Q: Given the third graph, if R2 500 grows to R3 250 in 6 years using simple interest, what is the annual simple interest rate?

$$i = \frac{A - P}{P \times n} = \frac{3\ 250 - 2\ 500}{2\ 500 \times 6} = 0,05 = 5\%$$

Q: A loan of R900 000 is to be repaid with monthly payments of R25 000 at 11,25% p.a. compounded monthly. How many months will it take to repay the loan?

$$n = \frac{\log(1-\frac{0,1125}{12} \times \frac{900\ 000}{25\ 000})}{\log(1+\frac{0,1125}{12})} = 44,6 \text{ months}$$

Q: If an investment of R50 000 earns 8% p.a. compounded quarterly for 5 years, what is the accumulated amount?

$$A = 50\ 000(1+\frac{0,08}{4}$$)^{4 \times 5}$$ = R74\ 207,37$$

Question 1

Refer to the first graph (Compound vs Simple Interest, 8% per year for 40 years). If you invest R500 at 8% per year for 40 years, what is the main reason the compound interest curve grows much faster than the simple interest curve?

Accepted Answer

Compound interest is calculated on both the initial principal and the accumulated interest, while simple interest is only calculated on the initial principal.

Question 2

Using the formula for compound interest, calculate the future value after 40 years if R500 is invested at 8% per annum, compounded annually.

Accepted Answer

$$500(1+0,08$$)^{40}$$ = R5\ 426$$

Question 3

Refer to the second graph (two investments over 5 years). If both investments start at R8 000, and after 5 years one is worth R12 000 and the other about R13 500, which investment used compound interest?

Accepted Answer

The investment that reaches about R13 500 after 5 years.

Question 4

Given the third graph, if R2 500 grows to R3 250 in 6 years using simple interest, what is the annual simple interest rate?

Accepted Answer

$$i = \frac{A - P}{P 	imes n} = \frac{3\ 250 - 2\ 500}{2\ 500 	imes 6} = 0,05 = 5\%$$

Question 5

In the fourth graph, two investments (A and B) start at R20 000. After 12 years, investment A is worth more than B. If A used compound interest and B used simple interest, which formula best describes the value of A after n years?

Accepted Answer

$$A = 20\ 000(1+i)^n$$

Question 6

A machine is purchased for R550 000. After 10 years, its replacement cost is R2 225 056,76 due to 15% annual inflation, and its scrap value is R238 913,65 due to 8% annual depreciation. What is the required sinking fund value to replace the machine?

Accepted Answer

$$R2\ 225\ 056,76 - R238\ 913,65 = R1\ 986\ 143,11$$

Question 7

If an investment of R10 000 grows to R21 589,25 in 10 years at a certain annual interest rate, compounded annually, what is the interest rate?

Accepted Answer

$$i = (\frac{21\ 589,25}{10\ 000}$$)^{1/10}$$ - 1 = 0,08 = 8\%$$

Question 8

A company wants to set up a sinking fund to replace equipment in 5 years. The equipment will cost R184 634,87 in 5 years, and the scrap value will be R53 244,64. If the sinking fund earns 12,5% p.a. compounded monthly, what is the monthly payment required (first payment immediately, last at end of 5 years)?

Accepted Answer

$$x = \frac{184\ 634,87 - 53\ 244,64}{\frac{(1+\frac{0,125}{12}$$)^{61}$$-1}{\frac{0,125}{12}}} = R1\ 552,43$$

Question 9

If a loan of R130 000 is to be repaid over 5 years at 13% p.a. compounded monthly, with the first payment due after 6 months, what is the amount used to calculate the monthly repayments?

Accepted Answer

$$A = 130\ 000(1+\frac{0,13}{12})^6$ = R137\ 195,90$$

Question 10

A loan of R900 000 is to be repaid with monthly payments of R25 000 at 11,25% p.a. compounded monthly. How many months will it take to repay the loan?

Accepted Answer

$$n = \frac{\log(1-\frac{0,1125}{12} 	imes \frac{900\ 000}{25\ 000})}{\log(1+\frac{0,1125}{12})} = 44,6 	ext{ months}$$

Question 11

If an investment of R50 000 earns 8% p.a. compounded quarterly for 5 years, what is the accumulated amount?

Accepted Answer

$$A = 50\ 000(1+\frac{0,08}{4}$$)^{4 	imes 5}$$ = R74\ 207,37$$

Question 12

A car is purchased for R250 000 and depreciates at 12% p.a. using the reducing-balance method. What is its value after 5 years?

Accepted Answer

$$A = 250\ 000(1-0,12)^5$ = R138\ 070,37$$

Question 13

If you want to double your investment in 6 years at an interest rate compounded monthly, what is the required annual interest rate?

Accepted Answer

$$i = 12 \left[ (2$$)^{1/72}$$ - 1 \right] \approx 0,1157 = 11,57\%$$

Question 14

A loan of R100 000 is to be repaid over 20 years at 18% p.a. compounded monthly, with payments starting three months after the loan is granted. What is the value of the loan after two months?

Accepted Answer

$$A = 100\ 000(1+\frac{0,18}{12})^2$ = R103\ 030,45$$

Question 15

If you invest R500 monthly for 10 years at 15% p.a. compounded monthly, what is the future value of the investment?

Accepted Answer

$$F = 500 \frac{(1+\frac{0,15}{12}$$)^{120}$$-1}{\frac{0,15}{12}} = R139\ 679,57$$

Financial Mathematics: Interest, Annuities, Depreciation, and Sinking Funds

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