17 questions arranged by DBE cognitive level — simple and compound decay, straight-line and reducing-balance depreciation, and nominal and effective interest rates, including 6 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 straight-line or reducing balance, ask: does this lose the same RAND amount every year, or the same PERCENTAGE of what's left?
Whenever a nominal rate and a compounding period are both given, convert them to match BEFORE substituting into any formula.
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. 11 are original "Equation Station SA Practice Question" items; 6 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 decay, compound decay, or straight-line depreciation formula, plus recognising the two depreciation methods.
Q1Equation Station SA Practice Question2 marks
Simple Decay
Simple Decay
R12 000 depreciates at 8% p.a. on a simple-decay (straight-line) basis for 3 years. Calculate its value at the end of the 3 years.
Memo
✓ This is simple (straight-line) decay, so the SAME rand amount is lost every year, based on the ORIGINAL value: \(A=P(1-in)=12\,000(1-0{,}08\times3)\)✓ \(\boxed{R9\,120{,}00}\)
Q2Equation Station SA Practice Question2 marks
Compound Decay
Compound Decay
R12 000 depreciates at 8% p.a. on a reducing balance for 3 years. Calculate its value at the end of the 3 years.
Memo
✓ This is compound (reducing-balance) decay, so each year's loss is a percentage of the PREVIOUS year's already-reduced value: \(A=P(1-i)^n=12\,000(0{,}92)^3\)✓ \(\boxed{R9\,344{,}26}\)
Q3Equation Station SA Practice Question2 marks
Straight-Line Depreciation
Straight-Line Depreciation
Machinery costing R45 000 depreciates on a straight-line basis at 10% p.a. Calculate its book value after 5 years.
Memo
✓ Straight-line depreciation subtracts the same fraction of the ORIGINAL cost every year: \(A=P(1-in)=45\,000(1-0{,}10\times5)\)✓ \(\boxed{R22\,500{,}00}\)
Q4Equation Station SA Practice Question1 mark
Depreciation Methods
Identify the Depreciation Method
An asset loses the same PERCENTAGE of its remaining (already-reduced) value every year. Which depreciation method is this?
Memo
✓ Losing a percentage of the already-reduced value each year means the base amount keeps shrinking — that is reducing-balance depreciation, which is exactly why it never reaches exactly R0.✓ \(\boxed{\text{Reducing-balance depreciation}}\)
29%
Level 2 | Routine Procedures
One Established Method
Reducing-balance depreciation, a nominal-to-effective conversion, solving for a depreciation rate, and quantifying the gap between the two decay methods.
Q5KwaZulu-Natal Provincial Topic Test, Finance, Growth & Decay, September 2025 (Q1.2)4 marks
Reducing-Balance Depreciation
A Six-Year Reducing-Balance Calculation
A school bought a school bus for R850 000. The bus depreciates at 12,5% p.a. on the reducing-balance method. Calculate the book value 6 years after it was bought.
Memo
✓ The bus depreciates on a reducing balance, so apply \(A=P(1-i)^n\) directly: \(A=850\,000(1-0{,}125)^6=850\,000(0{,}875)^6\)✓ \(\boxed{R381\,476{,}02}\)
Q6KwaZulu-Natal Provincial Topic Test, Finance, Growth & Decay, September 2025 (Q1.1)3 marks
Nominal & Effective Rates
Nominal to Effective
Calculate the effective annual interest rate if an investment earns interest at a rate of 9,5% p.a. compounded quarterly.
Memo
✓ Converting a nominal rate to an effective one accounts for the extra growth from compounding within the year — here, 4 times: \(1+i_{\text{eff}}=\left(1+\dfrac{0{,}095}{4}\right)^4\)✓ \(\boxed{i_{\text{eff}}\approx9{,}84\%}\)
Q7Equation Station SA Practice Question3 marks
Straight-Line Depreciation
Solving for the Depreciation Rate
A computer cost R18 000 and has a book value of R9 000 after 3 years, depreciated on a straight-line basis. Calculate the annual rate of depreciation.
Memo
✓ Substitute into the straight-line formula and isolate the bracket first: \(9\,000=18\,000(1-3i)\Rightarrow1-3i=0{,}5\)✓ Since \(i\) is a multiplier here, not trapped in an exponent, simple algebra finishes the job — just don't forget the final division by \(n=3\): \(i=\dfrac{0{,}5}{3}\)✓ \(\boxed{16{,}67\%}\)
Q8North West DBE, Paper 1, November 2025 (Q8.2.3)4 marks
Reducing-Balance Depreciation
Solving for the Rate Using an nth Root
A car bought for R175 000 sells 4 years later for R79 120. Calculate the annual reducing-balance depreciation rate.
Memo
✓ Substitute into the reducing-balance formula: \(79\,120=175\,000(1-i)^4\)✓ Since \(i\) is trapped inside a power of 4, isolate \((1-i)\) and take a 4th root — dividing by \(n\) would be wrong here: \(1-i=\left(\dfrac{79\,120}{175\,000}\right)^{\frac14}\)✓ \(\boxed{18{,}00\%}\)
Q9Equation Station SA Practice Question3 marks
Decay Comparison
Quantifying the Decay-Method Gap
R20 000 depreciates for 4 years at 12% p.a. Calculate how much MORE the reducing-balance value is than the straight-line value.
Memo
✓ Calculate the straight-line value first — a fixed rand amount lost every year: \(20\,000(1-0{,}12\times4)=R10\,400{,}00\)✓ Then the reducing-balance value — each year's loss shrinks along with the value itself: \(20\,000(0{,}88)^4\approx R11\,993{,}91\)✓ Reducing balance always retains MORE value beyond the first year: \(\boxed{R1\,593{,}91}\) more under reducing balance
24%
Level 3 | Complex Procedures
Multi-Step Methods
Comparing effective rates, the effect of compounding periods on decay, and working backward for an original amount.
Q10North West DBE, Paper 1, November 2025 (Q8.1.1 & Q8.1.2)4 marks
Nominal & Effective Rates
Comparing Two Investment Options
Option A pays 8,2% p.a. compounded daily (365 days). Option B pays 8,3% p.a. compounded monthly. Which option is the better investment?
Memo
✓ Convert Option A's nominal rate to an effective annual rate, accounting for its 365 daily compounds: \(\left(1+\dfrac{0{,}082}{365}\right)^{365}-1\approx8{,}54\%\)✓ Do the same for Option B, with its 12 monthly compounds: \(\left(1+\dfrac{0{,}083}{12}\right)^{12}-1\approx8{,}62\%\)✓ Comparing the two effective rates directly (never the nominal rates): \(\boxed{\text{Option B is better}}\), despite Option A compounding daily
Q11Equation Station SA Practice Question4 marks
Effect of Compounding Periods
The Effect of Compounding Frequency on Decay
Equipment costing R200 000 depreciates at a NOMINAL rate of 16% p.a. on a reducing balance for 3 years. Calculate its book value if the depreciation is calculated (a) monthly and (b) annually, and state which method leaves the equipment worth MORE.
Memo
✓ Convert the nominal rate and time period to match monthly compounding first: \(i=\dfrac{0{,}16}{12}\), \(n=36\), giving \(A=200\,000(1-i)^{36}\approx R123\,357{,}75\)✓ Then calculate the same nominal rate compounded only annually: \(A=200\,000(0{,}84)^3\approx R118\,540{,}80\)✓ Comparing the two shows depreciating the SAME nominal rate more often actually SLOWS the loss overall: \(\boxed{\text{Monthly compounding leaves it worth MORE}}\) — the opposite of what happens with growth
Q12KwaZulu-Natal DBE, Paper 1, November 2024 (Q8.2)4 marks
Compound Decay
Working Backward, in a Non-Financial Context
The number of bees in an apiary is declining each year at a rate of 2,5% of the total the previous year. There are currently 20 416 bees. Calculate how many bees there were 8 years ago.
Memo
✓ A shrinking population by a constant percentage each year is compound decay, exactly like a depreciating asset: \(A=P(1-i)^n\)✓ To go BACKWARD from today's value to a past value, divide by the decay factor instead of multiplying: \(P=\dfrac{20\,416}{(0{,}975)^8}\)✓ \(\boxed{25\,000}\) bees
Q13Equation Station SA Practice Question4 marks
Nominal & Effective Rates
Combining a Nominal Rate with Decay
Equipment costing R60 000 depreciates on a reducing balance at a NOMINAL rate of 15% p.a. compounded monthly. Calculate its value after 2 years.
Memo
✓ Convert the nominal rate to match its own stated monthly compounding before substituting anything: \(i=\dfrac{0{,}15}{12}=0{,}0125\), \(n=24\)✓ Now apply the reducing-balance formula with these converted values: \(A=60\,000(1-0{,}0125)^{24}\)✓ \(\boxed{R44\,365{,}13}\)
23%
Level 4 | Problem Solving
Combined Skills
A real multi-stage compounding problem, a genuine method decision, and depreciation combined with a trade-in and a three-way rate comparison.
Q14North West DBE, Paper 1, November 2025 (Q8.2.1 & Q8.2.2, adapted)7 marks
Compound Growth
A Multi-Stage Investment with a Withdrawal
R300 000 is invested at 9% p.a. compounded quarterly. After exactly 2 years, R175 000 is withdrawn. The remaining balance continues at 9% p.a. compounded quarterly until year 4, when the rate changes to 8,5% p.a. compounded monthly for a further 2 years. Calculate the balance after 6 years in total.
Memo
✓ Grow the original deposit for the first 2 years at 9% quarterly: \(300\,000(1{,}0225)^8\approx R358\,449{,}34\)✓ Subtract the withdrawal at the exact moment it happens — never before or after: \(R358\,449{,}34-R175\,000=R183\,449{,}34\)✓ Continue growing what remains at the SAME rate for the next 2 years, to reach year 4: \(\approx R219\,190{,}99\)✓ Only then switch to the new 8,5% monthly rate for the final 2 years: \(\boxed{R259\,652{,}50}\)
Q15Equation Station SA Practice Question5 marks
Decay Comparison
A Genuine Long-Term Method Decision
Two identical machines both cost R500 000 and both depreciate at 12% p.a. Machine A uses straight-line depreciation; Machine B uses reducing-balance depreciation. After 6 years, calculate how much MORE Machine B is worth than Machine A.
Memo
✓ Machine A's straight-line value after 6 years: \(500\,000(1-0{,}12\times6)=R140\,000{,}00\)✓ Machine B's reducing-balance value over the same 6 years: \(500\,000(0{,}88)^6\approx R232\,202{,}04\)✓ The question asks for the DIFFERENCE, not either value alone: \(\boxed{R92\,202{,}04}\) more for Machine B
Q16Equation Station SA Practice Question6 marks
Reducing-Balance Depreciation
Depreciation Combined with a Trade-In
A car costing R320 000 depreciates on a reducing balance at 19% p.a. After 4 years, it is traded in against a new car costing R450 000. Calculate the amount still to be paid after the trade-in value is deducted.
Memo
✓ Calculate the car's actual depreciated value — never assume the trade-in equals the original price: \(320\,000(0{,}81)^4\approx R137\,749{,}51\)✓ Subtract that real trade-in value from the new car's price: \(450\,000-137\,749{,}51\)✓ \(\boxed{R312\,250{,}49}\)
Q17Equation Station SA Practice Question4 marks
Nominal & Effective Rates
A Three-Way Rate Comparison
Option A: 10,2% p.a. compounded monthly. Option B: 10,4% p.a. compounded quarterly. Option C: 10,6% p.a. compounded semi-annually. Which option is the best investment?
Memo
✓ Convert each option to its effective annual rate before comparing anything — starting with Option A's monthly compounding: \(\left(1+\dfrac{0{,}102}{12}\right)^{12}-1\approx10{,}69\%\)✓ Option B, compounded quarterly: \(\left(1+\dfrac{0{,}104}{4}\right)^4-1\approx10{,}81\%\)✓ Option C, compounded semi-annually: \(\left(1+\dfrac{0{,}106}{2}\right)^2-1\approx10{,}88\%\)✓ A higher nominal rate can outweigh a lower compounding frequency: \(\boxed{\text{Option C is best}}\), despite compounding least often