Finance and growth (Grade 10) — Mathematics Grades 8–11

Calculate accumulated amounts using the simple growth and compound growth formulae.

Gr 10 Term 3 - Finance and growthTerm 3 - 2 weeks (CAPS)

Practise Finance and growth in the app →

What gets asked

Calculate the accumulated amount using simple growth, A = P(1 + in).
Calculate the accumulated amount using compound growth, A = P(1 + i)ⁿ.
Solve a hire purchase problem with a deposit and monthly instalments.
Decide whether a situation is simple growth or compound growth.

You must be able to

Write a percentage rate as a decimal before it goes into a formula.
Subtract a deposit from the cash price before working out the interest.
Keep money answers rounded to two decimal places, rounding only at the end.
Tell simple growth and compound growth apart from the wording of a problem.

Traps that cost marks

Using 8 instead of 0,08 for a rate of 8% when substituting into a formula. Always convert a percentage to a decimal first: 8% becomes 0,08.
Reporting only the interest earned as the final answer, and forgetting to add it to the principal. A is the total. Add the interest to the amount you started with: A = P + interest.
Multiplying by n instead of raising (1 + i) to the power n in a compound growth problem. Compound growth uses an exponent on (1 + i). It is not multiplied by n.

Worked example

R2 000 is invested at 8% p.a. simple interest for 3 years. Calculate the accumulated amount.

Identify the values: $P = 2000$ , $i = 0, 08$ , $n = 3$ .
Substitute into $A = P (1 + in)$ : $A = 2000 (1 + 0, 08 \times 3)$ .
$A = 2000 \times 1, 24 = 2480$ . The accumulated amount is R2 480.

Simple interest

Four letters, one formula, and the single step that sinks more marks than all the rest together.

principal: The amount you start with: the money invested, or the money still owed.
interest: The extra money charged or earned for the use of somebody else's money.
accumulated amount: The principal plus all the interest. It is the total, and it is written $A$ .
simple interest: Interest worked out on the original principal every year, and never on interest already earned.
hire purchase: Buying on credit: a deposit now, then equal monthly instalments on the rest.

Four things appear in every question. $P$ is the principal, $i$ is the rate, $n$ is the number of years and $A$ is the accumulated amount. Write down which is which before you touch the formula.

THE RATE IS NOT THE PERCENTAGE. A rate of $6%$ means $6$ rand in every $100$ , so $i$ is $6$ divided by $100$ , which is $0, 06$ . Not $6$ , and not $0, 6$ either. Slide the comma two places, every single time.

The formula is $A = P (1 + in)$ . That $1$ inside the bracket is the money you already had, so $A$ comes out as the whole total. Leave the $1$ out and you have found only the interest.

So the interest on its own is $A - P$ . A question asking what you earned wants that. A question asking what is in the account wants $A$ . They are never the same.

Working backwards, put in what you know and solve for the letter left over. If you solved for $i$ the answer is a decimal, and the question wanted a percentage, so multiply by $100$ .

Rules to remember

simple growth: $A = P (1 + in)$
interest earned: $A - P$
a rate of $6%$ means $i = 0, 06$

Examples

Thabo invests R2 000 at

8%

simple interest for

3

years. Find the accumulated amount.

Worked answer

List the values: $P = 2000$ , $i = 0, 08$ and $n = 3$ .
Substitute: $A = 2000 (1 + 0, 08 \times 3)$ .
Inside the bracket: $1 + 0, 08 \times 3 = 1, 24$ .
$A = 2000 \times 1, 24 = 2480$ .

Answer: R $2480$

The $1$ in the bracket carried the original R2 000 into the answer.

A spaza owner buys a R6 000 fridge on hire purchase. She pays a

10%

deposit, then repays the balance over

2

years at

12%

simple interest. Find her monthly instalment.

Worked answer

Deposit is $10%$ of $6000$ , which is $600$ , leaving a balance of $5400$ .
Interest is charged on the balance only: $5400 \times 0, 12 \times 2 = 1296$ .
Total to repay: $5400 + 1296 = 6696$ .
Two years is $24$ months, so $\frac{6 696}{24} = 279$ .

Answer: R $279$

The rate is per year, so $n$ was $2$ . The $24$ only shared the total out.

R4 500 grows to R5 400 in

4

years at simple interest. Find the rate.

Worked answer

Interest earned is $5400 - 4500 = 900$ .
Simple interest is principal times rate times years: $4500 \times i \times 4 = 900$ .
That is $18000 i = 900$ , so $i = 0, 05$ .
The question wants a percentage: $0, 05 \times 100 = 5$ .

Answer: $5%$

Answering $0, 05$ is right arithmetic with the wrong units on the end.

Traps

Using 8 instead of 0,08 for a rate of 8% when substituting into a formula. Always convert a percentage to a decimal first: 8% becomes 0,08.
Reporting only the interest earned as the final answer, and forgetting to add it to the principal. A is the total. Add the interest to the amount you started with: A = P + interest.
Charging the hire purchase interest on the full cash price. The deposit is paid at once. Interest runs on the balance left.
Handing in a rate as $0, 075$ when the question asked for a percentage. Multiply the decimal by $100$ and write the sign: $0, 075$ is $7, 5%$ .

Learn and practise “Simple interest” in the app →

Compound interest

When last year's interest starts earning interest of its own, the multiplying turns into a power.

compound interest: Interest worked out on the principal and on the interest already added. Interest earns interest.
period: One stretch of time the rate is applied over. In these questions it is one year.
inflation: The steady yearly rise in prices. The same money buys less than it did before.

Under simple interest the same principal is used every year. Under compound interest the interest is added on at the end of each period. The next period's interest is then worked out on that bigger amount.

That changes one thing in the formula. Simple growth is $A = P (1 + in)$ and compound growth is $A = P (1 + i)^{n}$ . The $n$ has moved upstairs. It is a power now, and it is no longer multiplied by anything.

The rate still goes in as a decimal. A rise of $6%$ makes $i = 0, 06$ , so the bracket is $1, 06$ . Writing $0, 6$ there would turn a $6%$ rise into a $60%$ one.

How do you know which formula? The words simple interest mean simple. Compounded yearly, interest added each year, inflation and population growth are all compound.

Count the periods on your fingers if you must. From the start of one year to the start of the next is one period. And keep the full number in the calculator until the last line, because rounding every year drifts the answer by rands.

Rules to remember

simple growth: $A = P (1 + in)$
compound growth: $A = P (1 + i)^{n}$

Examples

R2 000 is invested at

8%

per year, compounded yearly, for

3

years. Find the accumulated amount.

Worked answer

$P = 2000$ , $i = 0, 08$ and $n = 3$ .
Compound growth: $A = 2000 (1 + 0, 08)^{3}$ .
The bracket is $1, 08$ , and $1, 0 8^{3} = 1, 259712$ .
$A = 2000 \times 1, 259712$ , which is $2519, 424$ .

Answer: R $2519, 42$

Simple interest on the same money gave R2 480. The rest is interest earning interest.

A loaf of bread costs R18 today. Inflation runs at

6%

a year. Estimate the price in

3

years.

Worked answer

Inflation is compound growth, so the power formula is the one.
The rate is $6%$ , so $i = 0, 06$ and the bracket is $1, 06$ .
$1, 0 6^{3} = 1, 191016$ .
$18 \times 1, 191016 = 21, 438288$ .

Answer: about R $21, 44$

Prices rise on last year's price, not on today's, which is what the power does.

A township had

8000

people at the start of 2021, growing at

5%

a year. Estimate the number at the start of 2024.

Worked answer

From the start of 2021 to the start of 2024 is $3$ years.
$A = 8000 (1 + 0, 05)^{3}$ .
$1, 0 5^{3} = 1, 157625$ .
$8000 \times 1, 157625 = 9261$ .

Answer: about $9261$ people

People grow like money. Each year's growth is worked out on the new total.

Traps

Multiplying by n instead of raising (1 + i) to the power n in a compound growth problem. Compound growth uses an exponent on (1 + i). It is not multiplied by n.
Treating every investment that runs for more than a year as compound. Read the wording. Simple interest stays on the original principal, however many years pass.
Counting the years between 2021 and 2024 as four. Subtract the years: $2024 - 2021 = 3$ . Three periods of growth have passed, not four.
Rounding to the nearest rand at the end of every year. Each rounding is carried into the next year. Keep the full figure and round once, at the end.

Learn and practise “Compound interest” in the app →

Further growth problems

Comparing the two kinds of growth, running a formula backwards, and handling a rate that changes halfway.

growth factor: The number the money is multiplied by each year. At $9%$ the growth factor is $1, 09$ .
stage: A stretch of an investment over which the rate and the principal both stay put.
trial and improvement: Trying a value, seeing how far off it lands, and adjusting until the two sides meet.

Same money, same rate, same time: after exactly one year the two kinds of growth pay the same amount. The gap only opens in year two, which is the first year in which there is any interest to earn interest on.

When you compare, compare like with like. Interest against interest, or total against total. Setting one investment's interest beside the other's accumulated amount answers no question at all.

To find the principal, undo the whole power. Divide $A$ by the growth factor raised to the power $n$ . Dividing by the growth factor times $n$ is a different sum and gives a badly wrong answer.

To find $n$ there are no logarithms in Grade 10, so use trial and improvement. Multiply by the growth factor year after year, and count how many years it takes to pass $A$ .

When the rate changes partway, split the time into stages. Finish stage one, then use its accumulated amount as the principal of stage two. Never work the two stages out separately and add them.

Rules to remember

compound growth: $A = P (1 + i)^{n}$
finding the principal: $P = \frac{A}{( 1 + i ) ^{n}}$

Examples

R5 000 is invested for

2

years at

10%

. Find the interest under simple growth, and under compound growth.

Worked answer

Simple: the interest is $5000 \times 0, 1 \times 2 = 1000$ .
Compound: $A = 5000 \times 1, 1^{2} = 6050$ .
Its interest is $6050 - 5000 = 1050$ .
Compound pays R50 more over the two years.

Answer: R $1000$ simple, R $1050$ compound

That R50 is one year's interest on the R500 earned in year one.

How much must be invested now at

9%

compounded yearly to have R11 881 after

2

years?

Worked answer

Here $A = 11881$ , $i = 0, 09$ and $P$ is the unknown.
$11881 = P (1 + 0, 09)^{2}$ .
The growth factor squared is $1, 0 9^{2} = 1, 1881$ .
Divide: $P = \frac{11 881}{1 , 1881} = 10000$ .

Answer: R $10000$

Dividing by $1, 09 \times 2$ instead would give R5 450, nowhere near it.

R8 000 is invested at

10%

compounded yearly for

2

years. The rate then changes to

12%

for

1

more year. Find the final amount.

Worked answer

Stage one runs at the first rate: $8000 \times 1, 1^{2} = 9680$ .
That R9 680 is now the principal for stage two.
Stage two: $9680 \times 1, 12 = 10841, 6$ .
The final amount is R10 841,60.

Answer: R $10841, 60$

The first stage's total was carried forward, so the second stage grew all of it.

Traps

Saying compound growth always beats simple growth, whatever the time. Over one period they pay exactly the same. Compound only pulls ahead from period two.
Comparing the interest on one investment with the accumulated amount on the other. Compare interest with interest, or total with total. Say which one you are quoting.
Dividing the accumulated amount by the growth factor times $n$ . Divide by the growth factor to the power $n$ . For $2$ years at $9%$ that is $1, 1881$ .
Working each stage out from the original principal and adding the two answers. Stage one's total becomes stage two's principal. The stages run one after the other.

Learn and practise “Further growth problems” in the app →

Exchange rates

What the rand is worth, which way to point the calculator, and who wins when it slips.

exchange rate: What one country's money costs in another's. R $18$ to the dollar means a dollar costs R $18$ .
weaker rand: It takes more rand to buy the same dollar, so the rate number has gone up.
stronger rand: It takes fewer rand to buy the same dollar, so the rate number has come down.
imports: Goods a country buys in from abroad, paid for in the seller's money.
exports: Goods a country sells to buyers abroad, and is paid for in their money.

An exchange rate always carries units. R $18$ to the dollar means rand per dollar. Read the units before touching the calculator: they are the instruction.

With rand per dollar, going from dollars to rand means multiply. Going back, from rand to dollars, means divide. Never pick the operation that makes the answer look bigger.

A bigger rate number means a weaker rand, and that catches everybody. A smaller one means a stronger rand. Bigger normally feels better; here it means each rand buys less.

A weaker rand pushes the petrol price up, since oil is paid for in dollars. All imports cost more. Exports go the other way: exporters are paid in dollars, so a weak rand pays more.

To see how a cost has changed, convert each year on its own and compare. Work the percentage change on the original cost, never the new one.

Rules to remember

dollars to rand: multiply by the rate
rand to dollars: divide by the rate
percentage change: the change divided by the ORIGINAL amount, times $100$

Examples

An online shop lists a pair of takkies at

60

dollars. The rate is R

18

to the dollar. Find the price in rand.

Worked answer

The rate is rand per dollar, so multiply to get rand.
$60 \times 18 = 1080$ .
Dividing instead would give about R3, which buys no takkies.

Answer: R $1080$

The units of the rate chose the operation, not the size of the answer.

In January the rate was R

17

to the dollar. By June it was R

19

. Is the rand stronger or weaker, and what happens to the petrol price?

Worked answer

It now takes R19 to buy one dollar, where R17 was enough.
More rand for the same dollar means a weaker rand.
Oil is paid for in dollars, so petrol costs more rand.
Exporters are paid in dollars, so they now get more rand.

Answer: weaker, and the petrol price rises

The rate number rose, which looks like good news and is the opposite.

An imported part cost

100

dollars last year, at R

18

to the dollar. This year it costs

120

dollars at R

19, 50

. Find the percentage rise in rand.

Worked answer

Last year in rand: $100 \times 18 = 1800$ .
This year in rand: $120 \times 19, 50 = 2340$ .
The rise is $2340 - 1800 = 540$ .
On the original cost: $\frac{540}{1 800} \times 100 = 30$ .

Answer: $30%$

Both the dollar price and the rate moved, so each year needed its own conversion.

Traps

Choosing to multiply or divide by whichever looks bigger. Let the units decide. Rand per dollar times dollars gives rand.
Reading a move from R17 to R19 per dollar as the rand getting stronger. More rand for the same dollar is a weaker rand. It buys less than it did.
Saying a weak rand is bad for imports and exports alike. They move opposite ways. A weak rand makes imports dearer and exports more profitable.
Dividing the price rise by the new cost when finding a percentage change. The bottom of the fraction is the ORIGINAL cost: R1 800 here, not R2 340.

Learn and practise “Exchange rates” in the app →

Grade 10 Finance and growth

What gets asked

You must be able to

Traps that cost marks

Worked example

Simple interest

Rules to remember

Examples

Traps

Compound interest

Rules to remember

Examples

Traps

Further growth problems

Rules to remember

Examples

Traps

Exchange rates

Rules to remember

Examples

Traps

About this material