17 questions arranged by DBE cognitive level — simple and compound growth, hire purchase, inflation, population growth, and foreign exchange rates, including 5 real DBE/provincial exam questions. Work each one on paper first, then reveal the memo.
17
practice questions
4
cognitive levels
17
worked memos
100%
independently verified
How to use this bank.
Start at Level 1 and move up — don't jump to Level 4 first.
Before choosing simple or compound growth, ask: does this grow by the same RAND amount every year, or the same PERCENTAGE of an already-grown amount?
When a question asks for a number of years, always check the boundary years directly — never leave a raw decimal value as your final answer.
Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved from scratch before publication, cross-checked against its own working rather than assumed correct. 12 are original "Equation Station SA Practice Question" items; 5 are real questions from DBE and provincial exam papers, each independently re-derived and confirmed to match its official memo before publication, with the exact source cited on the question.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
24%
Level 1 | Knowledge
Direct Substitution
One direct application of the simple or compound growth formula, plus recognising which formula a scenario needs and how hire purchase interest works.
Q1Equation Station SA Practice Question2 marks
Simple Growth
Simple Growth
R4 000 is invested at 6% p.a. simple interest for 3 years. Calculate the value of the investment at the end of the 3 years.
Memo
✓ This is simple interest, so the SAME rand amount grows on the original principal every year — no compounding: \(A=P(1+in)=4\,000(1+0{,}06\times3)\)✓ \(\boxed{R4\,720{,}00}\)
Q2Equation Station SA Practice Question2 marks
Compound Growth
Compound Growth
R4 000 is invested at 6% p.a. compounded annually for 3 years. Calculate the value of the investment at the end of the 3 years.
Memo
✓ This is compound growth, so each year's interest is calculated on the PREVIOUS year's total, not the original amount — that's why n is an exponent: \(A=P(1+i)^n=4\,000(1{,}06)^3\)✓ \(\boxed{R4\,764{,}06}\)
Q3Equation Station SA Practice Question1 mark
Growth Formulas
Choose the Correct Formula
An investment grows so that every year's interest is calculated on the PREVIOUS year's total, not on the original amount. Which formula models this?
Memo
✓ Interest calculated on the previous total (not the original amount) means the interest itself is earning interest — that is the definition of compound growth, not simple growth.✓ \(\boxed{A=P(1+i)^n}\)
Q4Equation Station SA Practice Question1 mark
Hire Purchase
The Type of Interest Hire Purchase Uses
Hire purchase interest is always calculated as which type of interest, and on which amount?
Memo
✓ A hire purchase agreement is a loan, and by definition its interest is ALWAYS simple interest — but it is charged only on what is actually still owed, which is the cash price minus the deposit, never on the full price.✓ \(\boxed{\text{Simple interest, on the balance remaining after the deposit}}\)
29%
Level 2 | Routine Procedures
One Established Method
Hire purchase, inflation, population growth, foreign exchange, and quantifying the gap between simple and compound growth.
Q5Equation Station SA Practice Question3 marks
Hire Purchase
Hire Purchase Instalment
A television has a cash price of R8 500. A 20% deposit is required, with the balance repaid at 14% p.a. simple interest over 18 months, in equal monthly instalments. Calculate the monthly instalment.
Memo
✓ Subtract the deposit first — interest is only ever charged on what still needs to be financed: deposit \(=0{,}20\times8\,500=R1\,700\); balance financed \(=R6\,800\)✓ Hire purchase interest is simple interest, so convert 18 months to 1,5 years and apply it once to the whole balance: \(6\,800\times0{,}14\times1{,}5=R1\,428\)✓ Add the interest to the balance to get the total amount actually owed: \(6\,800+1\,428=R8\,228\)✓ Spread that total evenly over the 18 months: \(\boxed{R457{,}11}\) per month
Q6Equation Station SA Practice Question2 marks
Inflation
Projecting a Future Price
An item currently costs R450. If inflation is 5,5% p.a., calculate its expected price in 3 years' time.
Memo
✓ Prices under inflation compound exactly like an investment — each year's increase is a percentage of the ALREADY-inflated price, not the original: \(A=450(1{,}055)^3\)✓ \(\boxed{R528{,}41}\)
Q7North West DBE, Paper 1, November 2023 (Q4.1)3 marks
Population Growth
Projecting a Future Population
The population of Hartbeesfontein is 4 831 people in 2023. Calculate the expected population over the next 10 years, based on an annual growth rate of 1,3%.
Memo
✓ Population growth is always compound growth, since each year's growth is a percentage of an already-larger population: \(A=4\,831(1{,}013)^{10}\)✓ \(\boxed{5\,497}\) people
Q8Equation Station SA Practice Question2 marks
Foreign Exchange
Converting to Rand
The exchange rate is R18,50 to $1. Calculate the Rand value of $250.
Memo
✓ To convert a foreign amount into Rand, multiply by the Rand value of one unit of that currency: \(250\times18{,}50\)✓ \(\boxed{R4\,625{,}00}\)
Q9Equation Station SA Practice Question3 marks
Growth Comparison
Quantifying the Compounding Gap
R6 000 is invested for 2 years at 9% p.a. Calculate how much MORE the investment is worth under compound interest than under simple interest.
Memo
✓ Calculate the simple-interest value first — the original amount grows by the same rand amount every year: \(6\,000(1+0{,}09\times2)=R7\,080{,}00\)✓ Then the compound value — here, interest is now earning interest on itself for the first time in year 2: \(6\,000(1{,}09)^2=R7\,128{,}60\)✓ The gap is the difference between the two: \(\boxed{R48{,}60}\) more under compound interest
24%
Level 3 | Complex Procedures
Multi-Step Methods
Working backward from a current value to find a rate, a past price, or a time period.
Q10KwaZulu-Natal Provincial Common Test, Grade 10, September 2024 (Q1.3)4 marks
Inflation
Solving for the Inflation Rate
A cell phone has a cash price of R29 730,34 today. Due to inflation, its cash price is expected to rise to R31 968,11 after 3 years. Calculate the annual inflation rate.
Memo
✓ Set up the compound growth formula with the unknown rate inside the bracket: \(31\,968{,}11=29\,730{,}34(1+i)^3\)✓ Isolate \((1+i)\) by dividing, then undo the power of 3 with a cube root — never divide by \(n\) when the unknown is trapped inside an exponent: \(1+i=\left(\dfrac{31\,968{,}11}{29\,730{,}34}\right)^{\frac13}\)✓ \(\boxed{i\approx2{,}45\%}\)
Q11Equation Station SA Practice Question4 marks
Hire Purchase
Total Interest Paid
A motorbike has a cash price of R15 000. A 15% deposit is paid, and the balance is repaid at 12% p.a. simple interest over 3 years. Calculate the TOTAL interest paid over the 3 years.
Memo
✓ Subtract the deposit first, since interest is only ever charged on what remains: deposit \(=0{,}15\times15\,000=R2\,250\); balance \(=R12\,750\)✓ Hire purchase interest is simple interest, applied once over the full 3-year term: \(12\,750\times0{,}12\times3\)✓ \(\boxed{R4\,590{,}00}\)
Q12North West DBE, Paper 1, November 2023 (Q4.2)3 marks
Simple Growth
Solving for the Time Period
R3 600 is deposited at a simple interest rate of 6% per annum. Calculate the number of years the deposit will take to grow to R4 608 (round off to the nearest year).
Memo
✓ Set up the simple-growth equation with the unknown time period n: \(4\,608=3\,600(1+0{,}06n)\)✓ Rearrange to isolate n — since n is a multiplier here (not an exponent), no root or logarithm is needed, just algebra: \(n=\dfrac{4\,608/3\,600-1}{0{,}06}\approx4{,}67\)✓ A "number of years" answer must be a whole number, so round up to the year the target is actually reached: \(\boxed{5\text{ years}}\)
Q13Equation Station SA Practice Question3 marks
Foreign Exchange
Impact of a Changed Exchange Rate
An overseas trip costs $1 200. When it was booked, the exchange rate was R17,20 to $1. By the time it is paid for, the rate has changed to R18,60 to $1. Calculate how much MORE, in Rand, the trip now costs.
Memo
✓ Convert the trip's Dollar cost to Rand at the ORIGINAL exchange rate: \(1\,200\times17{,}20=R20\,640{,}00\)✓ Convert the same Dollar amount again at the NEW rate: \(1\,200\times18{,}60=R22\,320{,}00\)✓ The Rand weakened, so more Rand is now needed for the same $1\,200: \(\boxed{R1\,680{,}00}\) more
23%
Level 4 | Problem Solving
Combined Skills
A dramatic long-term comparison, real exam-style hire purchase and foreign-exchange problems, and comparing two populations.
Q14Equation Station SA Practice Question5 marks
Growth Comparison
A Dramatic Long-Term Comparison
R5 000 is invested for 15 years at 10% p.a. Calculate how much MORE the compound-interest investment is worth than the simple-interest investment.
Memo
✓ Calculate the simple-interest value first, over the full 15 years: \(5\,000(1+0{,}10\times15)=R12\,500{,}00\)✓ Then the compound value — the same principal and rate, but every year's growth compounds on the last: \(5\,000(1{,}10)^{15}\approx R20\,886{,}24\)✓ The longer the time period, the bigger this gap becomes: \(\boxed{R8\,386{,}24}\) more under compound interest, over 15 years
Q15Eastern Cape DBE, Paper 1, November 2020 (Q4.1)6 marks
Hire Purchase
Hire Purchase with a Compulsory Insurance Premium
Sylvia wants to buy a dishwasher priced at R9 899 by means of a hire purchase agreement: a 30% deposit; 12% p.a. simple interest on the balance; a compulsory monthly insurance premium of R65,30; the account settled over 36 months. Calculate her monthly instalment.
Memo
✓ Subtract the deposit first: deposit \(=0{,}30\times9\,899=R2\,969{,}70\); balance financed \(=R6\,929{,}30\)✓ Apply simple interest once, over the full 3-year term, to that balance: \(6\,929{,}30\times0{,}12\times3=R2\,494{,}55\)✓ Add the interest to the balance and spread the total over 36 months: \(6\,929{,}30+2\,494{,}55=R9\,423{,}85\), giving \(R261{,}77\) per month✓ The compulsory insurance is a SEPARATE monthly cost, added AFTER the loan instalment is worked out, never mixed into the interest calculation itself: \(261{,}77+65{,}30=\boxed{R327{,}07}\)
Q16Equation Station SA Practice Question6 marks
Population Growth
Comparing Two Growing Populations
Town A has a population of 50 000, growing at 3,2% p.a. Town B has a population of 38 000, growing at 4,5% p.a. After how many complete years will Town B's population first exceed Town A's?
Memo
✓ Set the two compound-growth expressions against each other and solve — the raw answer will not be a whole number: \(38\,000(1{,}045)^n>50\,000(1{,}032)^n\) gives \(n\approx21{,}92\)✓ Check the year just below the raw value directly, since a fractional year can't be a real answer: at \(n=21\), Town A \(\approx96\,882\) is still ahead of Town B \(\approx95\,769\)✓ Check the next year up: at \(n=22\), Town B \(\approx100\,079\) has now overtaken Town A \(\approx99\,982\)✓ The first year the overtaking actually happens is the answer: \(\boxed{22\text{ years}}\)
Q17Eastern Cape DBE, Paper 1, November 2020 (Q4.2, adapted)5 marks
Foreign Exchange
Comparing a Price in Two Currencies
George, from England, sees a machine for $6 800 in the USA. A similar machine costs £4 600 in England. Given $1=R16,24 and £1=R27,63, in which country is the machine cheaper for George to buy?
Memo
✓ Two prices in two different currencies can never be compared directly — convert the USA price to Rand first: \(6\,800\times16{,}24=R110\,432{,}00\)✓ Then convert that Rand amount into Pounds, the same currency as the England price: \(R110\,432{,}00\div27{,}63\approx\pounds3\,996{,}82\)✓ Now the two prices are in the same currency and can be compared directly: since \(\pounds3\,996{,}82<\pounds4\,600\), \(\boxed{\text{the machine is cheaper in the USA}}\)