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Study Guide For Grade

12

Mathematics

Term 4
Paper 1

Finance, Growth and Decay

Financial mathematics deals with money that increases or decreases over time. In every calculation, identify the principal amount, interest rate per payment or compounding period, and number of periods. Convert percentages to decimals and ensure that the interest period agrees with the time period.

The examination-style methods and values below follow the financial mathematics solutions in the supplied marking guidance.

Simple interest

Simple interest is calculated only on the original principal. The interest earned during each period is therefore constant.

The formula is:

A=P(1+in)A=P(1+in)

Here:

  • A is the accumulated or future amount.
  • P is the original principal.
  • i is the interest rate per period, written as a decimal.
  • n is the number of interest periods.

The simple interest earned is:

I=A−PI=A-P

Example

Calculate the accumulated amount when R12 000 is invested for 4 years at 8,5% simple interest per annum.

A=P(1+in)A=P(1+in)
A=12000(1+0,085(4))A=12\,000\left(1+0{,}085(4)\right)
A=12000(1,34)A=12\,000(1{,}34)
A=R16080A=\text{R}16\,080

The interest earned is:

I=A−PI=A-P
I=16080−12000I=16\,080-12\,000
I=R4080I=\text{R}4\,080

Common Mistake

Do not use simple interest when the question states that interest is compounded.

Compound interest

Compound interest is calculated on the principal and on interest already earned. This produces exponential growth.

The formula is:

A=P(1+i)nA=P(1+i)^n

Example

R40 000 is invested at 7,8% per annum, compounded annually, for 5 years. Calculate the accumulated amount.

A=P(1+i)nA=P(1+i)^n
A=40000(1+0,078)5A=40\,000(1+0{,}078)^5
A=40000(1,078)5A=40\,000(1{,}078)^5
A=R58230,94A=\text{R}58\,230{,}94

If compounding occurs several times per year, adjust both the interest rate and the number of periods.

i=inommi=\frac{i_{\text{nom}}}{m}
n=mtn=mt

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

Example

R25 000 is invested for 3 years at 9% per annum, compounded monthly.

i=0,0912i=\frac{0{,}09}{12}
n=12(3)=36n=12(3)=36
A=25000(1+0,0912)36A=25\,000\left(1+\frac{0{,}09}{12}\right)^{36}
A=R32716,13A=\text{R}32\,716{,}13

Growth and decay

Growth occurs when a quantity increases by a fixed percentage during each period. Examples include compound interest, population growth and inflation.

A=P(1+i)nA=P(1+i)^n

Decay occurs when a quantity decreases by a fixed percentage during each period.

A=P(1−i)nA=P(1-i)^n

Example

A town has a population of 85 000. The population grows by 2,4% per year. Calculate the expected population after 6 years.

A=P(1+i)nA=P(1+i)^n
A=85000(1+0,024)6A=85\,000(1+0{,}024)^6
A=98024,82A=98\,024{,}82

The expected population is approximately 98 025 people.

Example

A chemical sample has a mass of 500 g and loses 12% of its mass each hour. Calculate its mass after 4 hours.

A=P(1−i)nA=P(1-i)^n
A=500(1−0,12)4A=500(1-0{,}12)^4
A=500(0,88)4A=500(0{,}88)^4
A=299,85 gA=299{,}85\text{ g}

Remember

For growth, the multiplier is greater than 1. For decay, the multiplier lies between 0 and 1.

Depreciation

Depreciation is the decrease in the value of an asset over time. Two common methods are reducing-balance depreciation and straight-line depreciation.

Reducing-balance depreciation is calculated on the current value of the asset.

A=P(1−i)nA=P(1-i)^n

Example

A vehicle costing R320 000 depreciates at 14% per annum according to the reducing-balance method. Calculate its value after 5 years.

A=P(1−i)nA=P(1-i)^n
A=320000(1−0,14)5A=320\,000(1-0{,}14)^5
A=320000(0,86)5A=320\,000(0{,}86)^5
A=R150550,73A=\text{R}150\,550{,}73

Straight-line depreciation is calculated on the original value. The same rand amount is lost each year.

A=P(1−in)A=P(1-in)

Example

A machine costing R180 000 depreciates at 8% per annum on the straight-line method for 6 years.

A=P(1−in)A=P(1-in)
A=180000(1−0,08(6))A=180\,000\left(1-0{,}08(6)\right)
A=180000(0,52)A=180\,000(0{,}52)
A=R93600A=\text{R}93\,600

Exam Tip

Read carefully whether depreciation is calculated on the original value or the reducing balance.

Nominal and effective interest rates

A nominal interest rate is the quoted annual rate before the effect of compounding is included. An effective annual interest rate represents the actual percentage increase over one year.

If the nominal annual rate is compounded m times per year:

iperiod=inommi_{\text{period}}=\frac{i_{\text{nom}}}{m}

The effective annual rate is:

ieff=(1+inomm)m−1i_{\text{eff}}=\left(1+\frac{i_{\text{nom}}}{m}\right)^m-1

Example

Calculate the effective annual rate corresponding to a nominal rate of 10,8% per annum, compounded monthly.

ieff=(1+0,10812)12−1i_{\text{eff}}=\left(1+\frac{0{,}108}{12}\right)^{12}-1
ieff=(1,009)12−1i_{\text{eff}}=(1{,}009)^{12}-1
ieff=0,11351i_{\text{eff}}=0{,}11351
ieff=11,35%i_{\text{eff}}=11{,}35\%

To determine a nominal rate from an effective rate:

inom=m(1+ieffm−1)i_{\text{nom}}=m\left(\sqrt[m]{1+i_{\text{eff}}}-1\right)

Present and future value

Future value is the value to which an amount will grow. Present value is the amount that must be invested now to produce a given future amount.

Future value:

A=P(1+i)nA=P(1+i)^n

Present value:

P=A(1+i)−nP=A(1+i)^{-n}

or

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

Example

How much must be invested now to obtain R80 000 in 4 years at 9% per annum, compounded annually?

P=A(1+i)−nP=A(1+i)^{-n}
P=80000(1+0,09)−4P=80\,000(1+0{,}09)^{-4}
P=R56674,02P=\text{R}56\,674{,}02

This means that R56 674,02 invested now will grow to R80 000 after 4 years.

Annuities

An annuity is a sequence of equal payments made at equal time intervals. Examples include monthly savings, pension contributions and loan repayments.

The future value of an ordinary annuity is:

F=x[(1+i)n−1i]F=x\left[\frac{(1+i)^n-1}{i}\right]

Here, x is the regular payment, i is the rate per payment period and n is the number of payments.

If payments are made at the beginning of each period, the annuity has one additional period to earn interest:

F=x[(1+i)n−1i](1+i)F=x\left[\frac{(1+i)^n-1}{i}\right](1+i)

Example

R2 300 is deposited at the beginning of every quarter for 6 years. Interest is 5,8% per annum, compounded quarterly. Calculate the future value.

i=0,0584i=\frac{0{,}058}{4}
i=0,0145i=0{,}0145
n=6(4)=24n=6(4)=24
F=2300[(1+0,0145)24−10,0145](1+0,0145)F=2\,300\left[\frac{(1+0{,}0145)^{24}-1}{0{,}0145}\right](1+0{,}0145)
F=R66411,60F=\text{R}66\,411{,}60

The present value of an ordinary annuity is:

P=x[1−(1+i)−ni]P=x\left[\frac{1-(1+i)^{-n}}{i}\right]

Loans and investments

A loan is usually repaid through equal instalments. Interest is charged on the outstanding balance. The present value annuity formula is used to calculate the original loan, repayment or repayment period.

Example

A loan balance is R915 386,86. Monthly repayments of R10 000 are made, and interest is 6,8% per annum, compounded monthly. Determine the number of repayments needed.

P=x[1−(1+i)−ni]P=x\left[\frac{1-(1+i)^{-n}}{i}\right]
915386,86=10000[1−(1+0,06812)−n0,06812]915\,386{,}86=10\,000\left[\frac{1-\left(1+\frac{0{,}068}{12}\right)^{-n}}{\frac{0{,}068}{12}}\right]

After simplifying:

(1+0,06812)−n=0,4812\left(1+\frac{0{,}068}{12}\right)^{-n}=0{,}4812

Take logarithms:

−nlog⁡(1+0,06812)=log⁡(0,4812)-n\log\left(1+\frac{0{,}068}{12}\right)=\log(0{,}4812)
n=−log⁡(0,4812)log⁡(1+0,06812)n=\frac{-\log(0{,}4812)}{\log\left(1+\frac{0{,}068}{12}\right)}
n=129,419n=129{,}419

Since a fraction of a repayment period is not sufficient, round up to 130 repayments.

When comparing loans or investments:

  • Compare effective interest rates, not only nominal rates.
  • Check whether payments occur at the beginning or end of each period.
  • Use the same time unit for the interest rate and number of periods.
  • Keep full calculator values during calculations and round money to two decimal places only in the final answer.
  • For a loan term, round up when an additional payment is required.