ProblemsPercentage 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.
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Percentage to Use in FormulaProperty sold at a loss
For loss, subtract percentage of loss from 100% to find percentage to use in formula.
Example: 30% lossHouse 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.
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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
Percentage to Use in FormulaExample: Property sold at a loss
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Percentage to Use in FormulaProperty 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% profitMeans 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.
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Percentage to Use in FormulaProperty 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.
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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
Percentage to Use in FormulaExample: Property sold for a profit
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Profit or Loss ProblemsCalculating 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 lossask how much property is being sold for
(the Now value)
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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
Profit or Loss ProblemsExample: Calculating selling price
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Profit or Loss ProblemsCalculating profit or loss percentage
Other problems give original price and selling price, ask for percentage of profit or loss.
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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.
Profit or Loss ProblemsExample: Calculating profit
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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.
Profit or Loss ProblemsExample: Calculating loss
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Profit or Loss ProblemsAppreciation 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.
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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)
Profit or Loss ProblemsExample: Compound depreciation
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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.
Profit or Loss ProblemsExample, continued
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Profit or Loss ProblemsCompound 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%.
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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)
Profit or Loss ProblemsExample: Compound appreciation
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SummaryProfit 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.
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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).
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Capitalization ProblemsFormula: 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
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Capitalization ProblemsCapitalization 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.
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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 × RateSubstitute: $48,000 = Value × .12Isolate the unknown: Value = $48,000 ÷ .12Value = $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.
Capitalization ProblemsExample: Calculating value
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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%
Capitalization ProblemsExample: Finding the cap rate
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Capitalization ProblemsChanging 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.
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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
Capitalization ProblemsExample: Changing the cap rate
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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)
Capitalization ProblemsExample, continued
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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
Capitalization ProblemsExample: Changing the cap rate
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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
Capitalization ProblemsExample, continued
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Capitalization ProblemsCalculating net income
Instead of property’s annual net income, some problems give annual gross income and:
bad debt/vacancy factorlist of operating expenses
These must be subtracted from gross income to arrive at net income before capitalization formula can be applied.
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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?
Capitalization ProblemsExample: Calculating net income
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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)
Capitalization ProblemsExample, continued
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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
Capitalization ProblemsExample, continued
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Capitalization ProblemsCalculating 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.
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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 ÷ .0925Value = $1,065,297
Capitalization ProblemsExample: OER
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SummaryCapitalization 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
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Tax Assessment Problems
Some tax assessment problems can be solved using this variation on basic percentage formula (Part = Whole × %):
Tax = Assessed Value × Tax Rate
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Tax Assessment ProblemsAssessment 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
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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 RateSubstitute: Tax = $102,500 × .025Tax = $2,562.50
The property owner must pay $2,562.50.
Tax Assessment ProblemsExample: Assessment ratio
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Tax Assessment ProblemsTax 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.
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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)
Tax Assessment ProblemsExample: Tax rate per $100
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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)
Tax Assessment ProblemsExample: Tax rate per $1,000
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Tax Assessment ProblemsTax 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%)
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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
Tax Assessment ProblemsExample: Tax rate in mills
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SummaryTax Assessment Problems
• Tax = Assessed Value × Tax Rate
• Assessed value
• Assessment ratio
• Dollar amount per $100 or $1,000 of value
• Mills
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Seller’s Net Problems
These ask how much seller needs to sell property for in order to get specified net amount from sale.
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Seller’s Net ProblemsSteps for finding minimum price
1. Add together:seller’s desired netcosts 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.
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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
Seller’s Net ProblemsExample
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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
Seller’s Net ProblemsExample
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Step 2: Subtract commission rate from 100%
100% – 7% = 93%, or .93
Step 3: Calculate minimum sales price
$115,150 ÷ .93 = $123,817.20
Seller’s Net ProblemsExample, continued
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SummarySeller’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.
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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 taxesinsurance premiumsrentmortgage interest
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Proration ProblemsClosing 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.
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Proration Problems3 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
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Proration Problems365 days or 360 days
Problem will say whether to use:365-day year (calendar year), or360-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.
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Proration ProblemsProperty 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)
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Property Tax ProrationsPaid 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.
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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)
Property Tax ProrationsExample: Taxes in arrears
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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)
Property Tax ProrationsExample: Taxes paid in advance
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Proration ProblemsHazard 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.
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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
Insurance ProrationsExample
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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)
Insurance ProrationsExample, continued
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Proration ProblemsRent
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.
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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 = $464.52 per diem
Rent ProrationsExample
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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)
Rent ProrationsExample, continued
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Proration ProblemsMortgage interest
For interest prorations, remember that mortgage interest is almost always paid:
on a monthly basisin 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.
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Mortgage Interest ProrationsTwo 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 paymentbuyer’s prepaid interest (interim interest)
For both types, lender charges interest for day of closing.
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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
Mortgage Interest ProrationsExample: Seller’s final payment
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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
Mortgage Interest ProrationsExample, continued
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Mortgage Interest ProrationsBuyer’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 8Buyer’s first loan payment due June 1.June 1 payment includes May interest,
but not April interest.At closing, buyer must pay interest for
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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
Mortgage Interest ProrationsExample: Buyer’s prepaid interest
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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
Mortgage Interest ProrationsExample, continued
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SummaryProration 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
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