17 questions arranged by DBE cognitive level — simple and compound growth and decay, depreciation, nominal and effective rates, future and present value annuities, outstanding balances, and critically analysing loans and investments. 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.
Draw a timeline before setting up any annuity or loan equation. It is the single most useful tool in this whole topic.
When a question asks for a number of years, always check the boundary years directly — never leave a raw logarithm 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. 7 of the 17 questions carry real DBE/provincial exam citations (KwaZulu-Natal, Northern Cape and Limpopo, 2025); the rest are original "Equation Station SA Practice Question" items written to match the exact CAPS scope taught in the Summary Notes for this topic.
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 a growth or decay formula already stated, plus recognising which formula a scenario needs.
Q1Equation Station SA Practice Question2 marks
Simple Interest
Simple Growth
R5 000 is invested at 8% 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: \(A=P(1+in)=5\,000(1+0{,}08\times3)\)✓ \(\boxed{R6\,200{,}00}\)
Q2Equation Station SA Practice Question2 marks
Compound Interest
Compound Growth
R5 000 is invested at 8% 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 — that's why n is an exponent: \(A=P(1+i)^n=5\,000(1{,}08)^3\)✓ \(\boxed{R6\,298{,}56}\)
Q3Equation Station SA Practice Question2 marks
Straight-Line Depreciation
Straight-Line Depreciation
Equipment costing R60 000 depreciates on a straight-line basis at 12% p.a. Calculate its book value after 4 years.
Memo
✓ Straight-line depreciation subtracts a fixed fraction of the ORIGINAL cost every year: \(A=P(1-in)=60\,000(1-0{,}12\times4)\)✓ \(\boxed{R31\,200{,}00}\)
Q4Equation Station SA Practice Question1 mark
Growth & Decay Formulas
Choose the Correct Formula
A bank account pays interest calculated only on the original amount deposited, every year, for the life of the account. Write down the formula that models this account's value after \(n\) years.
Memo
✓ Interest on the original amount only, every year, is simple growth — interest never earns interest on itself, unlike compound growth.✓ \(\boxed{A=P(1+in)}\)
29%
Level 2 | Routine Procedures
One Established Method
Hire purchase, reducing-balance depreciation, a nominal-to-effective conversion, and both annuity formulas applied directly.
Q5Equation Station SA Practice Question4 marks
Hire Purchase
Hire Purchase Instalment
A washing machine has a cash price of R18 000. A shop requires a 15% deposit, with the balance repaid via hire purchase at 13,5% p.a. simple interest over 3 years, in equal monthly instalments. Calculate the monthly instalment.
Memo
✓ Subtract the deposit first — interest is only ever charged on what remains to be financed: deposit \(=0{,}15\times18\,000=R2\,700\); balance \(=R15\,300\)✓ Hire purchase interest is always simple interest, applied once over the full term: \(15\,300\times0{,}135\times3=R6\,196{,}50\)✓ Add the interest to the balance to get the total actually owed: \(15\,300+6\,196{,}50=R21\,496{,}50\)✓ Spread that total evenly over the 36 months: \(\boxed{R597{,}13}\) per month
Q6KwaZulu-Natal DBE, Paper 1, June 2025 (Q7.2)3 marks
Reducing-Balance Depreciation
Reducing-Balance Depreciation (Reverse)
Bongiwe's car has depreciated to a value of R230 476,05 after three years. Depreciation is calculated at a rate of 13% p.a., using the reducing-balance method. Calculate the price at which Bongiwe bought her car.
Memo
✓ Rearrange the reducing-balance formula to make the original price P the subject, by dividing rather than subtracting: \(A=P(1-i)^n\Rightarrow P=\dfrac{A}{(1-i)^n}=\dfrac{230\,476{,}05}{(0{,}87)^3}\)✓ \(\boxed{R350\,000{,}00}\)
Q7Northern Cape DBE, Paper 1, September 2025 (Q6.1.1)3 marks
Nominal & Effective Rates
Nominal to Effective
Thabo deposits R3 550 into a savings account which pays interest at a rate of 8% p.a., compounded half-yearly. Calculate the effective annual interest rate he receives on his savings.
Memo
✓ Converting to an effective rate accounts for the extra growth from 2 compounds within the year: \(1+i_{\text{eff}}=\left(1+\dfrac{0{,}08}{2}\right)^2\)✓ \(\boxed{i_{\text{eff}}=8{,}16\%}\)
Q8Equation Station SA Practice Question3 marks
Future Value Annuities
Future Value Annuity
R400 is deposited at the end of every month into an account earning 7,2% p.a. compounded monthly. Calculate the value of the account after 4 years.
Memo
✓ Convert the rate and time period to match the monthly deposit frequency first: \(i=\dfrac{0{,}072}{12}=0{,}006\), \(n=4\times12=48\)✓ Apply the future value annuity formula, derived from the sum of a geometric series of growing deposits: \(F=x\dfrac{(1+i)^n-1}{i}=400\times\dfrac{(1{,}006)^{48}-1}{0{,}006}\)✓ \(\boxed{R22\,174{,}00}\)
Q9Northern Cape DBE, Paper 1, September 2025 (Q6.3.1)3 marks
Present Value Annuities
Loan Instalment
Thembi was granted a loan of R35 000 at an interest rate of 18% p.a., compounded monthly. He agreed to repay the loan over 4 years, in equal monthly instalments. Calculate the monthly instalment.
Memo
✓ Convert the rate and time period to match monthly repayments: \(i=\dfrac{0{,}18}{12}=0{,}015\), \(n=4\times12=48\)✓ A loan is the present value of every instalment still to be paid, so set up that relationship first: \(P=x\dfrac{1-(1+i)^{-n}}{i}\)✓ Rearrange to solve for the instalment \(x\): \(x=\dfrac{Pi}{1-(1+i)^{-n}}\)✓ \(\boxed{R1\,028{,}12}\) per month
24%
Level 3 | Complex Procedures
Multi-Step Methods
Solving for the time period using logarithms in both a growth and a decay context, an outstanding balance, and a real effective-rate comparison.
Q10Equation Station SA Practice Question4 marks
Solving for n (Growth)
Solve for n: Growth
R20 000 is invested at 10,5% p.a. compounded annually. After how many complete years will the investment first be worth at least R35 000?
Memo
✓ Set up the compound growth formula with the unknown \(n\) trapped in the exponent, then use logarithms to bring it down: \(35\,000=20\,000(1{,}105)^n\Rightarrow n=\dfrac{\log(35\,000/20\,000)}{\log1{,}105}\approx5{,}60\)✓ Interest is only credited at the end of a full year, so check the boundary years directly rather than trusting the raw decimal: at \(n=5\), \(R32\,948{,}94\) (still short); at \(n=6\), \(R36\,408{,}57\) (target reached)✓ \(\boxed{6\text{ years}}\)
Q11Equation Station SA Practice Question4 marks
Solving for n (Decay)
Solve for n: Decay
A machine costing R150 000 depreciates on a reducing balance at 22% p.a. After how many complete years will its value first drop below R40 000?
Memo
✓ Set up the compound decay formula and solve for \(n\) with logarithms, exactly as for growth: \(40\,000=150\,000(0{,}78)^n\Rightarrow n=\dfrac{\log(40\,000/150\,000)}{\log0{,}78}\approx5{,}32\)✓ Check the boundary years directly rather than trusting the raw decimal: at \(n=5\), \(R43\,307{,}62\) (still above); at \(n=6\), \(R33\,779{,}94\) (below target)✓ \(\boxed{6\text{ years}}\)
Q12Northern Cape DBE, Paper 1, September 2025 (Q6.3.2)5 marks
Outstanding Balance
Outstanding Balance After Missed Payments
Continuing Thembi's loan (R35 000 at 18% p.a. compounded monthly, monthly instalment R1 028,12): Thembi was unable to pay the 21st and 22nd instalments. If he resumes payment at the end of the 23rd month, what is the outstanding balance at that point?
Memo
✓ The balance immediately after the 20th instalment is the present value of the 28 instalments still owed: \(\approx R23\,365{,}79\)✓ A missed instalment doesn't disappear — the balance keeps accruing interest for every month no payment is made, here 2 extra months: \(23\,365{,}79(1{,}015)^2\)✓ \(\boxed{R24\,072{,}02}\)
Q13KwaZulu-Natal DBE, Paper 1, June 2025 (Q7.1)5 marks
Nominal & Effective Rates
Comparing Two Rates
Nelisiwe received her yearly bonus and decided to invest the full amount. Bank A offers an interest rate of 8,5% p.a., compounded half-yearly. Bank B also offers an interest rate of 8,5% p.a., but compounded monthly. With which bank should she invest, and what effective annual rate does Bank B actually offer?
Memo
✓ Both banks quote the same nominal rate, so convert each to its true effective annual rate to compare fairly, starting with Bank A's half-yearly compounding: \(\left(1+\dfrac{0{,}085}{2}\right)^2-1\approx8{,}68\%\)✓ Then Bank B's monthly compounding: \(\left(1+\dfrac{0{,}085}{12}\right)^{12}-1\approx8{,}84\%\)✓ More frequent compounding of the SAME nominal rate always wins: \(\boxed{\text{Bank B is better}}\)
23%
Level 4 | Problem Solving
Combined Skills
An annuity followed by a further growth period, a genuine loan-option comparison, re-amortizing a real loan after missed payments, and identifying a pyramid scheme.
Q14Limpopo Department of Education, Mathematics Controlled Test 2, 21 August 2025 (Q5.4)6 marks
Future Value AnnuitiesTimeline
Annuity, Then a Further Growth Period
A businesswoman deposited R9 000 into an account at the end of every month for 60 months, earning interest at 7,5% p.a. compounded monthly. She then left the accumulated amount in the account for a further \(n\) months, with no further deposits, at the same interest rate. The total interest earned over the entire investment period was R190 214,14. Determine the value of \(n\).
Memo
✓ Find the value built up by the 60 monthly deposits using the future value annuity formula: \(F=9\,000\dfrac{(1{,}00625)^{60}-1}{0{,}00625}\approx R652\,743{,}95\)✓ Total interest is final value minus total deposited, so work backward from the stated interest to find the true final value: \(60\times9\,000+190\,214{,}14=R730\,214{,}14\)✓ With no further deposits, that final value is simply the annuity's value growing further at compound interest — solve for the extra time using logarithms: \(730\,214{,}14=652\,743{,}95(1{,}00625)^n\)✓ \(\boxed{n=18\text{ months}}\)
Q15Equation Station SA Practice Question5 marks
Critically Analysing Loans
Comparing Two Loan Options
A learner needs to borrow R200 000. Option A charges 15,5% p.a. compounded monthly, repaid over 5 years. Option B charges 12,5% p.a. compounded monthly, repaid over 7 years. Which option costs less in total interest over the full life of the loan?
Memo
✓ Work out Option A's full cost: instalment \(\approx R4\,810{,}64\) over 60 months, total \(\approx R288\,638{,}29\), interest \(\approx R88\,638{,}29\)✓ Then Option B's full cost, over its longer term: instalment \(\approx R3\,584{,}25\) over 84 months, total \(\approx R301\,076{,}81\), interest \(\approx R101\,076{,}81\)✓ A lower rate and a lower instalment are both traps if the comparison stops there — the shorter term means far fewer months of interest actually accumulate: \(\boxed{\text{Option A costs less overall}}\)
Q16Limpopo Department of Education, Mathematics Controlled Test 2, 21 August 2025 (Q5.3)7 marks
Present Value AnnuitiesMissed Payments
Re-Amortizing a Loan After Missed Payments
Exactly 10 months ago, a bank granted Jane a loan of R800 000 at an interest rate of 10,25% p.a., compounded monthly, to be repaid over 20 years by monthly instalments of R7 853,15, starting one month after the loan was granted. Due to financial difficulties, Jane missed the 7th, 8th and 9th instalments, but is able to resume paying from the end of the 10th month onwards. Calculate Jane's new (increased) monthly instalment, so that the loan is still settled within the original 20 years.
Memo
✓ Find the balance immediately after the 6th instalment, using the loan's own monthly rate: \(800\,000(1{,}008542)^6-7\,853{,}15\dfrac{(1{,}008542)^6-1}{0{,}008542}\approx R793\,748{,}94\)✓ A missed instalment doesn't disappear — that balance keeps growing for the 3 unpaid months before payments resume: \(793\,748{,}94(1{,}008542)^3\approx R814\,262{,}99\)✓ To still finish within the original 20-year term, re-amortize this new balance over the number of months genuinely remaining, counted from where payments actually resume: \(240-9=231\) months✓ \(\boxed{x\approx R8\,089{,}20}\)
Q17Equation Station SA Practice Question3 marks
Critically Analysing Loans
Recognising a Pyramid Scheme
A friend offers you an "investment": pay in now, and you are guaranteed a 40% return within one month, with your profit paid from money brought in by new investors you recruit, rather than from any real product, service, or trading activity. What should you conclude, and why?
Memo
✓ Returns paid from new recruits' money, rather than any real product, service, or trading activity, is the exact definition of a pyramid scheme✓ Because payouts depend on constantly recruiting new members, every cycle needs exponentially more people than the last — a mathematical guarantee of eventual collapse, not just bad luck✓ \(\boxed{\text{Such schemes are illegal in South Africa}}\), and most participants lose their money regardless of any paperwork used to dress it up