Finance, growth and decay-G11-Maths-T4P1

Finance, Growth and Decay

Finance calculations involve money invested, borrowed or reduced in value over time. Growth occurs when an amount increases, while decay occurs when it decreases. Always convert percentage rates to decimals before substituting into a formula.

Simple interest

Simple interest is calculated only on the original principal amount. The interest earned or charged is the same during every time period.

The accumulated amount is:

A = P (1 + i n)

The simple interest is:

I = P i n

In these formulas, P is the principal amount, A is the accumulated amount, i is the interest rate per time period written as a decimal, and n is the number of time periods.

Example

Calculate the accumulated value of an investment of R12 000 at 8% simple interest per annum for 3 years.

P = 12 000, i = \frac{8}{100} = 0.08, n = 3

A = P (1 + i n)

A = 12 000 (1 + 0.08 \times 3)

A = 12 000 (1.24)

A = R 14 880

The interest earned is:

I = A - P

I = 14 880 - 12 000 = R 2 880

Compound interest

Compound interest is calculated on the principal and on interest already added. Interest therefore earns further interest.

The accumulated amount is:

A = P {(1 + i)}^{n}

Example

R15 000 is invested at 9% compound interest per annum for 4 years. Calculate the accumulated amount.

P = 15 000, i = 0.09, n = 4

A = P {(1 + i)}^{n}

A = 15 000 {(1 + 0.09)}^{4}

A = 15 000 {(1.09)}^{4}

A = 21 173.72

A = R 21 173.72

Round money to the nearest cent unless instructed otherwise.

Compound growth

Compound growth occurs when a quantity increases by the same percentage during each time period. The growth factor is greater than one.

A = P {(1 + i)}^{n}

The original amount can be calculated by rearranging the formula:

P = \frac{A}{{(1 + i)}^{n}}

Example

The value of a property increases by 6% per annum. Find its value after 5 years if its current value is R850 000.

P = 850 000, i = 0.06, n = 5

A = 850 000 {(1.06)}^{5}

A = 1 137 292.39

A = R 1 137 292.39

Compound decay

Compound decay occurs when a quantity decreases by the same percentage during every time period. The decay factor is less than one.

A = P {(1 - i)}^{n}

Example

A machine is worth R80 000 and loses 12% of its value each year. Calculate its value after 3 years.

P = 80 000, i = 0.12, n = 3

A = P {(1 - i)}^{n}

A = 80 000 {(1 - 0.12)}^{3}

A = 80 000 {(0.88)}^{3}

A = R 54 517.76

Common Mistake

For decay, subtract the rate from one. Do not use a negative exponent or add the rate.

Depreciation

Depreciation is the decrease in the value of an asset over time. Vehicles, machinery and electronic equipment normally depreciate because of age, use and wear.

The original purchase price is called the cost price. The value after depreciation is called the book value. Two common methods are straight line depreciation and reducing balance depreciation.

Straight line depreciation

With straight line depreciation, the same rand amount is deducted every year. Depreciation is calculated on the original cost price.

The book value is:

A = P (1 - i n)

The depreciation per year is:

D = P i

Example

A vehicle costs R240 000 and depreciates at 15% per annum according to the straight line method. Calculate its book value after 4 years.

P = 240 000, i = 0.15, n = 4

A = P (1 - i n)

A = 240 000 (1 - 0.15 \times 4)

A = 240 000 (0.40)

A = R 96 000

The annual depreciation is:

D = 240 000 (0.15)

D = R 36 000

Reducing balance depreciation

With reducing balance depreciation, depreciation is calculated on the current book value. The rand amount of depreciation becomes smaller each year.

A = P {(1 - i)}^{n}

Example

Equipment costing R150 000 depreciates at 20% per annum on the reducing balance method. Find its book value after 3 years.

P = 150 000, i = 0.20, n = 3

A = 150 000 {(1 - 0.20)}^{3}

A = 150 000 {(0.80)}^{3}

A = R 76 800

Remember

Straight line depreciation is a simple decay model. Reducing balance depreciation is a compound decay model.

Inflation

Inflation is the general increase in the prices of goods and services over time. If inflation remains at a constant percentage, compound growth is used.

A = P {(1 + i)}^{n}

Example

A basket of goods currently costs R2 500. Estimate its cost after 4 years if inflation is 5.5% per annum.

P = 2 500, i = 0.055, n = 4

A = 2 500 {(1.055)}^{4}

A = 3 096.94

A = R 3 096.94

Inflation reduces the purchasing power of money. This means that the same amount of money buys fewer goods in the future.

Population growth

Population growth is usually modelled using compound growth when the population grows by a fixed percentage each year.

A = P {(1 + i)}^{n}

Example

A town has a population of 48 000 people. The population grows by 2.4% per annum. Estimate the population after 6 years.

P = 48 000, i = 0.024, n = 6

A = 48 000 {(1.024)}^{6}

A = 55 342.79

A population must be given as a whole number:

A \approx 55 343

Exchange rates

An exchange rate compares the values of two currencies. Decide whether to multiply or divide by considering the direction of the conversion.

Suppose the exchange rate is:

1 USD = R 18.50

To convert US dollars to rand, multiply by the exchange rate.

Example

Convert 350 US dollars to rand.

350 \times 18.50 = 6 475

350 USD = R 6 475

To convert rand to US dollars, divide by the exchange rate.

Example

Convert R9 250 to US dollars.

\frac{9 250}{18.50} = 500

R 9 250 = 500 USD

Exam Tip

Check whether the answer should be larger or smaller. When converting rand to a stronger currency, the numerical answer will normally be smaller.

Nominal interest rates

A nominal interest rate is the annual rate quoted before considering the effect of compounding during the year. Divide it by the number of compounding periods per year to obtain the rate per period.

i_{period} = \frac{i_{nominal}}{m}

Here, m is the number of compounding periods per year.

Example

Find the monthly interest rate for a nominal rate of 12% per annum compounded monthly.

i_{monthly} = \frac{0.12}{12}

i_{monthly} = 0.01

i_{monthly} = 1 %

Effective interest rates

The effective annual interest rate is the actual percentage increase over one year after all compounding has been included.

i_{effective} = {(1 + \frac{i_{nominal}}{m})}^{m} - 1

Example

Calculate the effective annual rate corresponding to 12% per annum compounded monthly.

i_{effective} = {(1 + \frac{0.12}{12})}^{12} - 1

i_{effective} = {(1.01)}^{12} - 1

i_{effective} = 0.126825

i_{effective} \approx 12.68 %

The effective rate is higher than the nominal rate because interest is compounded during the year.

Different compounding periods

Interest may be compounded annually, half-yearly, quarterly, monthly or daily.

The common values of m are:

Annually: one compounding period per year.

Half-yearly: two compounding periods per year.

Quarterly: four compounding periods per year.

Monthly: twelve compounding periods per year.

Daily: usually 365 compounding periods per year.

For a nominal annual rate compounded m times per year:

A = P {(1 + \frac{i}{m})}^{m n}

Example

R20 000 is invested at 10% per annum compounded quarterly for 3 years. Calculate the accumulated amount.

P = 20 000, i = 0.10, m = 4, n = 3

The rate per quarter is:

\frac{i}{m} = \frac{0.10}{4} = 0.025

The number of compounding periods is:

m n = 4 \times 3 = 12

A = 20 000 {(1 + \frac{0.10}{4})}^{4 \times 3}

A = 20 000 {(1.025)}^{12}

A = R 26 897.78

Exam Tip

The interest rate and the number of periods must refer to the same time interval. If interest is compounded monthly, use a monthly rate and the total number of months. Avoid rounding intermediate answers; round only the final monetary answer to the nearest cent.

Study Guide For Grade

11

Mathematics