Grade 10 Mathematics
Grade 10 · Term 3Mathematics

Finance & Growth

We apply the simple and compound growth formulae to solve real-world problems involving interest, hire purchase, inflation, population growth, and the effect of exchange rate fluctuations on everyday prices.

10.1 Simple & Compound Growth10.2 Inflation, Population Growth & Exchange Rates
Week 10

10.1 Simple & Compound Growth

  • Apply the simple growth formula $A = P(1 + in)$ to interest, hire purchase, and inflation
  • Apply the compound growth formula $A = P(1 + i)^n$ to compound interest and population growth
  • Compare simple and compound growth over different time periods
🌍

Real-World Connection

Saving R1 000 in a bank account matters: with simple interest at 8% per year, after 10 years you have R1 800. With compound interest at the same rate, you have R2 159 — R359 more, just from interest earning interest. This difference becomes enormous over decades: it is the mathematical foundation of every pension fund, mortgage, and investment.

Simple Growth (interest)

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

$A$ = accumulated amount; $P$ = principal; $i$ = interest rate (decimal); $n$ = number of periods (usually years)

Compound Growth

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

$A$ = accumulated amount; $P$ = principal; $i$ = interest rate per period (decimal); $n$ = number of periods

💡 Tip

Convert percentage rates to decimals before substituting: 8% → i = 0.08. For compound growth, always confirm the compounding period matches n (e.g., if interest is compounded monthly, n = months and i = annual rate ÷ 12).

ℹ️ Note

Hire purchase uses SIMPLE interest calculated on the full purchase price for the full period — not on the reducing balance. This makes hire purchase more expensive than a bank loan over the same period.

Worked Examples

Worked Example

Simple interest — savings account

Problem

R5 000 is invested at 7.5% simple interest per year. How much is it worth after 4 years?

Worked Example

Compound interest — compare with simple

Problem

R5 000 is invested at 7.5% compound interest per year for 4 years. How much more does it earn compared to simple interest?

Worked Example

Hire purchase calculation

Problem

A television costs R8 000. A hire purchase agreement requires a 10% deposit, and the balance is paid off over 2 years at 12% simple interest per year. Find the total amount paid.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
State the simple interest formula and identify each variable.
2
L1 · Knowledge3 marks
R2 000 is invested at 6% compound interest for 3 years. Calculate A.
3
L2 · Routine Procedures3 marks
How long will it take R3 000 to grow to R3 900 at 10% simple interest per year?
4
L2 · Routine Procedures3 marks
The population of a town is 50 000 and grows at 3% per year (compound). Find the population after 5 years.
5
L3 · Complex Procedures5 marks
A laptop costs R12 000. Hire purchase: 15% deposit, remainder over 3 years at 9% simple interest. Find the total paid and the monthly instalment.
6
L3 · Complex Procedures4 marks
Find the interest rate if R4 000 grows to R6 209.92 in 6 years with compound interest.
7
L4 · Problem Solving5 marks
R6 000 is invested at 8% compound interest. After how many years does the investment first exceed R10 000?
8
L4 · Problem Solving4 marks
Explain, with a calculation, why compound interest is always greater than simple interest for the same principal, rate and n>1n>1.
Week 11

10.2 Inflation, Population Growth & Exchange Rates

  • Apply compound growth formula to inflation and depreciation problems
  • Solve population growth and decay problems using $A = P(1 + i)^n$
  • Understand and apply exchange rates; analyse the effect of currency fluctuation on prices
🌍

Real-World Connection

South Africa imports most of its petrol. When the rand weakens from R18 to R20 per US dollar, the rand price of a barrel of oil rises by more than 11% — even if the dollar price of oil doesn't change. Exchange rate movements directly affect petrol prices, imported food costs, and the price of electronic goods. Understanding these links helps you make informed financial decisions.

Inflation / compound depreciation

A=P(1+i)n(inflation: i>0)A=P(1−i)n(depreciation: i>0)A = P(1 + i)^n \quad\text{(inflation: }i>0\text{)}\qquad A = P(1 - i)^n\quad\text{(depreciation: }i>0\text{)}

$i$ = inflation rate or depreciation rate (decimal per period); $n$ = number of periods

Definition

Exchange Rate

The price of one currency expressed in another. If the exchange rate is R18.50/USD, then R18.50 buys one US dollar. To convert rand to dollars: divide by rate. To convert dollars to rand: multiply by rate.

Rand value=Dollar value×exchange rate\text{Rand value} = \text{Dollar value} \times \text{exchange rate}

⚠️ Warning

A DEPRECIATION of the rand means the exchange rate NUMBER INCREASES (more rand per dollar). This makes imports MORE expensive — petrol, electronics, foreign travel. A STRENGTHENING rand means the number decreases — imports become cheaper.

Worked Examples

Worked Example

Inflation — cost of goods

Problem

A bag of groceries costs R450 today. If the inflation rate is 6.5% per year, what will it cost in 4 years?

Worked Example

Exchange rate — import pricing

Problem

A phone costs \350 (USD). (a) Find the rand price when R18.50 = \1. (b) The rand weakens to R21.00/\$1. Find the new rand price and the percentage increase.

Worked Example

Depreciation of a car

Problem

A car is bought for R180 000. It depreciates at 15% per year (compound). Find its value after 5 years.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
Convert R2 500 to US dollars if the exchange rate is R19.20 per USD.
2
L1 · Knowledge2 marks
Write the formula for compound depreciation and explain what happens when i=0.2i = 0.2.
3
L2 · Routine Procedures3 marks
Food prices increase at 5% inflation per year. How much will a R300 item cost in 3 years?
4
L2 · Routine Procedures3 marks
A population of 80 000 decreases at 2% per year. Find the population after 6 years.
5
L3 · Complex Procedures5 marks
A tourist from the UK has £500. Exchange rates: R23.10/£1 before a trip and R22.40/£1 when converting back. The tourist spent R8 000. How many pounds does she have left?
6
L3 · Complex Procedures4 marks
A machine costs R250 000 and depreciates at 20% per year. After how many years is it worth less than R100 000?
7
L4 · Problem Solving4 marks
A South African company imports goods worth \20 000. Exchange rate today: R18/\1. The rand depreciates by 8%. Find the new rand cost and the rand amount lost.
8
L4 · Problem Solving4 marks
Inflation is 7% per year. What salary would a worker earning R25 000/month today need in 5 years to maintain the same purchasing power?

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Finance & Growth Grade 10 Maths CAPS Notes & Examples | MathSciBuddy