Chapter 7 — Financial Mathematics
A Note on Curriculum Placement
Financial Mathematics is not listed as a separate competence area in the NBT MAT. It falls under Number Sense — but it is severely underrepresented in that chapter relative to how often it appears on the actual test. The reason is that the NBT regularly tests compound interest, depreciation, and hire purchase because these topics require both formula knowledge and careful arithmetic, making them excellent discriminators between students who understand the mathematics and students who merely have a calculator intuition.
This chapter exists because Chapter 2 (Number Sense) could not do these topics justice without becoming unmanageably long. If you are preparing for the NBT, treat Chapters 2 and 7 as a single block — the Number Sense competence area draws from both.
No calculator is permitted in the NBT. Every question in this chapter is designed so that the arithmetic resolves to a clean integer or simple fraction. The numbers are not realistic — real financial products rarely use 10 % or 20 % rates — but the mechanics are identical to real-world problems. Master the method with clean numbers, and the messier real-world versions become straightforward.
Formulas you must know: - Simple interest: $A = P(1 + rt)$ or $I = Prt$ - Compound interest: $A = P(1 + i)^n$ - Straight-line depreciation: $A = P(1 - rt)$ - Reducing-balance depreciation: $A = P(1 - i)^n$
Section 7.1 — Simple Interest
Simple interest grows by the same fixed amount each period because interest is always calculated on the original principal. If you borrow R1000 at 10 % p.a. simple interest, you pay R100 interest every year regardless of how long the loan has been running. This contrasts with compound interest, where the interest accrues on a growing balance.
Topic: Simple interest — calculate interest earned
R4 000 is invested at 5 % p.a. simple interest for 3 years. How much interest is earned?
A) R600
B) R700
C) R500
D) R1 200
Show the worked solution
Answer: A
Explanation
Simple interest is worked out on the original amount only, so the same rand amount is earned every year.
Year by year: 5 % of R4 000 = R200 a year. Three years: $3 \times 200 = \text{R}600$.
By formula: $I = Prt = 4\,000 \times 0.05 \times 3 = \text{R}600$
Why the others are wrong:
- D (R1 200) doubles the rate — that is $4\,000 \times 0.10 \times 3$.
- B (R700) and C (R500) are 3.5 and 2.5 years' worth. They punish estimating instead of computing. At exactly R200 a year, R600 is exact.
Takeaway: $I = Prt$. If the question asks for the interest, stop at $I$. If it asks for the amount, add $P$ back: $A = \text{R}4\,600$. The NBT sets both versions.
Topic: Simple interest — calculate accumulated amount
R2 000 is invested at 10 % p.a. simple interest for 4 years. What is the accumulated amount?
A) R2 400
B) R2 600
C) R2 800
D) R2 928
Show the worked solution
Answer: C
Explanation
The question wants the total: your R2 000 plus the interest. Use $A = P(1+rt)$.
Step 1 — Find the multiplier. 10 % for 4 years = 40 % growth, so multiply by 1.4.
Step 2 — Multiply. $$A = 2\,000 \times 1.4 = \text{R}2\,800$$
Check: R200 interest a year × 4 years = R800, and $2\,000+800 = \text{R}2\,800$ ✓
Why the others are wrong:
- A (R2 400) — only two years of interest.
- B (R2 600) — three years instead of four.
- D (R2 928) — the compound answer, $2\,000\times1.1^4$. The word "simple" is doing real work here.
Takeaway: Simple interest multiplies by $(1+rt)$. Compound multiplies by $(1+i)^n$. Check which word the question uses before you start.
Topic: Simple interest — find principal
A simple interest investment earns R300 interest at 15 % p.a. over 2 years. What was the principal?
A) R900
B) R1 000
C) R1 200
D) R800
Show the worked solution
Answer: B
Explanation
A reverse problem: you are given the interest and must find the principal. Rearrange first, then check forwards.
Step 1 — Rearrange. $P = \dfrac{I}{rt}$
Step 2 — Substitute. The denominator is $0.15 \times 2 = 0.30$: $$P = \frac{300}{0.30} = \text{R}1\,000$$
Step 3 — Check forwards. $1\,000 \times 0.15 \times 2 = 300$ ✓
Why the others are wrong:
- C (R1 200) mis-multiplied the denominator as $0.25$.
- A (R900) fails the check: $900\times0.30 = 270$.
- D (R800) fails too: $800\times0.30 = 240$.
Takeaway: On any find-the-principal/rate/time question, substitute your answer back into $I=Prt$. Five seconds, and it catches nearly every error.
Topic: Simple interest — find time
R5 000 is invested and earns R3 000 in simple interest at 12 % p.a. For how long was it invested?
A) 3 years
B) 4 years
C) 6 years
D) 5 years
Show the worked solution
Answer: D
Explanation
The unknown is the time. Easiest route: work out the interest per year first.
Step 1 — Interest per year. $5\,000 \times 0.12 = \text{R}600$
Step 2 — How many years of R600 make R3 000? $$t = \frac{3\,000}{600} = 5 \text{ years}$$
Why the others are wrong: each assumes a different rate.
- C (6 years) assumes R500 a year — a 10 % rate.
- B (4 years) assumes R750 a year — 15 %.
- A (3 years) assumes R1 000 a year — 20 %.
Takeaway: "Interest per year, then divide" turns every find-the-time question into one small division. Check: $5\,000 \times 0.12 \times 5 = 3\,000$ ✓
Topic: Simple interest — find rate
R8 000 grows to R9 200 in 3 years under simple interest. What is the annual interest rate?
A) 5 % p.a.
B) 6 % p.a.
C) 4 % p.a.
D) 7.5 % p.a.
Show the worked solution
Answer: A
Explanation
You are given the accumulated amount, not the interest, so split the interest off first.
Step 1 — Extract the interest. $I = 9\,200 - 8\,000 = \text{R}1\,200$
Step 2 — Substitute. $$r = \frac{I}{Pt} = \frac{1\,200}{8\,000 \times 3} = 0.05 = 5\%$$
Step 3 — Check. R8 000 at 5 % earns R400 a year; three years is R1 200 ✓
Why the others are wrong:
- D (7.5 %) used two years instead of three.
- C (4 %) divided by the final amount instead of the principal. Rates are always quoted on the principal.
- B (6 %) would earn R1 440, overshooting the actual R1 200.
Takeaway: Given $A$ rather than $I$? The first line of working is always $I = A - P$. Most wrong rate answers are born in that missed step.
Section 7.2 — Compound Interest
Compound interest earns "interest on interest" — each period's interest is added to the principal before the next period's interest is calculated. This exponential growth is what makes compound interest so powerful over long time periods, and why the NBT tests it regularly.
Topic: Compound interest — calculate $A$, one period
R10 000 is invested at 10 % p.a. compounded annually for 2 years. What is the accumulated amount?
A) R12 000
B) R12 200
C) R12 100
D) R13 100
Show the worked solution
Answer: C
Explanation
Compound interest is worked out on the current balance, not the original deposit. Watch it:
- Year 1: 10 % of R10 000 = R1 000 → R11 000
- Year 2: 10 % of R11 000 = R1 100 → R12 100
The second year earned R100 more — interest on the interest.
$$A = P(1+i)^n = 10\,000 \times 1.21 = \text{R}12\,100$$
Why the others are wrong:
- A (R12 000) is $10\,000\times1.20$, the simple interest answer. The R100 gap between A and C is compounding.
- B (R12 200) mis-squared $1.1$ as $1.22$. Write it out: $1.1\times1.1 = 1.21$.
- D (R13 100) is roughly the three-year value — an exponent miscount.
Takeaway: For 10 % over 2 years the factor is $1.1^2 = 1.21$, never $1.20$. Running one year at a time is often faster and safer than the formula.
Topic: Compound interest — three-year period
R1 000 is invested at 10 % p.a. compounded annually for 3 years. What is the accumulated amount?
A) R1 300
B) R1 331
C) R1 360
D) R1 221
Show the worked solution
Answer: B
Explanation
$$A = 1\,000 \times (1.1)^3 = \text{R}1\,331$$
Not sure of $1.1^3$? Build it: R1 000 → R1 100 → R1 210 → R1 331.
A pattern worth knowing — the digits of the powers of 1.1 are Pascal's triangle:
$$1.1^2 = 1.21 \qquad 1.1^3 = 1.331 \qquad 1.1^4 = 1.4641$$
The NBT uses 10 % rates precisely because these work without a calculator.
Why the others are wrong:
- A (R1 300) is the simple interest value, missing the R31 of interest-on-interest.
- D (R1 221) is a corrupted Pascal row — 1,2,2,1 instead of 1,3,3,1.
- C (R1 360) botched the expansion's middle term ($0.06$ instead of $0.03$).
Takeaway: $1.1^3 = 1.331$. Learn the ladder $1.21$, $1.331$, $1.4641$ — it turns these into ten-second questions.
Topic: Compound interest — find $n$
R10 000 grows to R14 641 at 10 % p.a. compounded annually. For how many years?
A) 4 years
B) 3 years
C) 2 years
D) 5 years
Show the worked solution
Answer: A
Explanation
Step 1 — isolate the growth factor. Divide both sides by the principal:
$$(1.1)^n = \frac{14\,641}{10\,000} = 1.4641$$
Step 2 — climb the ladder of powers.
$$1.1^1 = 1.1 \qquad 1.1^2 = 1.21 \qquad 1.1^3 = 1.331 \qquad 1.1^4 = 1.4641$$
The growth factor matches the fourth rung, so $n = 4$ years.
Why the others are wrong: all three are neighbouring rungs of the same ladder —
- B (3 years) would give $10\,000 \times 1.331 = \text{R}13\,310$: not enough.
- C (2 years) would give R12 100: far too little.
- D (5 years) would give $10\,000 \times 1.61051 \approx \text{R}16\,105$: too much.
If you know the ladder, no wrong option is even tempting; if you don't, every option looks equally plausible. That is the whole game with exponent-matching questions — they test recognition, not computation.
Takeaway: Divide $A$ by $P$ first, then match the growth factor against known powers. The values $1.21$, $1.331$, $1.4641$ should be instant recognitions by NBT day.
Topic: Compound vs simple — compare after 3 years
R1 000 is invested for 3 years at 10 % p.a. Which is correct?
A) Simple interest gives R1 300; compound interest also gives R1 300 — they are equal over 3 years
B) Simple interest always grows faster than compound interest
C) Simple interest gives R1 300; compound interest gives R1 210
D) Compound interest gives R1 331; this is R31 more than simple interest
Show the worked solution
Answer: D
Explanation
Work out both totals side by side — the comparison is the whole point.
Simple: the same R100 every year. $$A = 1000(1.3) = \text{R}1\,300$$
Compound: 10 % of a growing balance (R100, then R110, then R121). $$A = 1000(1.1)^3 = \text{R}1\,331$$
Compound wins by R31.
Why the others are wrong:
- A says they tie. They agree only after one period; after that compound always pulls ahead.
- B has it backwards — simple never beats compound at the same rate.
- C quotes R1 210, the two-year compound value, hidden inside a true-sounding sentence.
Takeaway: At equal rates compound always ≥ simple, and the gap widens each year. When options are sentences, check every number inside them, not just the story.
Topic: Compound interest — find $n$ given $A$
R10 000 grows to R12 100 at 10 % p.a. compounded annually. How many years?
A) 1 year
B) 3 years
C) 2 years
D) 4 years
Show the worked solution
Answer: C
Explanation
Step 1 — find the growth factor.
$$(1.1)^n = \frac{12\,100}{10\,000} = 1.21$$
Step 2 — recognise it. $1.21 = 1.1^2$, so $n = 2$ years. (Year-by-year confirmation: R10 000 → R11 000 → R12 100 ✓.)
Why the others are wrong: the wrong options are simply other rungs of the 10 % ladder —
- A (1 year) would leave the balance at R11 000.
- B (3 years) would grow it to R13 310.
- D (4 years) would grow it to R14 641.
Only 2 years lands exactly on R12 100. If two different rungs both feel plausible, it means the growth factor was never actually computed — always do the division $A/P$ first.
Takeaway: Divide $A$ by $P$ to get the growth factor, then match it to a known power of $(1+i)$. This "divide, then recognise" routine solves every find-$n$ compound question the NBT can ask.
Section 7.3 — Depreciation
Assets lose value over time. The two standard methods are:
- Straight-line (linear): the asset loses the same rand amount each year: $A = P(1 - rt)$.
- Reducing-balance (exponential): the asset loses the same percentage each year: $A = P(1 - i)^n$.
Reducing-balance always gives a higher residual value than straight-line for the same rate and period, because the depreciation amount shrinks each year (you're taking the percentage off a smaller and smaller base).
Topic: Straight-line depreciation — find value
A machine bought for R200 000 depreciates at 8 % p.a. straight-line. What is its value after 5 years?
A) R100 000
B) R120 000
C) R140 000
D) R160 000
Show the worked solution
Answer: B
Explanation
Straight-line depreciation is simple interest in reverse: the same rand amount is lost each year, always based on the original price.
Loss per year: $200\,000 \times 0.08 = \text{R}16\,000$
Over 5 years: $5 \times 16\,000 = \text{R}80\,000$ lost, leaving R120 000.
By formula: $A = P(1-rt) = 200\,000(0.60) = \text{R}120\,000$
Why the others are wrong — each uses the wrong rate:
- A (R100 000) writes off half — that needs 10 %, not 8 %.
- C (R140 000) writes off R12 000 a year — a 6 % rate.
- D (R160 000) writes off R8 000 a year — 4 %, exactly half the true rate.
Takeaway: Think "rand lost per year × years". After 5 years at 8 %, 40 % is gone and 60 % remains — which is all $(1-rt)$ says.
Topic: Reducing-balance depreciation — find value
Equipment costing R100 000 depreciates at 20 % p.a. on the reducing balance. Find its value after 3 years.
A) R51 200
B) R52 000
C) R60 000
D) R48 000
Show the worked solution
Answer: A
Explanation
Reducing balance is compound interest running downhill: each year the asset loses 20 % of its current worth, so the rand loss shrinks. Watch the chain:
$$100\,000 \to 80\,000 \to 64\,000 \to \text{R}51\,200$$
The losses are R20 000, R16 000, R12 800 — smaller each time, because 20 % comes off a smaller base.
$$A = P(1-i)^n = 100\,000(0.8)^3 = \text{R}51\,200$$
Why the others are wrong:
- C (R60 000) is straight-line thinking, and only two years of it.
- D (R48 000) reached R64 000 then subtracted R16 000 again, instead of recomputing 20 % on the new balance.
- B (R52 000) rounded $0.8^3$ to $0.52$ instead of $0.512$.
Takeaway: Multiply by 0.8 year by year — 100 → 80 → 64 → 51.2 is three quick steps and cannot go wrong.
Topic: Compare straight-line and reducing-balance
R100 000 asset. Both methods use 20 % p.a. After 3 years:
A) Both methods give R40 000 after 3 years
B) Straight-line gives R51 200; reducing balance gives R40 000
C) The two methods always produce the same residual value
D) Straight-line gives R40 000; reducing balance gives R51 200
Show the worked solution
Answer: D
Explanation
Run both methods and compare.
Straight-line removes a flat R20 000 (20 % of the original) each year: $$A = 100\,000(1 - 0.60) = \text{R}40\,000$$
Reducing-balance removes 20 % of the current value — R20 000, then R16 000, then R12 800: $$A = 100\,000(0.8)^3 = \text{R}51\,200$$
Reducing balance leaves more, because its deductions shrink.
Why the others are wrong:
- B has both numbers right but swapped — the classic compare-question error. Anchor it: shrinking deductions must leave the higher value.
- A gives both methods the straight-line answer, denying the very difference being tested.
- C says they always agree — true only after one year.
Takeaway: At the same rate, straight-line reaches zero in finite time; reducing balance decays forever without quite getting there. If your numbers are right but your letter is wrong, you fell for the swap.
Topic: Straight-line depreciation — find rate
Equipment worth R150 000 has a book value of R60 000 after 6 years using straight-line depreciation. What is the annual depreciation rate?
A) 6 % p.a.
B) 8 % p.a.
C) 10 % p.a.
D) 12 % p.a.
Show the worked solution
Answer: C
Explanation
A reverse problem. Find the loss, spread it evenly over the years, then express one year's loss as a percentage of the original.
Total loss: $150\,000 - 60\,000 = \text{R}90\,000$
Per year: $90\,000 \div 6 = \text{R}15\,000$
As a rate: $\dfrac{15\,000}{150\,000} = 10\%$
Algebraically: $60\,000 = 150\,000(1-6r) \Rightarrow r = 0.10$
Why the others are wrong:
- D (12 %) divided the 60 % loss by 5 years instead of 6.
- A (6 %) divided the remaining 40 % instead of the amount lost.
- B (8 %) fails the check: 8 % for 6 years leaves R78 000, not R60 000.
Takeaway: $r = (1 - A/P)/t$ — or in words, "total loss ÷ years ÷ original price", which is much harder to mangle under pressure.
Topic: Reducing-balance depreciation — find rate
Equipment bought for R100 000 has a book value of R64 000 after 2 years using reducing-balance depreciation. What is the annual depreciation rate?
A) 16 % p.a.
B) 20 % p.a.
C) 18 % p.a.
D) 36 % p.a.
Show the worked solution
Answer: B
Explanation
Reducing balance in reverse. Two years means the original was multiplied by $(1-i)^2$.
Step 1 — Isolate the factor. $(1-i)^2 = \dfrac{64\,000}{100\,000} = 0.64$
Step 2 — Take the square root. $1-i = 0.8$
Step 3 — Read the rate. $i = 0.20 = 20\%$
Check: 100 000 → 80 000 → 64 000 ✓
Why the others are wrong:
- D (36 %) is the total two-year loss, not the annual rate — true information answering the wrong question.
- C (18 %) halves that total. Averaging works for straight-line, but a compound loss is bigger in year one than year two.
- A (16 %) fails the check: $0.84^2 = 0.7056$ leaves R70 560.
Takeaway: Isolate $(1-i)^n$, take the $n$th root, subtract from 1. Never annualise a compound loss by dividing — that shortcut belongs to straight-line only.
Section 7.4 — Hire Purchase
Hire purchase (HP) allows a buyer to take possession of an item immediately and pay for it in instalments. The finance charge is calculated as simple interest on the original balance (after the deposit), not on the declining balance. This makes HP significantly more expensive than the stated interest rate suggests.
Topic: Hire purchase — calculate monthly payment
A TV costs R6 000 cash. HP terms: 10 % deposit, 24 monthly payments, with 10 % p.a. simple interest charged on the balance. What is the monthly payment?
A) R270
B) R300
C) R225
D) R312
Show the worked solution
Answer: A
Explanation
Hire purchase always follows four beats: deposit → balance → interest on that balance → divide by the number of payments. Note 24 monthly payments = 2 years of interest.
Step 1 — Deposit and balance. $$\text{Deposit} = 10\% \times 6\,000 = \text{R}600 \qquad \text{Balance} = \text{R}5\,400$$
Step 2 — Interest (simple, 2 years). $$I = 5\,400 \times 0.10 \times 2 = \text{R}1\,080$$
Step 3 — Total, then monthly. $$\frac{5\,400 + 1\,080}{24} = \frac{6\,480}{24} = \text{R}270$$
Why the others are wrong:
- C (R225) is $5\,400 \div 24$ — the interest forgotten entirely.
- B (R300) never subtracted the deposit, charging on the full R6 000.
- D (R312) implies R7 488 repaid — over R2 000 of interest on R5 400. Nothing in the question gives that.
Takeaway: HP interest is simple interest on the opening balance, for the full term. It does not shrink as you pay the debt down — which is exactly why HP is expensive.
Topic: Hire purchase — total cost
A fridge costs R5 000 cash. HP terms: 15 % deposit, 18 monthly payments of R280. What is the total HP cost?
A) R5 040
B) R5 400
C) R5 600
D) R5 790
Show the worked solution
Answer: D
Explanation
The instalment is given, so there is no interest to work out. The total is simply deposit + all instalments. This tests bookkeeping: find both parts and drop neither.
Deposit: $0.15 \times 5\,000 = \text{R}750$
Instalments: $18 \times 280 = \text{R}5\,040$
Total: $\text{R}5\,790$ — that is R790 above the cash price, which is the finance charge.
Why the others are wrong:
- A (R5 040) is the instalments alone — the deposit vanished. The most common HP error.
- B (R5 400) and C (R5 600) are plausible-looking numbers between the cash price and the true cost, but neither equals deposit + instalments.
Takeaway: Total HP cost = deposit + (payments × payment amount). Write both components down before adding — the wrong options sit exactly where a skipped component lands you.
Topic: Hire purchase — implied interest rate
An item costs R4 000. No deposit is required; 12 monthly payments of R400 are charged. What annual simple interest rate does this represent?
A) 16 %
B) 24 %
C) 20 %
D) 12 %
Show the worked solution
Answer: C
Explanation
Reverse-engineer the rate hiding inside the payment plan.
Paid in total: $12 \times 400 = \text{R}4\,800$
Interest: $4\,800 - 4\,000 = \text{R}800$
As an annual rate (the term is 1 year): $$r = \frac{800}{4\,000} = 20\%$$
Why the others are wrong:
- A (16 %) divided by the total paid instead of the amount financed. Rates are quoted on the principal.
- D (12 %) echoes the "12" in the question — a surface-number trap.
- B (24 %) would mean R960 of interest, but only R800 was charged.
Takeaway: Total paid − price = interest, then $r = \frac{I}{Pt}$. An innocent "R400 a month" hides a 20 % annual rate — the consumer lesson as much as the maths one.
Topic: Hire purchase — extra cost vs cash
An item costs R12 000 cash. HP: 25 % deposit plus 36 monthly payments of R350. How much more does the HP buyer pay compared to the cash buyer?
A) R2 600
B) R3 600
C) R4 200
D) R1 200
Show the worked solution
Answer: B
Explanation
Four lines — write every one down. The options are placed to catch anyone who skips a step.
- Deposit: $0.25 \times 12\,000 = \text{R}3\,000$
- Instalments: $36 \times 350 = \text{R}12\,600$
- Total HP: $\text{R}15\,600$
- Extra over cash: $15\,600 - 12\,000 = \text{R}3\,600$
Why the others are wrong:
- C (R4 200) is $350 \times 12$ — one year of instalments, answering a different question.
- A (R2 600) and D (R1 200) are what broken chains produce. Neither matches any clean method.
Takeaway: Extra cost = (deposit + all instalments) − cash price. Four short lines on paper beat holding the chain in your head.
Topic: Loan repayment — simple interest
A loan of R20 000 is taken at 8 % p.a. simple interest for 3 years. What is the total amount owed at the end?
A) R24 800
B) R25 000
C) R24 000
D) R25 250
Show the worked solution
Answer: A
Explanation
A loan is an investment with the roles reversed — the debt is what grows. Simple interest, so it grows by the same amount each year.
Per year: 8 % of R20 000 = R1 600. Three years adds R4 800, so you owe R24 800.
By formula: $A = 20\,000(1 + 0.24) = \text{R}24\,800$
Why the others are wrong:
- C (R24 000) used a multiplier of 1.20 — rounding $0.08\times3$ to "about 20 %". It is exactly 24 %.
- D (R25 250) is compound thinking; the question says simple.
- B (R25 000) is round-number bait matching no rate in the question.
Takeaway: $A = P(1+rt)$. When the options cluster close together, exact arithmetic — not estimation — is the tiebreaker.
Section 7.5 — Inflation and Real Value
Inflation erodes buying power: if prices rise at $\pi$ % p.a., an item costing $P$ today will cost $P(1+\pi)^n$ in $n$ years. The real return on an investment is approximately (nominal rate) − (inflation rate). If your savings account earns less than inflation, your money buys less each year even as the balance grows.
Topic: Future cost under inflation
An item costs R5 000 today. If inflation averages 20 % p.a., what will the item cost in 2 years?
A) R7 000
B) R6 000
C) R6 200
D) R7 200
Show the worked solution
Answer: D
Explanation
Inflation compounds exactly like interest — next year's increase is calculated on this year's already-raised price.
Year by year: R5 000 rises 20 % to R6 000; then R6 000 rises 20 % (that's R1 200 now, not R1 000) to R7 200.
Formula:
$$A = 5\,000 \times (1.2)^2 = 5\,000 \times 1.44 = \text{R}7\,200$$
Why the others are wrong:
- B (R6 000) applies only one year of inflation — it stops halfway.
- A (R7 000) is simple-growth thinking: $20\% \times 2 = 40\%$, so $5\,000 \times 1.4$. It misses the R200 that comes from year two's increase acting on year one's increase.
- C (R6 200) mangles the square: $(1.2)^2 = 1 + 2(0.2) + 0.04 = 1.44$, but dropping one of the two cross-terms gives $1.24$. Write out $1.2 \times 1.2$ instead of expanding in your head.
Takeaway: Future cost = $P(1 + \pi)^n$ — inflation is compound growth applied to prices. "Two years at 20 %" means ×1.44, never ×1.40.
Topic: Budget for a future purchase
A camera costs R3 000 today. Inflation is 10 % p.a. You plan to buy one in 3 years. How much should you budget?
A) R3 300
B) R3 900
C) R3 993
D) R4 000
Show the worked solution
Answer: C
Explanation
Three years of 10 % inflation means three multiplications by 1.1 — the same ladder as compound interest:
$$A = 3\,000 \times (1.1)^3 = 3\,000 \times 1.331 = \text{R}3\,993$$
Year-by-year check: R3 000 → R3 300 → R3 630 → R3 993 ✓.
Why the others are wrong:
- A (R3 300) budgets for one year of inflation, not three.
- B (R3 900) is the simple-growth answer ($10\% \times 3 = 30\%$, so $3\,000 \times 1.3$) — it ignores that each year's increase compounds on the last.
- D (R4 000) is a clean round number that corresponds to nothing: it would require total inflation of $33.3\%$, and $1.1^3$ is exactly 1.331. Tidy-looking options with no derivation behind them are decoys for estimators.
Takeaway: $1.1^3 = 1.331$ — the camera costs R993 more in three years. Budgeting questions are compound-growth questions wearing a shopping bag.
Topic: Approximate real return
An investment yields 12 % p.a. Inflation is 8 % p.a. What is the approximate real return?
A) 20 % p.a.
B) Approximately 4 % p.a.
C) 1.5 % p.a.
D) 8 % p.a.
Show the worked solution
Answer: B
Explanation
The real return asks: after prices have risen, how much more can your money actually buy? The working approximation is a subtraction:
$$\text{Real return} \approx \text{Nominal} - \text{Inflation} = 12\% - 8\% = 4\% \text{ p.a.}$$
Your balance grows by 12 %, but everything you might buy costs 8 % more — so your buying power gains only about 4 %.
Why the others are wrong:
- A (20 %) adds the rates. Inflation works against your return, not alongside it — adding makes no financial sense.
- C (1.5 %) divides them ($12 \div 8$). The ratio of the two rates measures nothing meaningful here.
- D (8 %) just echoes the inflation figure back at you — the surface-number reflex again.
Takeaway: Real ≈ nominal − inflation, and that approximation is always acceptable on the NBT. (The precise Fisher equation gives $(1.12/1.08) - 1 \approx 3.7\%$ — close to 4 %, and the exam never demands that precision.)
Topic: Real value of savings
A savings account earns 8 % p.a. Inflation is running at 10 % p.a. In real terms, the purchasing power of the savings is:
A) Decreasing — the real return is negative
B) Growing at 18 % p.a. in real terms
C) Growing at 2 % in real terms
D) Unaffected — inflation does not reduce the nominal balance
Show the worked solution
Answer: A
Explanation
Same subtraction as Q23, but the result is negative:
$$\text{Real return} \approx 8\% - 10\% = -2\%$$
The balance still grows in rands, but prices rise faster — so those rands buy about 2 % less each year.
Why the others are wrong:
- C (+2 %) subtracted backwards. The order is your rate minus inflation, and the sign is the whole answer.
- B (18 %) added the rates, which would mean high inflation helping your savings.
- D is the "money illusion" — judging wealth by the number on the balance. Inflation does not touch that number; it eats what the number can buy.
Takeaway: Inflation above your interest rate means a negative real return. The NBT tests this as a concept — know which way the subtraction goes.
Topic: Inflation — one-year price change
Groceries cost R500 per month. Food inflation is 6 % p.a. What monthly budget is needed in one year?
A) R506
B) R560
C) R530
D) R550
Show the worked solution
Answer: C
Explanation
One year, one increase — multiply once by $(1 + \pi)$:
$$A = 500 \times 1.06 = \text{R}530$$
(6 % of R500 is R30, because 1 % of R500 is R5.)
Why the others are wrong:
- A (R506) adds six rand instead of six per cent — R6 is 6 % of R100, not of R500. Percent-versus-rand confusion is the oldest trap in inflation questions.
- D (R550) adds 10 % — the round-number reflex overriding the actual rate in the question.
- B (R560) adds 12 %, double the true rate — the footprint of applying 6 % twice or slipping a decimal.
Takeaway: For one year, $A = P(1 + \pi)$. Convert the percentage properly — 6 % of R500 = R30 — so the budget must rise to R530. (That is an extra R360 a year, which is why food inflation hurts.)
Section 7.6 — Exchange Rates
An exchange rate quotes how many units of one currency equal one unit of another. To convert from the stronger currency (e.g. USD), multiply by the rate. To convert to the stronger currency, divide by the rate. Commission is typically charged as a percentage of the transaction value.
Topic: Convert ZAR to USD
The exchange rate is R18 = \$1. Convert R9 000 to US dollars.
A) \$50
B) \$162 000
C) \$450
D) \$500
Show the worked solution
Answer: D
Explanation
The rate "R18 = \$1" says each dollar costs R18. To find how many dollars R9 000 buys, ask: how many R18-chunks fit into R9 000?
$$\text{USD} = \frac{9\,000}{18} = \$500$$
Sense-check first, always: a dollar is worth many rand, so converting rand into dollars must give a much smaller number. Expecting "smaller" before you calculate makes the direction error impossible.
Why the others are wrong:
- B (\$162 000) multiplies instead of divides — the direction error the sense check exists to catch; R9 000 is nowhere near that many dollars.
- A (\$50) divides by 180 — a dropped decimal or extra zero turning R18 into R180 per dollar.
- C (\$450) divides by a "rounder" 20 instead of 18 — an estimation that never got upgraded back to exact arithmetic.
Takeaway: ZAR → USD: divide by the rand-per-dollar rate. Decide before computing whether the answer should be bigger or smaller than what you started with — that one habit eliminates half the options instantly.
Topic: Convert USD to ZAR
The exchange rate is R17.50 per US dollar. Convert \$200 to rand.
A) R3 000
B) R3 500
C) R4 000
D) R2 857
Show the worked solution
Answer: B
Explanation
Each dollar is worth R17.50, so 200 of them are worth:
$$\text{ZAR} = 200 \times 17.50 = \text{R}3\,500$$
No-calculator route: $200 \times 17.5 = 200 \times 17 + 200 \times 0.5 = 3\,400 + 100$. And the sense check: dollars into rand must give a bigger number — R3 500 for \$200 fits.
Why the others are wrong:
- A (R3 000) multiplies by 15, and C (R4 000) multiplies by 20 — both are the exchange rate rounded to a "nicer" number. The NBT picks rates like 17.50 precisely because they are exactly computable; rounding is choosing to get the question wrong.
- D (R2 857) implies a rate of $2\,857/200 \approx \text{R}14.29$ per dollar — a number that appears nowhere in the question, and one that would make \$200 worth less than the quoted rate allows.
Takeaway: USD → ZAR: multiply by the rand-per-dollar rate. Split awkward multiplications (×17.5 = ×17 + ×0.5) instead of rounding them away.
Topic: Exchange rate with commission
The exchange rate is R16 per US dollar. The bank charges 2 % commission. How much does it cost in rand to buy \$500?
A) R8 160
B) R8 000
C) R7 840
D) R8 320
Show the worked solution
Answer: A
Explanation
Two charges: the currency, then the bank's fee on top.
- Currency: $500 \times 16 = \text{R}8\,000$
- Commission: $2\% \times 8\,000 = \text{R}160$
- Total: $\text{R}8\,160$
Which way does the fee go? Ask who is paying whom. You are buying a service, so the fee adds.
Why the others are wrong:
- B (R8 000) forgot the commission entirely.
- C (R7 840) subtracted it, as if the bank paid you for the privilege.
- D (R8 320) charged it twice.
Takeaway: Fees always move the total against you. If your adjusted answer is better than the raw conversion, you pointed the fee the wrong way.
Topic: Profit from exchange rate movement
You bought \$1 000 when the rate was R15 per dollar (cost R15 000). You now sell when the rate is R18 per dollar. What is your profit in rand?
A) R1 000 loss
B) R1 500 profit
C) R3 000 profit
D) R3 000 loss
Show the worked solution
Answer: C
Explanation
Track the two rand amounts.
Bought: \$1 000 at R15 cost R15 000. Sold: \$1 000 at R18 brings in R18 000.
$$\text{Profit} = \text{R}3\,000$$
The intuition: you held dollars while each dollar gained three rand. Rate up (weaker rand) is good for the dollar-holder.
Why the others are wrong:
- D (R3 000 loss) has the right size, wrong sign — the "rand weakened so I lost" reflex. That weakening is exactly what made your dollars worth more.
- A (R1 000 loss) borrows the \$1 000 from the question and adds a sign error.
- B (R1 500) looks like a 10 % return, but no 10 % exists here. The gain is R3 per dollar on 1 000 dollars.
Takeaway: Profit = (sell rate − buy rate) × amount held. Decide who benefits before doing the arithmetic — that locks in the sign.
Topic: Finding the best exchange rate
You want to buy euros. Three banks quote: - Bank A: R20 per €1 - Bank B: R19.50 per €1 - Bank C: R20.50 per €1
Which bank gives you the best deal?
A) Bank A
B) Bank C
C) All three give the same number of euros
D) Bank B
Show the worked solution
Answer: D
Explanation
The quoted rate is the price of one euro. As the buyer, you want the lowest price — same as buying anything else.
Spend R1 000 at each bank:
- Bank B (R19.50) → €51.3
- Bank A (R20.00) → €50.0
- Bank C (R20.50) → €48.8
Bank B gives the most euros.
Why the others are wrong:
- B (Bank C) picks the highest rate. That is right when selling euros, backwards when buying.
- A (Bank A) picks the roundest number, not the best one.
- C says it makes no difference — the R1 000 test above disproves that.
Takeaway: Buying foreign currency, lowest rate wins. Selling, highest wins. If you blank, run R1 000 through two banks and compare.
Section 7.7 — Time Value of Money
Money today is worth more than the same amount in the future because today's money can earn interest. The present value (PV) of a future amount $A$ at rate $i$ over $n$ periods is $P = A/(1+i)^n$. When equal payments are made at regular intervals (an annuity), the future value is found by summing the compound-interest-grown contributions of each payment — a geometric series.
Topic: Future value of an annuity
R1 000 is deposited at the end of each year for 3 years at 10 % p.a. compound. What is the total accumulated value after 3 years?
A) R3 000
B) R3 310
C) R3 300
D) R3 460
Show the worked solution
Answer: B
Explanation
Separate deposits, each growing for a different length of time. Deposits land at the end of each year, so the first grows 2 years and the last grows 0.
| Payment | Years | Value at year 3 |
|---|---|---|
| Year 1 | 2 | $1000(1.1)^2 = 1210$ |
| Year 2 | 1 | $1000(1.1) = 1100$ |
| Year 3 | 0 | $1000$ |
| Total | R3 310 |
Why the others are wrong:
- A (R3 000) has no interest at all.
- C (R3 300) applies 10 % once to the whole R3 000, as if every deposit arrived on day one.
- D (R3 460) fits no timing. The only defensible answers are R3 310 (end-of-year) and R3 641 (start-of-year).
Takeaway: Build the table — one row per deposit, each with its own growth period. The last end-of-year payment earns nothing, and that is what examiners test.
Topic: Loan repayment — equal annual instalments
A loan of R50 000 is repaid in equal annual instalments over 4 years at 20 % p.a. simple interest. What is the annual instalment?
A) R12 500
B) R20 000
C) R22 500
D) R25 000
Show the worked solution
Answer: C
Explanation
Two stages: first find the total debt (principal plus all the simple interest), then share it equally across the payments.
Step 1 — total owed after 4 years.
$$A = 50\,000(1 + 0.20 \times 4) = 50\,000 \times 1.8 = \text{R}90\,000$$
Step 2 — split into 4 equal instalments.
$$\frac{90\,000}{4} = \text{R}22\,500$$
Why the others are wrong: the wrong options are a fence-post-error museum —
- A (R12 500) divides the bare R50 000 by 4: the interest never happened.
- B (R20 000) charges interest for only 3 years ($50\,000 \times 1.6 = 80\,000$, then ÷ 4) — an off-by-one in the year count.
- D (R25 000) charges interest for 5 years ($50\,000 \times 2.0 = 100\,000$, then ÷ 4) — the same fence-post error leaning the other way.
Takeaway: Total first, divide second. The wrong options differ only in how many years of interest were charged — which tells you exactly what the examiner expects students to fumble. Count the years deliberately.
Topic: Total interest on an annuity-style savings plan
R500 is deposited monthly for 24 months. Total deposited = R12 000. If the final accumulated amount is R13 200, what is the total interest earned?
A) R1 200
B) R1 300
C) R800
D) R1 800
Show the worked solution
Answer: A
Explanation
However complicated a savings plan looks, total interest is always the same one-liner: what you ended with minus what you put in.
Step 1 — total contributed. $24 \times 500 = \text{R}12\,000$. (It is stated in the question, but verify it anyway — mis-totalling the deposits is where this question is actually lost.)
Step 2 — subtract.
$$\text{Interest} = 13\,200 - 12\,000 = \text{R}1\,200$$
Why the others are wrong:
- B (R1 300), C (R800) and D (R1 800) are all neighbourhood decoys — none corresponds to any correct pairing of "final amount" and "total deposited". They punish two habits: subtracting in your head, and not writing $24 \times 500 = 12\,000$ down before subtracting. Do both steps on paper and the wrong options have nothing to grab.
Takeaway: Total interest = accumulated amount − total contributions, whatever the deposit pattern. The formula never changes; only the bookkeeping does.
Topic: Total interest on a lump-sum loan
R100 000 is borrowed at 12 % p.a. simple interest for 5 years. How much interest is paid in total?
A) R12 000
B) R48 000
C) R50 000
D) R60 000
Show the worked solution
Answer: D
Explanation
Per-year view: 12 % of R100 000 is R12 000 of interest per year. Five years:
$$I = Prt = 100\,000 \times 0.12 \times 5 = \text{R}60\,000$$
(Equivalently: the total owed is $100\,000 \times 1.6 = \text{R}160\,000$, of which R60 000 is interest.)
Why the others are wrong:
- A (R12 000) is one year's interest — the per-year figure presented as the total.
- B (R48 000) is four years' interest ($12\,000 \times 4$) — the off-by-one fence-post again.
- C (R50 000) is "half the loan" — a round number that corresponds to a 10 % rate, not the 12 % in the question.
Takeaway: $I = Prt$. Anchor yourself with the per-year amount (R12 000) and the wrong options unmask themselves as 1 year, 4 years, and "10 %-flavoured" impostors of the 5-year total.
Topic: Present value — how much to invest now
How much must be invested now at 10 % p.a. compound interest to have R14 641 in 4 years?
A) R11 000
B) R10 000
C) R12 100
D) R9 900
Show the worked solution
Answer: B
Explanation
Present value runs compound interest backwards: what must you plant today to grow into R14 641? Growth multiplies by $(1+i)^n$, so un-growing divides:
$$P = \frac{14\,641}{(1.1)^4} = \frac{14\,641}{1.4641} = \text{R}10\,000$$
The numbers are friendly by design — the Pascal digits 1, 4, 6, 4, 1 again.
Why the others are wrong:
- A (R11 000) divided by $1.1^3$ — one year short.
- C (R12 100) divided by $1.1^2$ — two years short.
- D (R9 900) is estimator's bait. Even the nearest principled error (multiplying by $0.9^4$) gives R9 606, not R9 900 — dividing by 1.1 and multiplying by 0.9 are different operations.
Takeaway: PV = future amount ÷ $(1+i)^n$, and check forwards: $10\,000 \times 1.4641 = 14\,641$ ✓. Un-growing is division by the growth factor, never subtraction of the percentage.
Mixed Practice Questions — M1 to M30
Topic: Simplifying expressions
Simplify $\dfrac{6x^2y}{2xy^2}$.
A) $\dfrac{3x}{y}$
B) $3xy$
C) $\dfrac{3y}{x}$
D) $3x^2y$
Show the worked solution
Answer: A
Topic: Number sense — scientific notation
Write $0.00045$ in scientific notation.
A) $4.5 \times 10^3$
B) $4.5 \times 10^{-3}$
C) $4.5 \times 10^{-4}$
D) $45 \times 10^{-5}$
Show the worked solution
Answer: C
Topic: Functions — domain
What is the domain of $f(x) = \dfrac{1}{x - 2}$?
A) $x \geq 2$
B) $x \neq 2$
C) $x > 2$
D) All real numbers
Show the worked solution
Answer: B
Topic: Trigonometry — exact value
$\cos 60° = $
A) $\dfrac{\sqrt{3}}{2}$
B) $\dfrac{\sqrt{2}}{2}$
C) $\dfrac{1}{2}$
D) $1$
Show the worked solution
Answer: C
Topic: Algebra — solving a linear equation
Solve $3x - 7 = 14$.
A) $7$
B) $\dfrac{7}{3}$
C) $21$
D) $\dfrac{21}{3}$
Show the worked solution
Answer: A
Topic: Geometry — exterior angle
An exterior angle of a triangle equals:
A) The sum of the interior angles
B) 90°
C) The sum of the two non-adjacent interior angles
D) The adjacent interior angle
Show the worked solution
Answer: C
Topic: Data — IQR
A data set has $Q_1 = 14$ and $Q_3 = 26$. The IQR is:
A) 40
B) 20
C) 12
D) 26
Show the worked solution
Answer: C
Topic: Compound interest
R5 000 is invested at 20 % p.a. compound interest for 3 years. The accumulated value is:
A) R8 000
B) R8 500
C) R8 600
D) R8 640
Show the worked solution
Answer: D
Topic: Algebra — quadratic formula
Using the quadratic formula, solve $x^2 - 6x + 8 = 0$.
A) $x = 2$ or $x = 4$
B) $x = -2$ or $x = -4$
C) $x = 1$ or $x = 8$
D) $x = 3$ only
Show the worked solution
Answer: A
Topic: Number sense — ratio
If $x:y = 2:5$ and $y = 35$, find $x$.
A) 7
B) 17.5
C) 14
D) 70
Show the worked solution
Answer: C
Topic: Functions — identifying a parabola
The graph of $f(x) = 2x^2 - 3$ has $y$-intercept at:
A) $(0,\ 2)$
B) $(0,\ -3)$
C) $(3,\ 0)$
D) $(-3,\ 0)$
Show the worked solution
Answer: B
Topic: Trigonometry — sine rule
In $\triangle ABC$: $\angle A = 30°$, $\angle B = 45°$, $a = 6$. Find $b$.
A) $3\sqrt{2}$
B) $4\sqrt{2}$
C) $6\sqrt{2}$
D) $6$
Show the worked solution
Answer: C
Topic: Data — probability
A bag has 4 red, 3 blue, and 5 green balls. $P(\text{green}) = $
A) $\dfrac{5}{7}$
B) $\dfrac{5}{9}$
C) $\dfrac{1}{5}$
D) $\dfrac{5}{12}$
Show the worked solution
Answer: D
Topic: Hire purchase
An item costs R6 000 cash. HP: 20 % deposit, 12 monthly payments at 15 % p.a. simple interest. Monthly payment?
A) R400
B) R460
C) R575
D) R500
Show the worked solution
Answer: B
Topic: Algebra — factorising a trinomial
Factorise $x^2 + 7x + 12$.
A) $(x + 3)(x + 4)$
B) $(x + 2)(x + 6)$
C) $(x + 1)(x + 12)$
D) $(x - 3)(x - 4)$
Show the worked solution
Answer: A
Topic: Functions — exponential graph
The graph of $y = 2^x$ passes through which of the following points?
A) $(-1,\ 0)$
B) $(0,\ 0)$
C) $(3,\ 9)$
D) $(3,\ 8)$
Show the worked solution
Answer: D
Topic: Straight-line depreciation
A vehicle bought for R320 000 depreciates at 10 % p.a. straight-line. Value after 4 years?
A) R192 000
B) R200 000
C) R196 000
D) R180 000
Show the worked solution
Answer: A
Topic: Geometry — area of a circle
A circle has radius 7 cm. Its area is:
A) $14\pi\ \text{cm}^2$
B) $49\pi\ \text{cm}^2$
C) $7\pi\ \text{cm}^2$
D) $28\pi\ \text{cm}^2$
Show the worked solution
Answer: B
Topic: Data — union of events
$P(A) = 0.5$, $P(B) = 0.4$, $P(A \cap B) = 0.15$. Find $P(A \cup B)$.
A) 0.75
B) 0.9
C) 0.65
D) 0.20
Show the worked solution
Answer: A
Topic: Compound interest — find P
An investment compounds at 10 % p.a. After 2 years the balance is R12 100. What was the original principal?
A) R11 000
B) R10 500
C) R9 900
D) R10 000
Show the worked solution
Answer: D
Topic: Algebra — laws of exponents
Simplify $\dfrac{2^5 \times 2^3}{2^6}$.
A) $4$
B) $8$
C) $2$
D) $16$
Show the worked solution
Answer: A
Topic: Exchange rates
R14 = $1. You have R7 000. How many US dollars can you buy (ignoring commission)?
A) $\$98\,000$
B) $\$700$
C) $\$500$
D) $\$7.14$
Show the worked solution
Answer: C
Topic: Functions — completing the square
Write $x^2 - 4x + 7$ in the form $(x - a)^2 + b$.
A) $(x - 4)^2 + 7$
B) $(x - 2)^2 + 11$
C) $(x - 2)^2 + 3$
D) $(x + 2)^2 + 3$
Show the worked solution
Answer: C
Topic: Geometry — Pythagoras
A ladder 13 m long leans against a wall. The base is 5 m from the wall. How high up the wall does the ladder reach?
A) 8 m
B) 14 m
C) 10 m
D) 12 m
Show the worked solution
Answer: D
Topic: Simple interest — find rate
R6 000 grows to R7 800 in 2 years under simple interest. The annual interest rate is:
A) 15 % p.a.
B) 18 % p.a.
C) 12 % p.a.
D) 20 % p.a.
Show the worked solution
Answer: A
Topic: Inflation
A house costs R500 000 today. If property prices inflate at 10 % p.a. for 3 years, its future price will be:
A) R650 000
B) R550 000
C) R665 500
D) R660 000
Show the worked solution
Answer: C
Topic: Algebra — simultaneous equations
Solve $x + y = 10$ and $x - y = 4$.
A) $x = 8,\ y = 2$
B) $x = 6,\ y = 4$
C) $x = 7,\ y = 3$
D) $x = 4,\ y = 6$
Show the worked solution
Answer: C
Topic: Reducing-balance depreciation
Equipment costing R81 000 depreciates at $\frac{1}{3}$ of its value each year (reducing balance). Value after 2 years?
A) R27 000
B) R54 000
C) R36 000
D) R40 500
Show the worked solution
Answer: C
Topic: Compound interest — present value
How much must be invested now at 10 % p.a. compound to have R12 100 in 2 years?
A) R10 000
B) R11 000
C) R9 900
D) R10 500
Show the worked solution
Answer: A
Topic: Data — mean and outlier
A data set $\{6, 8, 10, 12, 100\}$ has mean 27.2. After removing 100, the new mean is:
A) 12
B) 8
C) 9
D) 36
Show the worked solution
Answer: C