Simple interest. Simple interest is always calculated on:

1. the original amount invested (the principal) only
2. the amount at the start of each year, including past interest
3. the final amount after all the years
4. the interest earned so far

Answer: the original amount invested (the principal) only

Simple interest is a fixed amount each year, worked out on the PRINCIPAL only.

Tip: Compound interest is the one that adds interest onto interest.

Simple interest. R8 000 is invested at 9% per annum simple interest for 5 years. Find the value of the investment.

1. \(R11\,600\)
2. \(R12\,308{,}99\)
3. \(R3\,600\)
4. \(R8\,360\)

Answer: \(R11\,600\)

\(A = P(1 + i\,n) = 8\,000(1 + 0{,}09 \times 5)\)

\(A = 8\,000(1{,}45)\)

Tip: Simple interest keeps \(n\) OUTSIDE the bracket.

Compound interest. R8 000 is invested at 9% per annum compounded annually for 5 years. Find the value.

1. \(R12\,308{,}99\)
2. \(R11\,600\)
3. \(R3\,600\)
4. \(R10\,908{,}99\)

Answer: \(R12\,308{,}99\)

\(A = P(1 + i)^{n} = 8\,000(1{,}09)^{5}\)

Tip: Compound beats simple over time — interest earns interest.

Reducing-balance decay. On the diminishing-balance (reducing-balance) method, each year the value is multiplied by:

1. \((1 - i)\)
2. \((1 + i)\)
3. \((1 - i\,n)\)
4. \(i\)

Answer: \((1 - i)\)

Decay means the value shrinks, so use \(1 - i\); it is applied again each year on the reduced value.

Tip: Growth is PLUS, decay is MINUS.

Reducing-balance decay. A car worth R180 000 depreciates at 12% per annum on a reducing-balance basis. Find its value after 4 years.

1. \(R107\,945{,}16\)
2. \(R93\,600\)
3. \(R283\,224{,}96\)
4. \(R110\,764{,}80\)

Answer: \(R107\,945{,}16\)

\(A = P(1 - i)^{n} = 180\,000(1 - 0{,}12)^{4}\)

\(A = 180\,000(0{,}88)^{4}\)