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September 16 Chapter 12 Felde Principles - Robert Rooks 1
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September 16 Chapter 12 Felde Principles - Robert Rooks 2 Introduction Anyone involved in real estate should be familiar with some of the basic mathematical calculations that are essential to understanding real estate computations. Financial calculators, available today, make it unnecessary to be a “math expert” still, it is important to understand some of the simple calculations that are useful in the real estate business.
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September 16 Chapter 12 Felde Principles - Robert Rooks 3 Basic Math Review Most of the problems students encounter when trying to master real estate mathematics are that the student has forgotten the basic 5 th grade math, and when they do remember it they find it difficult to apply to the problems as they present themselves.
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September 16 Chapter 12 Felde Principles - Robert Rooks 4 Common Percentages, Decimals, and Fractions PercentageDecimalFraction 4 1/2 %0.0454.5/100 6 2/3 %0.0671/15 10%0.101/10 12 ½ %0.1251/8 16 2/3 %0.16671/6 25%0.25¼ 33 1/3 %0.33331/3 50%0.50½ 66 2/3 %0.6672/3 75%0.75¾ 100%1.001/1
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September 16 Chapter 12 Felde Principles - Robert Rooks 5 Converting Percentages To A Decimal To remove the percent (%) sign, you must move the decimal point two places, or numbers to the left. 4%= Or0.04 41/2%=04.5%Or0.045 105%= or1.05
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September 16 Chapter 12 Felde Principles - Robert Rooks 6 Dividing By Decimals When dividing decimals it is necessary to remove the decimal in the divisor. When you move the decimal in a divisor, you must move the decimal in the dividend an equal number of places. 1. 270 4 1/2 % would look like this;.045 / 270 1. 270 4 1/2 % would look like this;.045 / 270 045. / 270000. 6.000
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September 16 Chapter 12 Felde Principles - Robert Rooks 7 Multiplying Decimals When multiplying decimals you multiply the same way you multiply whole numbers. When you get the answer you count off the decimal places then apply that same number to the answer. The 16.55 X 3.5 = 57.925 or; 16.55 3.5. 57.925
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September 16 Chapter 12 Felde Principles - Robert Rooks 8 Paid X Percent = Made Formula This basic formula can be used in a wide variety of problems. In any problem that can be solved with this formula, you are always given two of the three items. It is your task to find which figures given represent “Paid”, “%” or “Made”. Once there are identified you know that you either multiply or divide. In the following sections we learn how to identify these items and use one of the following formulas:
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September 16 Chapter 12 Felde Principles - Robert Rooks 9 The circle can be used more efficiently if you apply these rules: 1.Always annualize figures. 2.Label the segments A, B, and C. 3.The Percentage always goes into the C segment. 4.When the percentage is more than 100% the large number goes in the A segment, and the small number goes into the B segment. 5.When the percentage is less than 100% the small number goes in the A segment, and the large number goes in the B segment
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September 16 Chapter 12 Felde Principles - Robert Rooks 10 These formulas may help. Made=PaidX % Paid=Made % %=Made Paid
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September 16 Chapter 12 Felde Principles - Robert Rooks 11 The formulas are often presented on this circle: Paid Rate % Made Divide Line Multiply Line AA BBCC ÷÷÷÷ XX
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September 16 Chapter 12 Felde Principles - Robert Rooks 12 When percentage is less than 100%: % Less than 100% Small Number Divide Line Multiply Line AA BBCC ÷÷÷÷ XX Large Number
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September 16 Chapter 12 Felde Principles - Robert Rooks 13 When percentage is more than 100%: % More than 100% Small Number Divide Line Multiply Line AA BBCC ÷÷÷÷ XX Large Number
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September 16 Chapter 12 Felde Principles - Robert Rooks 14 Investment Problems When you are asked to calculate the yield or rate of return on an investment or the amount of the investment needed to yield a stated number of dollars. Use one of the formulas as follows by substituting new names for “Paid”, “%”, and “Made”. 1.To find amount invested = Made % 2.To find rate of return=Rate of Return 3. To find income or yield=Paid X %
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September 16 Chapter 12 Felde Principles - Robert Rooks 15 Example 1 – An Investment Problem If an investor wants to earn $75 a month from their savings account and the account pays 5% simple interest, how much must they invest?
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September 16 Chapter 12 Felde Principles - Robert Rooks 16 Example 1 – Figuring The Circle 1.First of all the interest rate is less than 100%, so the small number will go into the segment A. 2.Secondly they have given us a monthly figure, $75 a month. This has to be annualized, or multiplied by 12. $75 X 12 = $900. 3.Third, the earnings, the $900 should be less than the principle. So, the $900 goes into Segment A.
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September 16 Chapter 12 Felde Principles - Robert Rooks 17 Example 1 – Applying the Circle $18,000 5% $900.00 AA BB CC
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September 16 Chapter 12 Felde Principles - Robert Rooks 18 Example #2 An investor purchased 2 lots that were 60’ wide by 150’ deep, paying$18,000 for each lot. The investor then split the parcels into 3 lots measuring 40’ X 150’. If the investor sold each new lot for $15,000, what was the investor’s return? 3 – 40’ X 150’ lots X $15,000 ea. = $45,000 Selling Price 2 – 60’ X 150’ lots X $18,000 ea. = $36,000 Purchase Price Profit Made= $9,000 $9,000 $36,000= 25%
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September 16 Chapter 12 Felde Principles - Robert Rooks 19 Example #2 Paid Rate % Made $45,000 - $36,000 = $9,000 25% $36,000 AA BB BB AA CC CC
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September 16 Chapter 12 Felde Principles - Robert Rooks 20 Example #3 An investor purchased a 5 year straight note with a face amount of $4,000 at a 20% discount. If the interest rate on the note was 6%, what was the annual rate of yield to the purchaser on the amount invested? Face amount of note $4,000 20% Discount X 20% Amount of Discount $ 800 Face Value of Note $ 4,000 Less 20% Discount($ 800) Purchase Price of Note $ 3,200
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September 16 Chapter 12 Felde Principles - Robert Rooks 21 Example #3 Continued Annual Return on note: Note $4,000 Interest Rate X 6% Annual Interest $ 240 Return on Investment: Interest + Discount Divided by Investment $240 + ($800 5 Yrs.) $3,200 = $ ?? ($240 +$160)=$400 $3,200= $12.5%
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September 16 Chapter 12 Felde Principles - Robert Rooks 22 Appraisal Problem It is possible to establish the value of an apartment building or other income producing property by capitalizing the net income. This follows the same basic thinking used in the investment problems in which a stated amount of investment (purchase price) will develop a certain amount of income, depending upon the rate of return desired. The rate of return is called a capitalization rate. Use one of the following formulas as follows by substituting new names for “Paid”, “%”, and “Made”.
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September 16 Chapter 12 Felde Principles - Robert Rooks 23 Appraisal Problem Analyzed Paid=Purchase Price %=Capitalization Rate Made=Net income (or loss) To find purchase price = Made % To find Capitalization Rate=Made paid To find Income or Loss = Paid X %
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September 16 Chapter 12 Felde Principles - Robert Rooks 24 Appraisal Example #1 An apartment building produces a net income of $132,000. If a prospective purchaser wants a return of 8% on their purchase price, what should they pay for the property? Made = $132,000 8% Building Value Building Value Made = $132,000 8% $1,650,000 CC
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September 16 Chapter 12 Felde Principles - Robert Rooks 25 Appraisal Example #2 An apartment owner has a property that is immediately adjacent to a new proposed freeway route. He estimates that the noise created by the freeway will result in a reduction of rents and he will lose $2,100 each month. Assuming a capitalization rate of 12%, what will be the overall loss in value of his property? Solution: Paid=Made % Rent lost=$2,100X12=$25,200/Year $25,200 12%=$210,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 26 Same Problem Using Circles Rent Loss = $25,000 12% Loss = $25,000 12% Buildings Loss in Value $210,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 27 Appraisal Example #3 Mrs. Thompson owns a 6 unit apartment building and presently has all units rented for $1,250 per month each. If the rents drop to $1,100 per month per unit, the overall loss in value using a 9% capitalization rate would be? Solution:Paid=Made (or loss) % Present Rent $1,250 Reduced Rent($1,100) Loss $ 150 Per Unit
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September 16 Chapter 12 Felde Principles - Robert Rooks 28 Appraisal Example #3 Continued $ 150 per mo. X 12 Months = $1,800 Per Unit Per Year $ 1,800 Per UnitX6 Units = $10,800 Per Year $10,800 9% =$120,000 Loss in Value Rent Loss = $10,800 9% Loss = $10,800 9% Buildings Loss in Value $120,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 29 Example #4 Appraisal Problem An appraiser is attempting to establish the “going market” capitalization rate for a specific geographical area. She finds a 20 unit apartment complex that recently sold for $1,500,000 with an annual net income of $120,000. Based on this information, what is the capitalization rate on this property? Solution: Made Paid= % $120,000 $1,500,000= 8%
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September 16 Chapter 12 Felde Principles - Robert Rooks 30 Problem #4 Circles Finding the Capitalization Rate Net Income = $120,000 8% Net Operating Income $120,000 8% Buildings Value $1,500,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 31 Commission Problem You will have to calculate the amount of commission to be earned on a sale of property. This is not an unusual type of problem and the “Paid X % = Made” formula can be applied as follows: Paid=The Selling Price %=Rate of Commission Made=Amount of Commission
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September 16 Chapter 12 Felde Principles - Robert Rooks 32 Commission Problem To find a selling price –Made % To find commission Rate –Made Paid To find Commission–Paid X % Paid % Made
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September 16 Chapter 12 Felde Principles - Robert Rooks 33 Example 1 You sold a residence for $365,000. Your office pays 60% to the selling salesperson and the office keeps the remainder. The listing was based on a 6% Commission. What did you receive? $365,000 6% Made X 60% $365,000 6% = $21,900 6% = $21,900 $21,900 X 60% = $13,140
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September 16 Chapter 12 Felde Principles - Robert Rooks 34 Example 1 Alternative You could solve example 1 with a simple 2 step process; 1.$365,000X 6%=$21,400 2.$ 21,400X60%=$13,140
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September 16 Chapter 12 Felde Principles - Robert Rooks 35 Example 2 Your office listed 10 acres of recreational property for $260,000 with an agreed commission of 10%. You present an offer to the seller for 10% less than the listed price. The seller agrees to accept the offer if your office will reduce their commission by 25%. If the broker agrees to the reduction, what is the total commission paid? Break this down into questions: 1.How much is your offer? 2.How much is the reduced commission going to be? 3.How much total commission will be paid?
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September 16 Chapter 12 Felde Principles - Robert Rooks 36 Example 2 – Question 1 Your offer would be: $260,000X10%=$26,000 Then you subtract $26,000 from $260,000 $260,000-$26,000=$234,000 $234,000 is the price you have offered. An Alternative would be to multiply $260,000 by 90% and you would have the offering price. $260,000X90%=$234,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 37 Example 1 – Question 2 How much is your reduced commission going to be? One Step Old CommissionX75%=New Commission 10%X75%= 7.5% Commission
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September 16 Chapter 12 Felde Principles - Robert Rooks 38 Example 1 – Question 3 Find your commission: $234,000X7.5%=$17,550 Your reduced commission is $17,550 you can check that several ways, one way is to multiply $234,000 X 10% = $23,400 - $17,550 = $5,850 Check: $5,850 $23,400=25%
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September 16 Chapter 12 Felde Principles - Robert Rooks 39 Example 3 You as a broker negotiate a lease for 3,000 square feet of warehouse storage space at a monthly rental rate of $0.50 per square foot. If your commission is 8% of the first years gross rent, what is your commission. Rent is; 1.3,000 Sq. Ft. X $0.50= $1,500 Month 2.$1,500 X 12 Months= $18,000 per year 3.$18,000X8%=$1,440 Commission
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September 16 Chapter 12 Felde Principles - Robert Rooks 40 D.Interest and Loan Problems One of the best applications of this formula is for interest problems. Always work on an annual basis, first. Usually the interest payment is for only one month. The answer will involve the 1 st month. Paid=Loan Amount %=Rate of Interest Made=Annual Interest Rate Earned
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September 16 Chapter 12 Felde Principles - Robert Rooks 41 D.Interest and Loan Problems continued 1.To find the loan amount=Made % 2.To find Interest Rate=Made Paid 3.To find Interest=PaidX%
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September 16 Chapter 12 Felde Principles - Robert Rooks 42 Loans Example 1 If you borrowed $5,000 for 90 days using a straight note (interest only) and paid $100 in interest, what was the interest rate on the note? Solution:%=Made Paid Paid=Loan Amount %=? Made=$100 per quarter = $100 X 4=$400
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September 16 Chapter 12 Felde Principles - Robert Rooks 43 Loans Example 1 Page 2 Once you know the annual interest, you can determine the annual interest rate. Annual Interest Principal=Interest Rate $400 $5,000=8% Remember just because the loan term is for one quarter of a year, the annual interest has to be decided on. However, you could divide $100 by $5,000 and multiply by four.
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September 16 Chapter 12 Felde Principles - Robert Rooks 44 Loans Example 2 If you loaned a client $2,500, interest only, (straight note), and you earned $150 in 8 months, what was the interest rate? Solution:%=Made Paid 1.Figure the total interest for each month and annualize the monthly figure. $150 8 Months=$18.75 per month
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September 16 Chapter 12 Felde Principles - Robert Rooks 45 Loans Example 2 Page 2 Convert the monthly payment to an annual payment: $18.75X12=$225 The annual return. $225 $2,500=9% The interest rate on your loan is 9% per annum
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September 16 Chapter 12 Felde Principles - Robert Rooks 46 Example #3 If a $60,000 straight note, interest only, written for a three year term, earns $1,640 interest per quarter, what is the interest rate? Solution:%=Made Paid First convert the quarterly interest to annual interest, or; $1,640 X 4 = $6,560 per annum (year) Then:$6,560 $60,000=10.9%
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September 16 Chapter 12 Felde Principles - Robert Rooks 47 Example 3 Using the Circle Our problem on the circle would look as follows: Paid Rate Made $60,000 10.9% $1,640 X 4 = $6,560 Annually $1,640 X 4 = $6,560 Annually
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September 16 Chapter 12 Felde Principles - Robert Rooks 48 Example #4 If you received $420 interest on a straight note that is in the amount of $14,000 for a term of 90 days, what is the interest on the note. Solution:%=Made Paid The problem again is to convert the Made from a quarterly return to an annual return, then; Quarterly ReturnX4=Annual Return $420X4=$1,680
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September 16 Chapter 12 Felde Principles - Robert Rooks 49 Example #4 Page 2 Now you are ready to divide the annual interest by the note amount to come out with an annual interest rate: Annual Interest Note Amount=Interest Rate $1,680 $14,000=$12%
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September 16 Chapter 12 Felde Principles - Robert Rooks 50 Example #4 Page 3 On the circle we would have: Paid Rate Made $420 X 4 = $1,680 Annually $420 X 4 = $1,680 Annually $14,000 12%
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September 16 Chapter 12 Felde Principles - Robert Rooks 51 NOTE The classic formula for interest that you learned in High School is stated as follows; I=Px Rx T Interest=PrincipalxRatexTime
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September 16 Chapter 12 Felde Principles - Robert Rooks 52 Monthly Short Cuts When you are faced with the problem of figuring monthly interest rates, you can multiply the face value by the interest rate, then divide by 12 or you can divide the interest rate by 12 and multiply by that number. $6,000 at 10% interest for 9 years. How much is the 1 st months interest? 10% 12 Months = 0.0083333, then $6,000 X 0.0083333=$50
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September 16 Chapter 12 Felde Principles - Robert Rooks 53 Acreage and Area Problems A number of problems edge in on the study of geometry, but not to the extent that we used it in High School, no theorems or postulates. Area=Length X Width Depth=Area Width Width=Area Depth Area=Width XDepth
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September 16 Chapter 12 Felde Principles - Robert Rooks 54 Example 1 - Area A rectangular lot contains 10,800 square feet of land. If the depth of the lot is 180 feet, what is the width? Solution:Width=Area Depth ?=10,800 Sq. Ft. 180 feet 60 Feet=10,800 Sq. Ft. 180 feet The width then is 60 feet wide
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September 16 Chapter 12 Felde Principles - Robert Rooks 55 Example 1 – Area With Circle Using the circle of power you would come up with the following sequence for Area Width Depth Area 10,800 Square Feet of Land 60 Ft.Wide 180 Ft. Deep 180 Ft. Deep
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September 16 Chapter 12 Felde Principles - Robert Rooks 56 Example 2 If you bought a commercial lot that measured 50 feet by 100 feet and paid $3.00 per square foot, what would the cost be per front foot? (the front footage is on the width, 50 feet) Solution:WidthXDepth=Square Footage 50’X 100’=5,000 sq. ft. 5,000 Sq. Ft. x $3.00/Sq. Ft.=$15,000 $15,000 50 Front Feet=$300 The cost per front foot is $300
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September 16 Chapter 12 Felde Principles - Robert Rooks 57 Example 3 A rectangular parcel of land contains three acres. If the depth of the parcel is 1,139 feet, what is its width? Solution:Width=Area Depth 43,560 Square Feet in an Acre Total Square Footage = 43,560 sq. ft. X 3 = 130,680 sq. ft. 130,680 Sq. Ft. 1,139 Ft.=114.7 Feet Wide The answer then is 114.7 Feet Wide
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September 16 Chapter 12 Felde Principles - Robert Rooks 58 Additional Factors The following are helpful and sometimes necessary when attempting to solve your problems: 1.ACRE of LAND=43,560 Square Feet 2.ONE SQUARE YARD=9 Square Feet 3.ONE SQUARE ACRE=About 209’ X 209’ 4.LOT DIMENSIONS – First dimension given is the width, the second dimension given is the depth.
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September 16 Chapter 12 Felde Principles - Robert Rooks 59 Cost Problems COST PROBLEMS – One of the most difficult problems encountered is the so called “cost” problem. In these problems you are given a selling price and are asked to calculate either the dollar amount of profit or the original cost before a given rate of profit. These can be solved by using the “Magic Circle.”
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September 16 Chapter 12 Felde Principles - Robert Rooks 60 Cost Problems & the Magic Circle Costs Cost 100% + Rate of Profit Selling Price
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September 16 Chapter 12 Felde Principles - Robert Rooks 61 Example #1 Cost Problems An investor sold his property for $35,000 which represented a 10% profit over what he paid for it. What was his cost? Formula: Cost=Selling Price 100% + Rate of Profit $32,000=$35,200 110% The investor’s Cost was $32,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 62 Example #1 Cost Problems Cost 100% + Rate of Profit Selling Price $32,000 100% + 10% = 110% $35,200
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September 16 Chapter 12 Felde Principles - Robert Rooks 63 Example #2 Sally sold two acres of land for $48,300. If she made 15% profit over her original cost, what was her profit? Formula:Cost = Selling Price 100% + Profit $42,000=$48,300 115% Selling Price – Cost=Profit $48,300–$42,000=$ 6,300 $48,300 115%=$42,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 64 Example #2 Cost 100% + Rate of Profit Selling Price $42,000 100% + 15% = 115% $48,300 Then simply subtract the Selling Price from the purchase price to find the profit. $48,300 – $42,000 = $6,300 Then simply subtract the Selling Price from the purchase price to find the profit. $48,300 – $42,000 = $6,300
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September 16 Chapter 12 Felde Principles - Robert Rooks 65 Example #3 If a house sold for $56,950 and this price was 11% more than the cost of the house, what was the original cost? Formula: Cost = Selling Price 100% + Profit $51,306.30=$56,950 111%
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September 16 Chapter 12 Felde Principles - Robert Rooks 66 Example #3 Cost 100% + Rate of Profit Selling Price $51,306 100% + 11% = 111% $56,950
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September 16 Chapter 12 Felde Principles - Robert Rooks 67 Selling Price Problems One other difficult type of problem is used in which you are given a net amount received by a seller after escrow or by a holder of a note after selling it at a discount and then asked to establish the selling price or face amount of the loan. These problems can be solved using many methods, one simple method is the “Magic Circle”.
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September 16 Chapter 12 Felde Principles - Robert Rooks 68 Selling Price Problems 2 A seller received a check for $23,500 from a broker who had sold the property and deducted a 6% commission from the selling price. What is the selling price? Here you would subtract the percentage gone, the 6% from 100%; 100%-6%=94% Then multiply the selling price by the adjusted net of 94% $23,500 94%=$25,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 69 Selling Price Problems Amount or Selling Price 100% - Discount Net $25,000 100% - 6% = 94% ??? $23,500 ??? $23,500
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September 16 Chapter 12 Felde Principles - Robert Rooks 70 Example 4 Note Discounts First and second trust deeds created in purchase sale transactions are often discounted and sold to create immediate cash. This is true in the secondary mortgage market where 1 st Trust Deeds and Mortgages are sold to investors. The discount raises the return on the Trust Deed or Mortgage. This creates a whole area of possible income for real estate brokers and investors.
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September 16 Chapter 12 Felde Principles - Robert Rooks 71 Example 4 Note Discounts Page 2 If a bank sold a Trust Deed or Mortgage Note at a 4% discount, and the bank netted $240,000, what was the face value of the note? The logic here is; 100% - Discount X ?=$240,000 100% - 4% = 96% X ?=$240,000 ?=$240,000/96% $240,000/96%=$250,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 72 Example 4 Note Discounts Page 3 Amount or Selling Price 100% - Discount Net $250,000 ????? 100% - 4% = 96% $240,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 73 Seller Net Problems Problems that we really run into in the business are problems concerning what will the seller wants to net upon the sale of their property. We have an owner who wants to net $120,000 at the close of escrow. They have a 1 st Trust Deed in the amount of $110,000. The expenses of sale will amount to 2% of the selling price, and the commission will be 6%. What must you sell the property for to net the seller $120,000?
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September 16 Chapter 12 Felde Principles - Robert Rooks 74 Seller Net Problems Page 2 First find the net before expenses of sale. Net+Loan=Net Before Expenses $120,000+$110,000=$230,000 Net Equals Selling Price times 100% Less expenses of 8% Net Equals Selling Price times 100% - 8% = 92% 92% X Selling Price=$230,000 Selling Price=$230,000/92%=$250,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 75 Seller Net Problems Page 3 Amount or Selling Price 100% - Discount Net $250,000 ????? 100% - 8% = 92% $230,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 76 Amortization Chart Term Years 9%10%11%12%12 1/2 % 13%13 1/2 % 14%14 ½ %15% 10 Years 12.6713.2213.7814.3514.6414.9315.2315.5315.8316.13 15 Years 10.1510.7511.3712.0012.3312.6512.9913.3213.6614.00 20 Years 9.009.6610.3211.0111.3611.2712.0812.4412.8013.17 25 Years 8.409.099.8010.5310.9011.2811.6612.0412.4312.81 30 Years 8.058.789.5210.2910.6711.0611.4511.8512.2512.64
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September 16 Chapter 12 Felde Principles - Robert Rooks 77 If you needed to know how much the monthly payment on $1,000 for 30 Years at 10% interest you would do the following. Term Years 9%10%11%12% 12 1/2 % 13%13 1/2 % 14%14 ½ %15% 10 Years 12.6713.2213.7814.3514.6414.9315.2315.5315.8316.13 15 Years 10.1510.7511.3712.0012.3312.6512.9913.3213.6614.00 20 Years 9.009.6610.3211.0111.3611.2712.0812.4412.8013.17 25 Years 8.409.099.8010.5310.9011.2811.6612.0412.4312.81 30 Years 8.058.789.5210.2910.6711.0611.4511.8512.2512.64 Point of intersection gives monthly payment on $1,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 78 Typical Problems Using Amortization Charts What is the monthly payment to pay off a $1,000 loan at 11% interest over a 15 year period? From the fifteen year line, go to your right until you get to the 11% Column, you will find 11.37, that means $11.37 per month for 15 years to pay off $1,000 at 11% interest.
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September 16 Chapter 12 Felde Principles - Robert Rooks 79 What is the monthly payment required to amortize a $12,500 loan at 11% interest for 25 Years. First – find out what the monthly payment for $1,000 at 11% for 25 years is. Follow the 25 Year Row to the 12 1/2 % Column and you will find $10.90 per month per $1,000 of loan. Second determine how many $1,000 bills are in your loan, in $12,500 there are 12 1/2 $1,000 bills. Third multiply the number of $1,000 bills by the monthly rate – 12 1/2 X $10.90=$136.25 Per Month
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September 16 Chapter 12 Felde Principles - Robert Rooks 80 What is the total interest paid over the life of a 30 year loan of $1,000 at 11% Interest? The table Row intersection at 30 Years with the 11% Column shows $9.52 per month per $1,000 of value. $9.52 X 360 Months= $3,427.20 Total Paid Subtract the original amount and you will have the interest. $3,427.20-$1,000=$2,427.20 Total Interest
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September 16 Chapter 12 Felde Principles - Robert Rooks 81
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September 16 Chapter 12 Felde Principles - Robert Rooks 82 1.An owner took out a 2 nd trust deed and note on her home, in the amount of $2,500. If the loan was payable $30.00 per month including interest at 8%, the amount of the first month's payment that would be interest is: a.$30.00 b.$13.33 c.$16.67 d.$20.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 83 2.If you want to collect $37.50 per month income on an investment that represents a 9% return, the amount to be invested must be: a.$4,500.00 b.$5,000.00 c.$ 416.67 d.$4,166.67
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September 16 Chapter 12 Felde Principles - Robert Rooks 84 3.The maximum number of 50’ X 110’ lots that could be obtained from an acre of land would be: a. 5 lots b. 7 lots c. 8 lots d.10 lots
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September 16 Chapter 12 Felde Principles - Robert Rooks 85 4.An agent sold a home for $280,000. If the broker was to receive a 6% commission and the agent was to be paid 40% of this amount, the agent would receive: a.$ 6,350 b.$ 6,720 c.$10,080 d.$11,250
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September 16 Chapter 12 Felde Principles - Robert Rooks 86 5.If a rectangular lot had a 45ft. Frontage and contained 1,080 square yards, what would the depth of the lot? a. 24 feet b. 72 feet c.144 feet d.216 feet
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September 16 Chapter 12 Felde Principles - Robert Rooks 87 6.If you were to divide an acre of land into four lots of equal size and each lot had a depth of 200’, the front footage of each lot would be most nearly: a. 50 feet b.100 feet c. 55 feet d.218 feet
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September 16 Chapter 12 Felde Principles - Robert Rooks 88 7.If you borrowed $5,000 for 90 days on a straight note and paid $100.00 in interest for the loan, the interest rate would have to be: a. 6% b. 7% c.10% d. 8%
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September 16 Chapter 12 Felde Principles - Robert Rooks 89 8.One month’s interest on a one year straight note amounted to $125.00. If the interest rate on the loan was 5%, what was the face amount of the note? a.$ 10,000.00 b.$ 20,000.00 c.$ 30,000.00 d.$ 40,000.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 90 9.How many acres of land are there in a parcel of land that is 240’ wide and 1089’ deep? a.4 acres b.5 acres c.6 acres d.7 acres
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September 16 Chapter 12 Felde Principles - Robert Rooks 91 10.If a home sold for $264,000 and the selling price was 10% more than what was paid for the property the original cost of the property was: a.$237,600 b.$240,000 c.$242,000 d.$246,000
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September 16 Chapter 12 Felde Principles - Robert Rooks 92 11.An apartment complex produces an annual net income of $87,500. If this represents an 8% return on the value of the property, the property’s value is: a.$ 700,000 b.$ 853,450 c.$ 925,850 d.$1,093,750
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September 16 Chapter 12 Felde Principles - Robert Rooks 93 12.In order to earn $75.00 per month from an investment that yields a 5% return, you would have to invest. a.$ 1,500.00 b.$ 9,000.00 c.$15,000.00 d.$18,000.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 94 13.If the total interest paid on a five year straight note amounted to $4,320.00, and the interest rate was 7.2%, what was most nearly the amount of the loan? a.$ 8,000.00 b.$11,500.00 c.$11,667.00 d.$60,000.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 95 14.An investor purchased a 4 acre parcel of land for $20,000 and subdivided it into 8 one half acre lots. If the investor sold each lot for $3,750, the rate of profit would be: a.25% b.33 1/3 % c.40% d.50%
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September 16 Chapter 12 Felde Principles - Robert Rooks 96 15.Assume that you wish to construct a building that is to have interior dimensions of 24’ X 30’. If it were to have 6” walls, how many square feet of land would be required for the building. a.744 b.747.25 c.775 d.803.25
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September 16 Chapter 12 Felde Principles - Robert Rooks 97 16.Parker received a check in the amount of $152,000. If this was the net amount from the sale of his property after the broker had deducted a 5% commission, the selling price was: a.$156,000 b.$158,000 c.$160,000 d.$167,200
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September 16 Chapter 12 Felde Principles - Robert Rooks 98 17.If an investor receives $125.00 per month income on an investment of $25,000, the rate of return is: a.5% b.6% c.7% d.7 1/2 %
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September 16 Chapter 12 Felde Principles - Robert Rooks 99 18.A parcel of land contains 16,500 sq. ft.. If the width of the parcel is 110’, the depth is: a.135’ b.140’ c.145’ d.150’
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September 16 Chapter 12 Felde Principles - Robert Rooks 100 19.An investor sold an apartment for $35,200.00, realizing a 10% profit over what he had paid for it. The amount of profit was: a.$3,520.00 b.$3,811.00 c.$3,200.00 d.$3,872.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 101 20.A retailer purchased stock for her store for $9,300.00. She sold the goods for 33 1/3 % more than what she paid for them, but lost 15% of the total selling price due to bad debts. The net profit on the sale was. a.$10,540.00 b.$ 3,100.00 c.15% d.$ 1,240.00
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September 16 Chapter 12 Felde Principles - Robert Rooks 102
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