DSC 1630
ASSIGNMENT 3 SEMESTER 2 2022
Use this document as a guide to learn, compare and reference
, 1 Elise buys a house and makes a down payment of 35% of the price of the house. She manages to
secure a loan at an interest rate of 13,25% per year, compounded monthly, for a period of 20 years. Her
monthly payment is R10 437,26.
The size of the down payment (to the nearest rand) is
AR527 959.
B.R570 375.
C.R307 125.
D.R472 500.
. The present value of an ordinary annuity can be calculated by the formula P = Ra n i . Thus the present
value of the loan is:
P = 10 437,26a 20×12 0,1325 12
P = 877 500,4207 . . . The value of the loan to the nearest rand is R887 500.
Now we know that 877 500 = 65% of H = 65 100 × H (1)
X = 35% of H = 35 100 × H (2)
From (1) 65 100 × H = 877 500
i.e. H = 100 65 × 877 500.
Substituting this value of H into (2) we get X = 35 100 × 100 65 × 877 500
i.e. X = 35 65 × 877 500 = 472 500. The down payment was R472 500.
2The following is an extract from the amortisation schedule of a home loan:
Outstanding Interest Outstanding
principal at the due at Monthly Principal principal at the
Month beginning of
month payment repaid end of the
the month end month
R2 R2
147 R8 155,83 C R6 141,56
080,54 014,27
R2 R2
148 R6 141,56 R49,90 D
080,54 030,64
R2 R2
149 D R33,40 R2 063,78
080,54 047,14
R2 R2
150 R2 063,78 R16,77 0
080,54 063,77
The value of C is
A.R49,50.
B.R166,33.
C.R66,27.
D.R41,65.
The value of C, which is the interest due at month end 147, can be calculated as:
Monthly payment at month 147 − principal repaid at month 147 Value of C = 2 080,54 − 2 014,27
= 66,27
The value of C is R66,27
1/8
ASSIGNMENT 3 SEMESTER 2 2022
Use this document as a guide to learn, compare and reference
, 1 Elise buys a house and makes a down payment of 35% of the price of the house. She manages to
secure a loan at an interest rate of 13,25% per year, compounded monthly, for a period of 20 years. Her
monthly payment is R10 437,26.
The size of the down payment (to the nearest rand) is
AR527 959.
B.R570 375.
C.R307 125.
D.R472 500.
. The present value of an ordinary annuity can be calculated by the formula P = Ra n i . Thus the present
value of the loan is:
P = 10 437,26a 20×12 0,1325 12
P = 877 500,4207 . . . The value of the loan to the nearest rand is R887 500.
Now we know that 877 500 = 65% of H = 65 100 × H (1)
X = 35% of H = 35 100 × H (2)
From (1) 65 100 × H = 877 500
i.e. H = 100 65 × 877 500.
Substituting this value of H into (2) we get X = 35 100 × 100 65 × 877 500
i.e. X = 35 65 × 877 500 = 472 500. The down payment was R472 500.
2The following is an extract from the amortisation schedule of a home loan:
Outstanding Interest Outstanding
principal at the due at Monthly Principal principal at the
Month beginning of
month payment repaid end of the
the month end month
R2 R2
147 R8 155,83 C R6 141,56
080,54 014,27
R2 R2
148 R6 141,56 R49,90 D
080,54 030,64
R2 R2
149 D R33,40 R2 063,78
080,54 047,14
R2 R2
150 R2 063,78 R16,77 0
080,54 063,77
The value of C is
A.R49,50.
B.R166,33.
C.R66,27.
D.R41,65.
The value of C, which is the interest due at month end 147, can be calculated as:
Monthly payment at month 147 − principal repaid at month 147 Value of C = 2 080,54 − 2 014,27
= 66,27
The value of C is R66,27
1/8
