Chapter 7 — Financial Mathematics

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.

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.

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.

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: $5\,000 \times 0.12 \times 5 = 3\,000$ ✓

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

Answer: B

Explanation

Four lines — write every one down. The options are placed to catch anyone who skips a step.

1. Deposit: $0.25 \times 12\,000 = \text{R}3\,000$
2. Instalments: $36 \times 350 = \text{R}12\,600$
3. Total HP: $\text{R}15\,600$
4. 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.

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.

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.

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.

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.)

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.

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.

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.

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.

Answer: A

Explanation

Two charges: the currency, then the bank's fee on top.

1. Currency: $500 \times 16 = \text{R}8\,000$
2. Commission: $2\% \times 8\,000 = \text{R}160$
3. 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.

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.

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.

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.

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.

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.

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.

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.

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

Answer: A

Explanation

Step 1 — Divide the numbers: $6 \div 2 = 3$.

Step 2 — Subtract the exponents for each letter: $x^{2-1} = x$ and $y^{1-2} = y^{-1}$.

Step 3 — $3xy^{-1} = \dfrac{3x}{y}$.

Why the others are wrong:

• B added the exponents instead of subtracting them.

• C swapped which letter ends up on the bottom.

• D left $x^2$ and did nothing with $y$.

Takeaway: Divide the coefficients, subtract the exponents. A negative exponent means the letter belongs in the denominator.

Answer: C

Explanation

Step 1 — Scientific notation is $a \times 10^n$ with $1 \leq a < 10$.

Step 2 — Move the point four places right to get 4.5, so $n = -4$.

Step 3 — $4.5 \times 10^{-4}$.

Why the others are wrong:

• A positive exponent — that would make the number bigger, not smaller.

• B moved the point only three places.

• D $45$ is not between 1 and 10, so this is not standard form.

Takeaway: Count the places you move the decimal point. Moving RIGHT (making the digits bigger) means a NEGATIVE exponent.

Answer: B

Explanation

Step 1 — A fraction is undefined when its denominator is zero.

Step 2 — $x - 2 = 0$ when $x = 2$.

Step 3 — So every real number except 2: $x \neq 2$.

Why the others are wrong:

• A excludes valid values below 2 and still includes 2.

• C excludes everything below 2 for no reason.

• D includes $x = 2$, where the function is undefined.

Takeaway: A denominator only ever forbids the values that make it ZERO — not a whole half of the number line.

Answer: C

Explanation

Step 1 — $60°$ is a special angle.

Step 2 — $\cos 60° = \dfrac12$.

Step 3 — (Compare $\sin 60° = \dfrac{\sqrt3}{2}$.)

Why the others are wrong:

• A $\dfrac{\sqrt3}{2}$ is $\sin 60°$ or $\cos 30°$.

• B $\dfrac{\sqrt2}{2}$ is $\cos 45°$.

• D $\cos 0° = 1$.

Takeaway: As the angle grows from $0°$ to $90°$, cosine SHRINKS from 1 to 0. A large angle cannot have a large cosine.

Answer: A

Explanation

Step 1 — Add 7 to both sides: $3x = 21$.

Step 2 — Divide by 3.

Step 3 — $x = 7$.

Why the others are wrong:

• B divided 7 by 3 instead of adding it across first.

• C stopped at $3x = 21$.

• D divided 14 by 3 without moving the 7.

Takeaway: Undo the addition before the multiplication — the reverse of the order the expression was built in.

Answer: C

Explanation

Step 1 — An exterior angle and its adjacent interior angle make a straight line.

Step 2 — The three interior angles add to $180°$ as well.

Step 3 — So the exterior angle equals the sum of the two NON-ADJACENT interior angles.

Why the others are wrong:

• A the three interior angles total $180°$, which is not one exterior angle.

• B only true for a specific right-angled case.

• D the exterior angle is SUPPLEMENTARY to the adjacent one, not equal.

Takeaway: Exterior $=$ sum of the two OPPOSITE interior angles. It follows from both sums being $180°$, and it saves a step every time.

Answer: C

Explanation

Step 1 — The interquartile range is $Q_3 - Q_1$.

Step 2 — $26 - 14$.

Step 3 — $= 12$.

Why the others are wrong:

• A added the quartiles.

• B used a different pair of values.

• D reported $Q_3$ alone.

Takeaway: IQR is one subtraction: upper quartile minus lower quartile. It measures the middle half of the data and ignores the extremes.

Answer: D

Explanation

Step 1 — Compound interest: $A = P(1 + i)^n$.

Step 2 — $A = 5000(1.2)^3$.

Step 3 — $1.2^3 = 1.728$, so $A = 8640$.

Why the others are wrong:

• A used SIMPLE interest: $5000(1 + 0.2 \times 3) = 8000$.

• B used $1.2^2$ and added a year's interest.

• C rounded $1.728$ to $1.72$.

Takeaway: Compound is a POWER, simple is a multiple. Over three years at 20% the difference is R640 — small enough to look plausible and wrong.

Answer: A

Explanation

Step 1 — $a=1$, $b=-6$, $c=8$, so $\Delta = 36 - 32 = 4$.

Step 2 — $x = \dfrac{6 \pm \sqrt4}{2} = \dfrac{6 \pm 2}{2}$.

Step 3 — $x = 4$ or $x = 2$.

Why the others are wrong:

• B both signs flipped; these roots multiply to $+8$ but add to $-6$.

• C used factors 1 and 8, which add to 9.

• D would need $\Delta = 0$.

Takeaway: Check with the sum and product: the roots must add to 6 and multiply to 8. 2 and 4 do; $-2$ and $-4$ multiply correctly but add to $-6$.

Answer: C

Explanation

Step 1 — $\dfrac{x}{y} = \dfrac25$ with $y = 35$.

Step 2 — $\dfrac{x}{35} = \dfrac25$.

Step 3 — $x = 35 \times \dfrac25 = 14$.

Why the others are wrong:

• A used $\dfrac15$ of 35.

• B used a half.

• D multiplied by 2 without dividing by 5.

Takeaway: One part is $35 \div 5 = 7$, so two parts is 14. Working in PARTS is usually quicker than cross-multiplying.

Answer: B

Explanation

Step 1 — The $y$-intercept is the value when $x = 0$.

Step 2 — $f(0) = 2(0)^2 - 3 = -3$.

Step 3 — The point is $(0,\ -3)$.

Why the others are wrong:

• A read off the coefficient 2 instead of the constant.

• C swapped the coordinates.

• D swapped the coordinates and the sign.

Takeaway: For a $y$-intercept put $x = 0$; the answer is always the constant term. The point has $x = 0$, so it is $(0,\ c)$.

Answer: C

Explanation

Step 1 — Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$.

Step 2 — $b = \dfrac{6\sin 45°}{\sin 30°} = \dfrac{6 \cdot \frac{\sqrt2}{2}}{\frac12}$.

Step 3 — $= 6\sqrt2$.

Why the others are wrong:

• A divided by 2 twice.

• B used $a = 4$.

• D used $\sin 45° = \sin 30°$.

Takeaway: Dividing by $\dfrac12$ DOUBLES. That is why the answer is larger than the side you started with.

Answer: D

Explanation

Step 1 — Total balls: $4 + 3 + 5 = 12$.

Step 2 — Green ones: 5.

Step 3 — $\dfrac{5}{12}$.

Why the others are wrong:

• A used 7 as the total — the non-green balls.

• B used 9 as the total.

• C used $\dfrac{1}{5}$, ignoring the counts entirely.

Takeaway: The denominator is EVERY ball, including the green ones. Counting only the others is the standard slip.

Answer: B

Explanation

Step 1 — Deposit: $20\%$ of $6000 = 1200$, leaving $4800$ to finance.

Step 2 — Simple interest for one year: $4800 \times 0.15 = 720$.

Step 3 — Total owed $= 4800 + 720 = 5520$, over 12 months: $5520 \div 12 = 460$.

Why the others are wrong:

• A divided the balance by 12 with no interest.

• C financed the full R6 000 instead of the balance after the deposit.

• D used a deposit of R600.

Takeaway: Deposit first, then interest on what is LEFT. Charging interest on the full price is the most common hire-purchase error.

Answer: A

Explanation

Step 1 — Two numbers multiplying to 12 and adding to 7.

Step 2 — $3 \times 4 = 12$ and $3 + 4 = 7$ ✓

Step 3 — $(x+3)(x+4)$.

Why the others are wrong:

• B $2 \times 6 = 12$ but $2 + 6 = 8$.

• C $1 \times 12 = 12$ but $1 + 12 = 13$.

• D $(x-3)(x-4)$ gives $+12$ but $-7x$.

Takeaway: List the factor pairs of the constant, then pick the pair with the right SUM. Both signs positive here, because both the product and the sum are.

Answer: D

Explanation

Step 1 — Substitute each $x$ and see what $y$ comes out.

Step 2 — $2^3 = 8$, so $(3,\ 8)$ is on the graph ✓

Step 3 — (An exponential never reaches 0, which rules out two options at a glance.)

Why the others are wrong:

• A $2^{-1} = \dfrac12$, not 0 — and $y = 2^x$ is never 0.

• B $2^0 = 1$, so the graph passes through $(0, 1)$.

• C $2^3 = 8$, not 9 — that would be $3^2$.

Takeaway: $y = 2^x$ is always positive, so any point with $y = 0$ is impossible. Check the survivors by substituting.

Answer: A

Explanation

Step 1 — Straight-line depreciation takes the same amount off every year.

Step 2 — $10\%$ of the ORIGINAL $320\,000$ is $32\,000$ per year.

Step 3 — After 4 years: $320\,000 - 4(32\,000) = 192\,000$.

Why the others are wrong:

• B used 3.75 years, or rounded.

• C used a rate slightly under 10%.

• D used reducing balance for part of the time.

Takeaway: Straight-line always uses the ORIGINAL value; reducing balance uses the current one. The same rate gives different answers, and the question says which to use.

Answer: B

Explanation

Step 1 — Area of a circle is $\pi r^2$.

Step 2 — $r = 7$, so $r^2 = 49$.

Step 3 — $49\pi$ cm².

Why the others are wrong:

• A used $2\pi r$, the circumference.

• C used $\pi r$.

• D used $4\pi r$.

Takeaway: $\pi r^2$ for area. Squaring 7 gives 49 — if the answer looks like twice the radius, you have used the circumference formula.

Answer: A

Explanation

Step 1 — The addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Step 2 — $= 0.5 + 0.4 - 0.15$.

Step 3 — $= 0.75$.

Why the others are wrong:

• B added without subtracting the overlap.

• C subtracted the overlap twice.

• D reported the overlap plus a little.

Takeaway: Subtract the intersection ONCE — adding $P(A)$ and $P(B)$ counts the overlap twice. Only for mutually exclusive events is there nothing to subtract.

Answer: D

Explanation

Step 1 — $A = P(1+i)^n$, so $P = \dfrac{A}{(1+i)^n}$.

Step 2 — $P = \dfrac{12\,100}{1.1^2} = \dfrac{12\,100}{1.21}$.

Step 3 — $= 10\,000$.

Why the others are wrong:

• A divided by 1.1 once instead of twice.

• B subtracted the interest instead of dividing.

• C subtracted 10% of the final amount.

Takeaway: To go BACKWARDS in time, divide by $(1+i)^n$. Subtracting the interest does not work, because the interest was charged on a smaller amount.

Answer: A

Explanation

Step 1 — Multiplying powers of the same base ADDS exponents: $2^5 \times 2^3 = 2^8$.

Step 2 — Dividing SUBTRACTS: $2^8 \div 2^6 = 2^2$.

Step 3 — $2^2 = 4$.

Why the others are wrong:

• B stopped at $2^3$.

• C reached $2^1$.

• D reached $2^4$.

Takeaway: Add the exponents on top, subtract the one underneath: $5 + 3 - 6 = 2$. Never multiply the exponents.

Answer: C

Explanation

Step 1 — Each dollar costs R14.

Step 2 — Divide the rands by the rate: $7000 \div 14$.

Step 3 — $= 500$ dollars.

Why the others are wrong:

• A multiplied instead of dividing.

• B divided by 10.

• D divided the rate by the amount.

Takeaway: Ask which way makes the number smaller. R7 000 cannot buy 98 000 of a more valuable currency, so it must be a division.

Answer: C

Explanation

Step 1 — Half the coefficient of $x$: $-4 \div 2 = -2$.

Step 2 — $(x-2)^2 = x^2 - 4x + 4$, which is 4 too much.

Step 3 — So $x^2 - 4x + 7 = (x-2)^2 + 3$.

Why the others are wrong:

• A used the whole coefficient instead of half of it.

• B added 4 instead of subtracting it.

• D wrong sign inside the bracket.

Takeaway: Halve the middle coefficient for the bracket, then ADJUST the constant by what the bracket added. Expand to check.

Answer: D

Explanation

Step 1 — The ladder is the hypotenuse: $13^2 = 169$.

Step 2 — The base is one leg: $5^2 = 25$.

Step 3 — Height $= \sqrt{169 - 25} = \sqrt{144} = 12$ m.

Why the others are wrong:

• A used the 8-15-17 triple.

• B added the two lengths.

• C estimated.

Takeaway: The ladder is always the longest side, so you SUBTRACT. 5-12-13 again — the triples are worth knowing.

Answer: A

Explanation

Step 1 — Simple interest: $A = P(1 + in)$.

Step 2 — $7800 = 6000(1 + 2i)$, so $1 + 2i = 1.3$.

Step 3 — $2i = 0.3$, giving $i = 0.15 = 15\%$.

Why the others are wrong:

• B divided the total interest by one year instead of two.

• C used $\dfrac{1800}{15000}$.

• D rounded up from 15%.

Takeaway: Find the total interest first (R1 800), then divide by the principal AND by the number of years. Forgetting the years doubles the rate.

Answer: C

Explanation

Step 1 — Inflation compounds: $A = P(1+i)^n$.

Step 2 — $500\,000(1.1)^3$.

Step 3 — $1.1^3 = 1.331$, so $A = 665\,500$.

Why the others are wrong:

• A used simple growth: $500\,000 \times 1.3$.

• B applied one year only.

• D used $1.32$ instead of $1.331$.

Takeaway: Inflation is compound growth. $1.1^3 = 1.331$, not $1.3$ — the extra R5 500 is the interest earned on interest.