1. CHAPTER 1 ARITHMETIC AND GEOMETRIC SEQUENCES
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CHAPTER 1
ARITHMETIC
AND
GEOMETRIC SEQUENCES
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2. 1.1 SEQUENCE (PROGRESSION)
A list of numbers arranged in a specified order.
Two sequences namely:
- arithmetic
- geometric
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3. 1.2 ARITHMETIC SEQUENCE
One in which the difference between any term and the preceding term is the same throughout.
Example:
Arithmetic sequence Common difference
(a) 4, 8, 12, 16, …… 4
(b) ½, 1, 1½, 2, …… ½
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4. 1.3 Nth TERM AND SUM OF FIRST N TERMS OF AN ARITHMETIC SEQUENCE
If first term of an arithmetic sequence is a and common difference is d, then the arithmetic sequence is written as:
a, a + d, a + 2d, a + 3d, ……
To find the nth term:
Tn = a + (n – 1)d
where:
Tn = nth term
a = first term
n = number terms
d = common difference
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5. Given the following arithmetic sequence: 2,10,18,…… find
Example 1 pg 3
Given the following arithmetic sequence: 2,10,18,…… find
The tenth term,
The sum if the first ten terms.
Example 2 pg 4
Given the arithmetic sequence: 30,23,16,9,2…… find the 12th term and the sum of the first 12 terms.
Example 3 pg 4
Find the number of terms in the following arithmetic sequence: 12,17,22,…., 67. hence, find the sum of all the terms.
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6. Example 4 pg 5
Find the minimum number if terms that must be taken from the following sequence: 8,16,24,32,….. So that the sum is more than 120.
Example 5 pg 5
Find the first term and the common difference if an arithmetic progression if the fourth term is 33 and the tenth term is 120.
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7. Example 6 pg 6 Ishak starts with monthly salary of RM 1250 for the first year and receives an annual increment of RM 80. how much is his monthly salary for the nth year service? How much will he receive monthly for his tenth year of service.
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8. 1.4 GEOMETRIC SEQUENCE
A sequence of numbers and the ratio between any term. It is obtained by multiplying the first term by the ratio to get the second term and so on.
The ratio is known as common ratio and can be obtained by dividing any term by the term before it.
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9. Geometric sequence Common ratio (r)
Example:
Geometric sequence Common ratio (r)
(a) 1, 2, 4, 8, 16, 32, …… (4/2) or (8/4)
(b) -3, 6, -12, 24, …… (6/-3) or (-12/6)
(c) 0.1, 0.01, 0.001, …… ½ (1 – ½) or (1½ - 1)
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10. 1.5 Nth TERM AND SUM OF FIRST N TERMS OF AN GEOMETRIC SEQUENCE
If a geometric sequence is a, ar, ar2, ar3, ……, arn-1, thus:
Sn = a(1 – rn) / (1 – r) for r < 1
Sn = a(rn - 1) / (r – 1) for r > 1
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11. Example 1 pg 8 Given the following geometric sequence: 5, 15, 45, 135, ……, find a) the eighth term and the tenth term, b) the sum if the first eight terms. Example 2 pg 8 Find the number if terms in the following sequence: 2, 6, 18, ….., calculate the sum of all the terms.
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12. b) the sum of the first ten terms.
Example 3 pg 8
Maimunah saves RM 1000 in a saving account that pays 8% compounded annually. Find the amount in her account at the end of 5 years.
Example 4 pg 8
The third term of a geometrics progression is 360 and the sixth term is find
a) the first term
b) the sum of the first ten terms.
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13. CHAPTER 2 SIMPLE INTEREST
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14. 2.1 INTEREST
“Interest” comes from the Latin word intereo which means “to be lost”.
When developed into the concept of borrowing money, the lender is likely to lose his money when he pays back the money with interest.
Nowadays, interest is not only paid but gained if we make an investment.
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15. 2.2 SIMPLE INTEREST FORMULA
The simple interest amount is calculated by the following formula:
I = Prt
Where: I = Simple interest
P = Principal
r = Interest rate (in decimals)
t = Time / Period (in years)
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16. 2.3 SIMPLE AMOUNT FORMULA
Simple amount is the sum of the original principal and the interest earned.
Therefore, the simple amount formula is given as:
S = P(1 + rt)
Where: S = Maturity value
P = Principal
r = Interest rate (in decimals)
t = Term / Period (in years)
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17. Example 1 pg 22 RM is invested for 4 years 9 months in a bank earning a simple interest rate of 10% per annum. Find the simple amount at the end if the investment period. Example 2 pg 23 Raihan invested RM 5000 in an investment find for three years. At the end of the investment period, his investment will be worth RM find the simple rate that is offered.
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18. Example 3 pg 23
Twenty four month ago, a sum of money was invested. Now the investment is worth RM if the investment is extended for another twenty four months, it will become RM Find the original principle and the simple interest that was offered.
Example 4 pg 24
Muthu invested RM in two accounts, some at 10% per annum and the rest at 7% per annum. His total interest for one year was RM 820. find the amount invested at each rate.
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19. 2.4 FOUR BASIC CONCEPTS
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There are four different methods for determining terms (t):
1. Exact time : It is the exact number if days between two given dates.
2. Approximate time : Time computed on the assumption that each month has 30 days.
3. Exact interest : Interest calculated based on 365 days a year or 366 days for a leap year.
4. Ordinary interest : Interest is calculated based on 360 a year.
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20. Example 1 pg 25 Find a) exact time, b) approximate time From 15 March to 29 August of the same years. Example 2 pg 25 RM 1000 was invested on 15 March if the simple interest rate offered was 10% per annum, find the interest received on 29 August 2005 using a) Exact time and exact simple interest. b) Exact time and ordinary simple interest. c) Approximate time and exact simple interest. d) Approximate time and ordinary simple interest.
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21. REV 00
2.5 PRESENT VALUE
Present value may be debt or an investment amount that is lent or invested today, and that will be mature in a specific time together with interest.
By transposing the maturity value formula, we have the present value formula as follows:
P = S / (1 + rt) or
P = S (1 + rt)-1
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22. REV 00
2.6 EQUATION OF VALUE
Every value of money has an attached date, the date on which it is due.
An equation that states the equivalence of two sets of dated values at a stated date is called an equation of value or equivalence.
The stated date is called the focal date, the comparison date or the valuation date.
To set up and solve an equation of value, the following procedure should be carried out:
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23. 1. Draw a time diagram with all the dated values.
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1. Draw a time diagram with all the dated values.
2. Select the focal date.
3. Pull all the dated values to the focal date
using the stated interest rate.
4. Set up the equation of value and then
solve.
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24. Example 1 pg 26 Find the present value at 8% simple interest of a debt RM 3000 due in ten months. Example 2 pg 27 A debt of RM 800 due in four months and another of RM 1000 due in nine months are to be settled by a single payment at the end of six months. Find the size of this payment using a) the present as the focal date, b) the date of settlement as the focal date, Assuming money is worth 6% per annum simple interest.
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25. a) the present as the focal date
Example 3 pg 28
A debt of RM 500 due two months ago and RM 900 due in nine months are to be settled by two equal payments, one at the end of three months and another at the end of six months. Find the size of the payment using
a) the present as the focal date
b) the end of six months as a focal date
Assuming money is worth 10% per annum simple interest.
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26. CHAPTER 3 COMPOUND INTEREST
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CHAPTER 3
COMPOUND INTEREST
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27. REV 00
3.1 TIME VALUE OF MONEY
Money has time value, that is a ringgit today is worth more than a ringgit tomorrow.
Money has time value because of its investment opportunities.
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28. REV 00
3.2 COMPOUND INTEREST
In compounding, after the interest is calculated, it is then added to the principal and becomes an adjusted principal.
Processes are repeated until the end of the loan or investment term.
Normally used with long-term loan or investment, and the interest is calculated more than once during the loan or investment term.
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29. REV 00
The interest earned is called compound interest, and the final sum at the end of the period of borrowing is called the compound amount.
Therefore, compound interest is the difference between the original principal and the amount.
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30. REV 00
3.3 SOME IMPORTANT TERMS
Some of the common terms used in relation to compound interest are:
1. Original principal
2. Nominal interest rate
3. Interest period or conversion period
4. Frequency of conversions
5. Periodic interest rate
6. Number of interest periods in the investment
period
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31. 3.4 COMPOUND INTEREST FORMULA
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3.4 COMPOUND INTEREST FORMULA
The method used in finding compound amount at the end of the nth period is as follow:
S = P(1 + i)n
Where:
P = Principal / Present Value
S = Future Value
n = Number of Periods (number of years multiplied by number of times compounded per year)
i = Interest rate per compound period
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32. Example 1 pg 41 Find the future value of RM 1000 which was invested for a) 4 years at 4% compounded annually, b) 5 years 6 months at 14% compounded semi annually, c) 2 years 3 months at 4% compounded quarterly d0 5 years 7 months at 5% compounded monthly e) 2 years 8 months at 9% compounded every 2 months f) 250 days at 10% compounded daily.
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33. Example 2 pg 42
RM 9000 is invested for 7 years 3 months. This investment is offered 12% compounded monthly for the first 4 years and 12% compounded quarterly for the rest of the period. Calculate the future value of this investment.
Example 3 pg 42
Lolita saved RM 5000 in a saving account which pays 12% interest compounded monthly. Eight months later she saved another RM Find he amount in the account two years after her first saving.
Example 4 pg 43
What is the nominal rate compounded monthly that will make RM 1000 become RM 2000 in five years?
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34. 3.5 EFFECTIVE, NOMINAL AND EQUIVALENT RATES
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3.5 EFFECTIVE, NOMINAL AND EQUIVALENT RATES
Effective rate : Simple rate that will produce the same accumulated amount as the nominal rate is compounded each period after one year.
Nominal rate : Stated annual interest rate at which interest is compounding more than once a year.
Equivalent rate : Two different rates that yield the same value at the end of one year.
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35. 3.6 RELATIONSHIP BETWEEN EFFECTIVE AND NOMINAL RATES
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3.6 RELATIONSHIP BETWEEN EFFECTIVE AND NOMINAL RATES
The relationship between the nominal rate and effective rate is derived as follows:
Assume a sum RM P is invested for one year. Then the future value after one year:
(a) At r% effective = P(1 + r)
(b) At k% compounded m times a year = P(1 + k/m)m
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36. Example 1 pg 41 Find the effective rate which is equivalent to 16% compounded semi annually. Example 2 pg 45 Find the nominal rate, compounded monthly which is equivalent to 9% effective rate. Example 3 pg 45 Kang wishes to borrow some money to finance some business expansion. He has received two difference quotes: Bank A: Charged 15.2% compounded annually Bank B: Charged 14.5% compounded monthly. Which bank provides a better deal?
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37. 3.7 RELATIONSHIP BETWEEN TWO NOMINAL RATES
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3.7 RELATIONSHIP BETWEEN TWO NOMINAL RATES
The relationship between two nominal rates is given as follows:
(1 + k/m)m + (1 + K/M)M
Where:
k and K are two different annual rates with respectively two different frequencies of conversions, m and M.
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38. Example 1 pg 46 Find K% compounded quarterly which is equivalent to 6% compounded monthly.
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39. REV 00
3.8 PRESENT VALUE
Present value or discounted value is the value which will yield the sum (S) after certain time and at a specific interest rate.
We can find present value by transposing the formula as below:
S = P(1 + i)n
P = S / (1 + i)n or P = S(1 + i)-n
transpose
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40. Example 1 pg 47 A debts of RM 3000 will mature in three years’ time
Example 1 pg 47 A debts of RM 3000 will mature in three years’ time. Find a) the present value of this debts b) the value of this debt at the end of the first year c) the value of this debts at the end if four years. Assuming money is worth 14% compounded semi annually. Example 2 pg 49 A debt of RM 7000 matures at the end of the second year and another of RM 8000 at the end of six years. If the debtor wishes to pay his debts by making one payment at the end of the fifth year, find the amount he mist pay if money is worth 6% compounded semi annually using a) the present as the focal date b) the end of the fifth year as the focal date.
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41. Example 3 pg 50 A debt of RM 7000 matures at the end of the second year and another RM 8000 at the end of six years. If the debtor wishes to pay his debts making two equal payments at the end of the fourth year and the seven year, what are these payments assuming money is worth 6% compounded semi annually. Example 4 pg 51 Roland invested RM at 12% compounded monthly. This investment will be given to his three children when they reach 20 years old. Now his three children are 15, 16 and 19 years old. If his three children will receive equal amounts, find the amount each will receive.
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42. REV 00
3.9 EQUATION OF VALUE
An equation that expresses the equivalence of two sets of obligations at a focal date.
In other words, it expresses the following:
What is owed = What is owned at the focal date or
What is given = What is received at the focal date
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43. 3.10 CONTINUOUS COMPOUNDING
We have been discussing compounding of interest on discrete time intervals (daily, monthly, etc).
If compounding of interest is done on a continuous basis, then we will have a different picture of the future value as shown below:
Continuous compounding
Discrete compounding
Future value
Future value
Time
Time
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44. CHAPTER 4 ANNUITY
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45. 4.1 FUTURE VALUE OF ORDINARY ANNUITY CERTAIN
Future value (accumulated value) of an ordinary annuity certain is the sum of all the future values of the periodic payments.
The derivation of the formula of future value of ordinary annuity certain are as follow:
Periodic payments = RM R
Interest rate per interest period = i%
Term of investment = n interest periods
Future value of annuity at end of n interest periods = RM S
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46. Formula: S = R[(1 + i)n – 1 / i] Interest earn: I = S- nR DDG 1113
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47. Example 1 pg 64 RM 100 is deposited every month for 2 years 7 months at 12% compounded monthly. What is the future value of this annuity at the end if the investment period? How much interest is earned? Example 2 pg 64 RM 100 is deposited every 3 months for 2 years 9 months at 8% compounded quarterly. What is the future value of this annuity at the end if the investment period? How much interest is earned?
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48. Example 3 pg 64
RM 100 was invested every month in an account that pays 12% compounded monthly for two years. After the two years, no more deposit was made. Find the amount of the account at the end of five years and the interest earned.
Example 4 pg 64
Lily invested RM 100 every month for 5 years in an investment scheme. She was offered 5% compounded monthly for the first 3 years and 9 % compounded monthly for the rest of the period. Find the accumulated amount at the end of five years. Hence, determine the interest earned.
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49. Example 5 pg 66
The table below shows the monthly deposits that were made into an investment account that pays 12% compounded monthly.
Find the value of this investment at the end of find also the interest earned.
Year
Monthly deposit
2003
RM 100
2004
RM 200
2005
RM 300
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50. 4.2 PRESENT VALUE OF ORDINARY ANNUITY CERTAIN
Consist of the sum of all the present values of periodic payments.
The deviation of the formula of present value of ordinary annuity certain is illustrated in the following:
Periodic payments = RM R
Interest rate per interest period= i%
Term of investment = n interest periods
Future value of annuity at end of n interest periods = RM A
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51. Formula:
A = R[1 - (1 + i)-n / i]
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52. 4.2 PRESENT VALUE OF ORDINARY ANNUITY CERTAIN
Consist of the sum of all the present values of periodic payments.
The deviation of the formula of present value of ordinary annuity certain is illustrated in the following:
Periodic payments = RM R
Interest rate per interest period= i%
Term of investment = n interest periods
Future value of annuity at end of n interest periods = RM A
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53. Example 1 pg 70
Raymond has to pay RM 300 every month for 24 months to settle a loan 12% compounded monthly. a) What is the original value of the loan? b) What is the total interest that he has to pay? Example 2 pg 70 John won an annuity that pays RM 1000 every three months for three years. What is the present value of this annuity if money is worth 16% compounded quarterly?
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54. Example 3 pg 71
James intends to give a scholarship worth RM 5000 every year for six years. How much he deposit now into an account that pays 7% per annum to provide this scholarship?
Example 4 pg 72
Under a contract, Jenny has to pay RM 100 at the beginning of each month for 15 monhts. What is the present value of the contract if money is worth 12% compounded monthly? Find the interest paid by Jenny?
Example 5 pg 73
Find the present value of an annuity of RM 500 every year for 5 years if the first payment is made in 2 years. Assuming money is worth 6% compounded annually.
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55. Solving for R, n and i
Example 1 pg 75 Find the amount to be invested every three months at 10% compounded quarterly to accumulate RM in three years. Find the interest earned. Example 2 pg 75 Maria invested RM in an account that pays 6% compounded monthly. She intends to withdraw an equal amount every month for two years and when she makes her last withdrawal, her account will have zero balance. Find the size of these withdrawals.
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56. Example 3 pg 75
A RM used car is bought for RM 2000 down, 14 payments of RM 500 a month and a final 15th payment. If interest charged is 9% compounded monthly, find the size of the final payment.
Example 4 pg 78
Joanne purchase a shop and mortgaged it for RM the mortgages required repayment in equal monthly payments over ten years at 16% compounded monthly. Just immediate making the 80th payment, she had the loan refinanced at 14% compounded monthly. What is the new monthly payment if the number of payments remained the same?
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57. Example 5 pg 78 Jimmy has to pay RM 443
Example 5 pg 78 Jimmy has to pay RM every month to settle a loan of RM at 6% compounded monthly. Find the number of payments he has to make. Example 6 pg 78 Roger borrowed RM 100,000 at 12% compounded monthly. How many monthly payment of RM 2000 should roger make? What would be the concluding size of the final payment?
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58. 4.4 AMORTIZATION SCHEDULE
An interest bearing a debt is said to be amortized when all the principal and interest are discharged by a sequence of equal payments at equal intervals of time.
4.4 AMORTIZATION SCHEDULE
A table showing the distribution of principal and interest payments for the various periodic payments.
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59. Example 1
A loan of RM 1000 at 12% compounded monthly is to be amortised by 18 monthly payment. a) Calculate the monthly payment b) Construct an amortisation schedule. Solution From A= Ra n 1% 1000 = Ra 18 1% 1000= R(16.398) = RM 60.98
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60. DDG 1113 BUSINESS MATHEMATICS
Period
Beginning Balance
Ending Balance
Monthly payment
Total Paid
Interest
Total Principle paid
Total Interest paid 1
1000
949.02
60.98
10.00
50.98 2
897.53
121.96
9.49
102.47
19.49 3
845.52
182.95
8.98
154.48
28.47 4
792.99
243.93
8.46
207.01
36.92 5
739.94
304.91
7.93
260.06
44.85 6
686.36
365.89
7.40
313.64
52.25 7
632.24
426.87
6.86
367.76
59.11 8
577.58
487.86
6.32
422.42
65.44 9
522.37
548.84
5.78
477.63
71.21 10
466.61
609.82
5.22
533.39
76.44 11
410.30
670.80
4.67
589.7
81.10 12
353.42
731.78
4.10
646.58
85.20 13
295.97
792.77
3.53
704.03
88.74 14
237.95
853.75
2.96
762.05
91.70 15
179.35
914.73
2.38
820.65
94.08 16
120.16
975.71
1.79
879.84
95.87 17
60.38
1.20
939.62
97.07 18
0.00
0.64
97.68
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61. 4.5 SINKING FUND
When a loan is settled by the sinking fund method, the creditor will only receive the periodic interest due.
The face value of the loan will only be settled at the end of the term.
In order to pay the face value, debtor will create a separate fund in which he will make periodic deposits over the term of the loan.
The series of deposits made will amount to the original loan.
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62. 4.6 ANNUITY WITH CONTINUOUS COMPOUNDING
Future value of annuity:
S = R[ekt – 1 / ek/p – 1]
Present value of annuity:
A = R[1 – e-kt / ek/p – 1]
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63. S = Future value of annuity
Where:
S = Future value of annuity
A = Present value of annuity
R = Periodic payment or deposit
e = natural logarithm
k = annual continuous compounding rate
t = time in years
p = number of payments in 1 year
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64. The future value of this annuity at the end of four years.
Example 1
James wins an annuity that pays RM1000 at the end of every six months for four years. If money is worth 10% per annum continuous compounding what is
The future value of this annuity at the end of four years.
The present value of this annuity
S= R[e kt -1 / e k/p -1] From A= R[1-e-kt / e k/p -1]
S= 1000[e10%(4) -1/ e 10%/2 -1] = 1000[1-e-10%(4)/ e 10%/2 -1]
= RM = RM
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65. TRADE AND CASH DISCOUNT
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CHAPTER 5
TRADE AND CASH DISCOUNT
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66. REV 00
5.1 TRADE DISCOUNT
Trade discount is a reduction from the list price offered by wholesalers to retailers so they can resell the merchandise at a profit.
The amount of discount that retailers receive from wholesalers is known as trade discount amount.
Trade discount = List price – Net price
Trade discount rate is normally a manufacturer quotes a discount rate in percentage to the retailer. The trade discount rate must be calculated on the list price.
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67. The Formula for calculating the net price is NP= L ( 1-r)
Example 1 pg 94
The list price of a leather belt is RM 180. a trade discount of 30% is offered. What is the net price of the belt?
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68. Example 2 pg 95
Weendy jean offers a discount of 32 ¼ % on all the jeans it sells. What is the net price of a pair of jeans that is listed at RM 420? Example 3 pg 95 The net price of a camera with 40% trade discount is RM 480. what is the list price.
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69. Find the net prices of the item for the two shops.
Example 4 pg 95
A bill of RM 1200 including a prepaid handling charge of RM 200 is offered a trade discount of 15%. What is the net price?
Example 5 pg 95
Blue Danube sells an item for RM 100 less 20% while Yellow River sells the same item for RM 120 less 40%
Find the net prices of the item for the two shops.
What further discount percentage must offered by the shop that sells at a higher net price in order to meet the competitor’s price?
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70. REV 00
5.2 CHAIN DISCOUNT
A trade discount in a series of two or more successive discounts.
Wholesaler lists the chain discount as a group, for example 15%, 10%, 5%.
These discounts might also be given in circumstances such as when a large quantity is ordered.
NP = L(1 - r1) (1 – r2) (1 – r3)
Where:
NP = Net price
L = List price
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71. Example 1 pg 97
A computer is advertised for RM 4800 less 20% and 10%. Find a) The net price b) The total discount. Example 2 pg 98 A television set with a catalogue price of RM2500 is offered a chain discount lf 30%, 10% and 5%. Calculate the net price. Example 3 pg 98 A washing machine is advertised at RM 2000 less RM 40%, 12% and 2 ½ %. Find the net price.
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72. 5.3 SINGLE DISCOUNT EQUIVALENT
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5.3 SINGLE DISCOUNT EQUIVALENT
A single discount equivalent is a single discount which is equivalent to a chain discount.
The single discount equivalent, r for a chain discount of r1, r2 and r3 is given by:
r = 1 – (1 - r1) (1 – r2) (1 – r3)
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73. Example 1 pg 99
A product is advertised at RM 1500 less 20%, 10% and 5%. Find a) The single discount equivalent, b) The net price. Example 2 pg 99 Find the single discount equivalent of 10% and 3%
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74. REV 00
5.4 CASH DISCOUNT
Cash discounts are often stated under the heading of invoice as, for example 3/10, 2/20, n/30.
This means, if the buyer pays the invoice within 10 days of the invoice date, the buyer is entitled to receive a 3% discount or, if payment is made within 20 days from the invoice date, the buyer will receive a 2% discount.
N/30 (sometimes written net 30) means the credit period is 30 days.
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75. Example 1 pg100
An invoice dated 2 January 2005 for RM 4010 was offered cash discount terms of 1/10, n/30. if the invoice was paid on 11 January 2005, what was the payment? Example 2 pg 101 An invoice dated 10 April 2005 for RM 2300 was offered cash discount terms if 3/10, 2/20, n/60. find the payment if the invoice was paid on 28 April 2005? Example 3pg 101 The total of an invoice with cash discount terms if 3/10, n/30 amounts to RM 2090 which includes a prepaid freight charge of RM 50. find the amount that is needed to pay the invoice within the cash discount period.
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76. Borrowing to take advantage of the cash discount
Not many businesses have sufficient cash in hand to take advantage if the cash discounts offered. Many if these companies borrow form banks using short term loans to take advantage of the offer.
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77. Example 1 pg 102
On 20 may, Mei Lan purchased some goods invoiced at RM 3000 with cash discount terms of 3/10, n/30. in order to pay the invoice on 30 May, she borrowed the money for 20 days at 9% per annum simple interest. How much did she save by borrowing to the advantage of the discount?
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78. 5.5 PARTIAL PAYMENT OF INVOICE
REV 00
5.5 PARTIAL PAYMENT OF INVOICE
If a buyer pays only part of the invoice within the discount period, he receives a proportionate fraction of the cash discount that is offered.
He will only receive the full amount of the cash discount if he settles all the payment
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79. Example 1pg 102
An invoice amounting to RM 3000 and dated 15 July 2004 offered cash discount terms of 10/15, n/30. find the amount outstanding if the buyer paid RM 1000 on 20 July 2004 Amount paid = (credit given) x (1- discount rate)
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80. Trade and Cash Discount
Example 1 pg 103 An invoice of RM and dated 18 April 2005 was offered 25% trade discount and cash discount terms of 9/10, n/30. find a) The trade discount offered, b) The cash discount offered c) The net payment if the invoice was paid on 28 April 2005
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81. Example 2pg 104
An invoice of RM 9000 dated 19 April 2005 was offered 13% trade discount and cash discount terms of 3/10, n/30. find the net payment if the invoice was paid on 30 April 2005.
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82. CHAPTER 6 MARKUP AND MARKDOWN
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83. 6.1 MARKUP
In order to gain profit from selling, the company must sell their product at a higher price than the product cost.
The difference between a product’s cost and selling price is refereed to as markup.
It can be either in money value or percentage.
The rate of mark-up is known as markup percentage.
Markup = Selling Price – Cost Price
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84. R= C + M 6.2 MARKUP PERCENT Markup is usually expressed as a percent.
It can be expressed as:
a) Markup percent based on retail price (selling price) = %Mr = M/R x 100%
b) Markup percent based on cost price = %Mc = M/C x 100%
R= C + M
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85. Example 1 pg 114
The cost price of an antique table is RM what is the retail price if the seller wants a 20% mark-up based on
a) Cost price
b) Retail price.
Example 2 pg 115
Mariam’s shop purchase 90 shirts at a cost of RM 20 each. The shop expects that 10% of the shirts will be sold at a reduced price of RM each. If the shop is to maintain a 75% mark-up on cost on the entire purchase, find the regular price of the shirts.
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86. Example 3 pg 115
A retailer purchased 200 kg of cucumber at 50 cents per kilogram. A 5% spoilage is expected. If he plans to make a 40% mark-up based on overall cost, what is the selling price of the cucumber?
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87. 6.3 CONVERSION OF MARKUP PERCENT
Markup percent based on retail price:
Since R = C + M
1 + %Mc = 100% + %Mc
Hence :
%Mr = %Mc / 1 + %Mc
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88. Markup percent based on cost price: Since R = C + M
100% = 1 - %MR + %MR
Hence :
%Mc = %MR / 1 - %MR
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89. Example 1 pg 116
a) The mark up percent based in cost price of an item is 20%. What is the mark up based on retail price.
b) The mark up percent based on retail price of an item is 15%. What is its mark up percent based on cost price.
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90. 6.4 MARKDOWN
Sometimes retailer may reduce the marked price due to special promotions, festive seasons or the items being obsolete.
Markdown is the reduction from the selling price or marked price, normally in terms of percentage.
Markdown Amount = Old Selling Price – New Selling Price
Markdown percent based on old price, %MD = MD/OP x 100%
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91. a) the regular price of the oven b) the cost of the oven
Example 1 pg 117
The markdown percent on a TV set is 10%. If the new retail price is RM 900, find the old retail price.
Example 2 pg 117
During a clearance sale, an appliance department marked down a microwave oven by 12%, making the selling price RM 400. at this selling price, the department make a 30% markup on the selling price. Find
a) the regular price of the oven
b) the cost of the oven
c) the mark up per cent of the oven at the regular price.
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92. The actual selling price The list price.
Example 3 pg 118
A retailer wants to sell an item that costs RM 200 at a price less 15% discount that will give him a 28% mark up based on cost. Find
The actual selling price
The list price.
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93. 6.5 PROFIT AND LOSS
Business does not always generate income or profit.
If the business is not managed properly, it may increase operating expense and indirectly increase cost of the product.
The actual cost of the product is the purchase price plus operating expenses.
In accounting, operating expenses include cost of goods sold, official rental, advertising, salary, commission and so on.
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94. In general, there are three possibilities in business:
a) M = OE : Break-even
b) M > OE : Profit
c) M < OE : Loss
The mark up equation, retail price= Cost + markup
Retail price= cost+ net profit+ Operation expenses
R= C+ NP+ OE
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95. FORMULA: Break-even Price, BEP = Cost Price + OE
Retail Price = Cost + Net Profit + OE
Net Profit / (Loss) = Retail Price – BEP
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96. f) the net profit or loss of the retail price was RM 280.00
Example 1 pg 119
A retailer bought a radio for RM 200. buying expenses amounted to RM 20. operating expenses incurred were 20% of the cost price. If the retailer made a 25% net profit based in cost, find
a) the retail price
b) the gross profit
c) the net profit
d) the breakeven price
e) the maximum markdown that could be offered so that there is no profit or loss
f) the net profit or loss of the retail price was RM
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97. Example 2 pg 119
A dealer bought a hi-fi set fir RM 2000 less 10% and 5%. He sold it at a discount of 20%. If the gross profit earned by the dealer is 20% in the net retail price, find the list price of the hi-fi set.
Example 3 pg 119
An item costing RM 200 was listed in a catalogue at RM 400 with a trade discount of 20%. After some time, the cost of the item decreased to RM 180. if the dealer wants to maintain the same mark-up percent as before the cost reduction, find the extra trade discount that may be given to a customer.
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98. d) the net loss or profit if the retail price is RM 1200.
Example 4 pg 121
A retailer purchased 12 watched for RM the operating expenses incurred for the sale of the watched were 20% if the cost. The retailer made a 30% net profit based on cost/ for each watch, find
a) the selling price.
b) the breakeven price,
c) the maximum markdown percent that could be offered without incurring any loss.
d) the net loss or profit if the retail price is RM 1200.
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99. Example 5 pg 122
A retailer bought a computer for RM the estimated operating expenses incurred for the sale of the computer are 5% of the retail price. If the retailer wants a 20% net profit based in the retail price, find
a) the retail price
b) the net profit
c) the mark- up
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100. CHAPTER 7 PROMISSORY NOTE
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100

101. 7.1 PROMISSORY NOTES
A written document made by one person or party to pay a stated sum of money a specified future date to another person or party.
Are negotiable documents and can be of two types; interest bearing notes and non-interest bearing notes.
The main features of a promissory note are as follows:
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101

102. The maker is the person that signs the note. 2. Payee
The payee is the person to whom the payment is to be made.
3. Date of the note
The date of the note is the date on which the note is made.
4. Term of the note
The term of the note is the length of time until the note is due for payment.
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102

103. The face value of the note is the amount stated on the note.
6. Maturity value
The maturity value of the note is the total sum of money which the payee will receive on the maturity date. The maturity value of a non-interest bearing note is the face value while the maturity value of an interest-bearing note is the face value plus day interest that is due.
7. Maturity date
The maturity date of the note is the date on which the maturity value is due.
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103

104. EXAMPLE OF PROMISSORY NOTE
RM APRIL 2005
Sixty days after date I promise to pay the
order of Mohammed Ali
Ringgit Malaysia: Two thousand five hundred only
for value received with interest at the rate of 8.00%
per annum until paid.
No Due: 19 June 2005
Mat Jenin
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105. In the promissory note above, a) who is the maker of the note,
Example 1 pg 135
In the promissory note above,
a) who is the maker of the note,
b) who is the payee of the note
Calculate the maturity value of the date.
Example 2 pg 135
a promissory note dated 22 February 2005 reads ‘ three months from date, I promise to pay RM with interest at 9% per annum.’ Find
a) the maturity date of the note.
b) the maturity value of the note.
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106. Example 3 pg 136
The maturity value of a 60 day interest bearing promissory note is RM 450. if the interest rate is 6% per annum, what is the face value of the note?
Example 4 pg 136
The interest on a 90 day promissory note is RM 46. if the interest rate is 7% per annum, find the face value of the note.
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107. 7.2 BANK DISCOUNT
Bank discount is computed in much the same way as simple interest except that it is based on the final amount (to be paid back) or maturity value.
D = Sdt
Where:
D = Bank discount
S = Amount of maturity value
d = Discount value
t = term of discount in years
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107

108. Bank proceeds = Maturity Value – Bank Discount P = S – D P = S – Sdt
P = S(1 – dt)
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108

109. Example 1
Sharifah borrows RM 8000 for three months from a lender who charges a discount rate of 10%. find a. the discount b. the proceeds. Example 2 If Tong needs RM 4000 now, how much should he borrow from his bank for 1 ½ years at 12% bank discount
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110. Example 3
Sheela receives an invoice of RM 2000 with cash discount terms of 3/10, n/40
How much should be borrowed for 30 days from a bank that charges a 9% discount rate to take advantage of the cash discount.
How much will be saved by borrowing the money to take advantage of the cash discount.
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111. 7.3 SIMPLE INTEREST RATE EQUIVALENT TO BANK DISCOUNT RATE
At interest rate, r% and a discount rate, d% are said to be equivalent if the two rates give the same present value for an amount due in the future.
If the present value is S, then the present value of S, at r% simple interest rate is S(1 + rt)-1 and the present value of S, at d% bank discount rate is S(1 – dt).
Equating the two present value, we get:
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111

112. 1/1 + rt = 1 – dt Solving for d, we get dt = 1 – 1/1 + rt
S(1 + rt)-1 = S(1 – dt)
1/1 + rt = 1 – dt
Solving for d, we get
dt = 1 – 1/1 + rt
dt = rt / 1 +rt
d = r / 1 + rt
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112

113. Solving for r, we get 1 + rt = 1 / 1 – dt rt = 1 / 1 – dt -1
rt = dt / 1 – dt
r = d / 1 - dt
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113

114. Example 1 pg 139
A bank discounts a RM 4000 note due in six months using a bank discount rate of 12%. Find the equivalent simple interest rate that is charged by the bank. Example 2 pg 140 What discount rate should a leader charge to earn an interest rate of 20% on nine-month loan?
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115. 7.4 DISCOUNTING PROMISSORY NOTES
A promissory note can be sold to a bank before its maturity date if the holder is in need of cash.
Selling the note to the bank is called discounting the note.
The amount received on the date of discounting is called the proceeds.
The proceeds of a promissory note are computed as follows:
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115

116. Maturity value = Face Value + Interest
Find the maturity value of the note. For non-interest bearing note, it is the face value. If the note is interest bearing, then
Maturity value = Face Value + Interest
Find the bank discount, D with the formula
D = Sdt
Compute the proceeds
Proceeds = Maturity Value – Bank Discount
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116

117. Example 1 pg 141
Marina , a businesswoman, received a promissory note for RM 1500 with interest at 10% per annum that was due in 60 days. The note was dated 10 April The note was discounted on 15 April 2005 at a bank that charges 12% discount. Determine a. the maturity date b. the maturity value c. the discount period d. the proceed.
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118. Example 2 pg 141
On 22 February 2005, Rahman received a 90 day promissory note with a simple interest rate of 8% per annum. On 13 April 2005, he discounted the note at 7%. The proceeds he received were RM find a. the maturity date of the note b. the maturity value of the note c. the face value of the note d. the simple interest rate earned by the bank which is equivalent to the discount rate.
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119. Example 3
Tai Sing Auto company had a note dated 15 December 2005 for RM 4800 with interest at 8% per annum. The term of the note was three months. If the company discount the note on 30 January 2006 at a bank that charged a discount rate of 7%, what were the proceeds?
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120. CHAPTER 8 INSTALMENT PURCHASES
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121. 8.1 INSTALMENT PURCHASES
Many items like electrical appliances can either be purchased on cash term or instalment basis.
In an instalment purchase normally a down payment is made, to be followed by a series of regular payments (usually monthly or weekly).
All hire purchase sales in Malaysia are controlled by the Hire Purchase Act (1967).
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122. 8.2 INTEREST CHARGE BASED ON ORIGINAL BALANCE
Simple interest formula is used to calculate the interest charged.
Original Balance = Cash Price – Down Payment
Installment Price = Cash price + Total Interest or
Installment Price = Down Payment + Total Monthly Payment
Monthly Payment = Original Total Balance Interest
Number of payments
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123. b. the installment price c. the monthly payment Example 2 pg 154
Jenny bought a refrigerator listed at RM 800 cash trough an installment plan. She paid RM 100 as a down payment. The balance was settled by making ten monthly installments. If the interest rate charged was 8.5% per annum on the original balance, find
a. the total interest
b. the installment price
c. the monthly payment
Example 2 pg 154
Marianna bought an electric appliance through an instalment plan in which she paid RM 200 down. She had to make 12 monthly payments of RM 120 each to settle the unpaid balance. If the dealer charged her an interest of 5% per annum on the original balance, find the cash price of the item.
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124. 8.3 INTEREST CHARGE BASED ON REDUCING BALANCE
Interest charged on reducing balance is an annual rate which is applied only to he balance due at the time of each payment.
Two methods of reducing balance namely:
Annuity method
Constant ration method
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125. Annuity method:
It is also called as amortization method. If A is the amount of loan borrowed, I the interest rate per interest period and n the number of interest periods or the number of installment repayments, then
A = R [1 – (1 + i)-n ]/ i
Where R is the instalment payment for each period. Solving for R, we get:
R = Ai / (1 – (1 + i)-n )
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126. b) the total interest charged c) the instalment price
Example 1 pg 155
A washing machine is selling for RM 2000 cash. Through an instalment purchase, the buyer has to pay RM 400 down and ten monthly instalments. If the interest charged is 8% per annum on reducing balance, find
a) the monthly payment
b) the total interest charged
c) the instalment price
By using the annuity method
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127. Example 2 pg 156
Jasmin purchased some equipment from a wholesaler. The wholesaler offered her terms under which 12% of the purchase price will be added to the purchase price and the debt would be settled by 12 monthly payments. She can borrow from a finance company which charges 15% compounded monthly and repay the finance company by making 12 monthly payments. Which alternative should Jasmin choose?
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128. Constant Ratio formula
Frequently used to approximate the actual annual percentage rate APR or effective rate.
r = 2MI / B(n + 1)
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129. r = Annual interest rate
Where
r = Annual interest rate
M = 12 for monthly instalments and 52 for weekly instalments
I = Total interest charged for instalment plan
B = Original balance outstanding or principal of original debt
N = Total number of instalments
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130. The Constant Ratio formula can also be used to calculate total interest charged if interest rate on reducing balance is given, that is:
I = B(n + 1)r / 2M
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131. a) the total interest charged b) the monthly payment
Example 1 pg 158
A washing machine is being sold for RM 2000 cash. Through an insatlment purchase, the buyer has to pay RM 400 down and 10 monthly instalments. If the interest charged is 8% per annum on the reducing balance, find
a) the total interest charged
b) the monthly payment
c) the instalment price
By using the Constant Ration Formula
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132. b) total interest charged c) flat rate (simple interest rate) charged
Example 2 pg 159
Zaleha purchased a RM 8000 piano through an instalment plan. She has to pay RM 2000 down and 18 monthly payments of RM 350 each. Find the
a) instalment price
b) total interest charged
c) flat rate (simple interest rate) charged
d) approximate APR by using the Constant Ratio Formula
Example 3 pg 160
Nelly bought a hi-fi set listed at RM 800 cash through an instalment plan in which she had to make six monhtly instalment payments at 10% per annum simple interest. By using the Constant Ratio Formula, what was the approximate effective rate charged by the dealer.
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133. 8.4 UNEQUAL INSTALMENT PAYMENTS AND REPAYMENTS SCHEDULES
Some instalment purchases may not require equal payments.
There may be a variation on the method which total interest for the whole term of payment is distributed equally.
And it must be noted that this type of variation is identical to the Constant Ratio method.
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134. Example 1 pg 161
Diana bought a RM 4600 stereo set on an instalment basis in which an interest of 1% per month in any outstanding balance was charged. She made a RM 1000 down payment. For the balance, she had to pay RM 600 every month (principal payment) plus any interest due. Construct an repayment schedule to show the monthly payments.
Example 2 pg 162
Rosita bought a television worth RM 3000 on an instalment basis in which she was charged 1 ¼ % per month on any outstanding balance. She made a RM 1000 down payment and paid RM 600 every month. Find the number of payments she made and the value of the final payment. Construct a repayment schedule.
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135. 8.5 RULE OF 78 IN HIRE PURCHASE ACT (1967)
In Malaysia under the Hire And Purchase Act (1967) which controls the hire purchase agreement, the Rule of 78 (which is sometimes called the sum of digits method) is used to calculate the balance outstanding of any hire purchase agreement.
The Rule of 78 states that the outstanding balance of a hire purchase loan with a flat (simple) interest rate is given by:
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136. N = Number of payments yet to be settled I = Total interest charged
B = RN – I[ …… + N / …… + n]
B= RM – I
Where:
R = Monthly payment
N = Number of payments yet to be settled
I = Total interest charged
n = Total number of payments
N (N+1)
n ( n+1)
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137. a) the total interest charged b) the monthly payment
Example 1 pg 164
The finance charge in one year hire purchase loan is RM 390. find the interest that was unearned by the lender if the loan was settled two months early.
Example 2 pg 164
A loan of RM at a flat rate of 10% per annum was repaid by making 24 monthly instalments. Find
a) the total interest charged
b) the monthly payment
c) the outstanding loan just after the tenth payment using Rule of 78
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138. Example 3 pg 165
Juliet purchased a car listed at RM form Car Finance Bhd. trough a hire purchase agreement in which she had to pay RM down ad 24 monthly instalments of RM 2000 each. After one year of payment defaults, the car was legally repossessed and sold for RM find the amount of refund that she would be receive form the company for the amount of the money she would have to pay.
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139. CHAPTER 9
DEPRECIATION
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140. 9.1 DEPRECIATION
Depreciation is an accounting procedure for allocating the cost of capital assets, such as buildings, machinery tools and vehicles over their useful life.
Can also be viewed as decline in value of assets because of age, wear and tear or decreasing efficiency.
Many properties such as buildings, machinery, vehicles, and equipment depreciate in value as they get older.
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141. Several terms are commonly used in calculation relating to depreciation:
Original cost
The original cost of an asset is the amount of money paid for an asset plus many sales taxes, delivery charges, installation charges and other costs incurred.
Salvage value
The salvage value (scrap value or trade-in value) is the value of an asset at the end of its useful life.
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142. 3. Useful life
The useful life is the expectancy of the asset or the number of years the asset is expected to last.
4. Total depreciation
The total depreciation or the wearing value of an asset is the difference between cost and scrap value.
5. Annual depreciation
The annual depreciation is the amount of depreciation in a year. It may or may not be equal from year to year.
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143. The accumulated depreciation is the total depreciation to date.
Book value
The book value or carrying value of an asset is the value of the asset as shown in the accounting record. It is the difference between the original cost and the accumulated depreciation charged to that date.
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144. Three methods of depreciation are commonly used. These methods are:
- Straight line method
- Declining balance method
- Sum of years digits method
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145. 9.2 STRAIGHT LINE METHOD
The simplest of the three methods and probably the most common method used.
The total amount of depreciation is spread evenly to each accounting period through the useful life of the asset.
Annual depreciation
= Cost – Salvage Value / Useful life
= Total depreciation / Useful life
Annual rate of depreciation
= Annual depreciation / Total depreciation x 100%
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146. Book value = Cost – Accumulated depreciation
= 1 / Useful life x 100%
Book value = Cost – Accumulated depreciation
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147. a) calculate the annual depreciation
Example 1 pg 181
Lau company bought a lorry for RM The lorry is expected to last five years and its salvage value at the end if five years is RM using the straight line method,
a) calculate the annual depreciation
b) calculate the annual rate of depreciation
c) Calculate the book value of the lorry at the end of the third years
d) Prepare a depreciation schedule
Example 2 pg 182
The book value of an asset after the third year and fifth year using the straight line method are RM 700 and RM 5000 respectively. What is the annual depreciation of the asset?
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148. 9.3 DECLINING BALANCE METHOD
Declining balance method is an accelerated in which higher depreciation charges are deducted in the early life of the asset.
BV = C (1 – r)n
Where:
BV = Book value
C = Cost of asset
r = Rate of depreciation
n = Number of years
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149. where r= annual rate of depreciation n= useful life in years
The book value at the end of the useful life is the salvage value, S. Hence, the annual rate of depreciation is given by
r= 1- n√ s/c
where r= annual rate of depreciation
n= useful life in years
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150. a) find the annual rate of depreciation
Example 1 pg 183
The cost of a fishing boat is RM The declining balance method is used for computing depreciation. If the depreciation rate is 15%, compute the book value and accumulated depreciation of the boat at the end of five years.
Example 2 pg 183
Given
Cost of asset = RM 15000
Useful life = 4 years
Scrap value= RM 3000
a) find the annual rate of depreciation
b) construct the depreciation schedule
Using the declining method
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151. 9.4 SUM OF YEARS DIGITS METHOD
Based on the sum of the digits representing the number of years of useful life of the asset.
If an asset has a useful life of 3 years, the sum of digits is S = = 6
It can be calculated with the formula:
S = n(n + 1) / 2
Where:
S = Sum of years digits
n = Useful life
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152. Example 1 pg 185
A machine is purchase for RM Its life expectancy is five years with a zero trade in value. Prepare a depreciation schedule using the sum of year digits methods.
Example 2 pg 186
A computer is purchased for RM It is estimated that its salvage value at the end of eight years will be RM 600. Find the depreciation and the book value of the computer for the third year using the sum of year digits method.
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