Grade 12 Mathematics: Finance, growth & decay

a) $51020,30=45000(1+i{)}^{18}$ ✓
$(1+i{)}^{18}=1,133784$, so $1+i=(1,133784{)}^{\frac{1}{18}}=1,007$ ✓
$i=0,007$ per month, so the nominal rate is $12\times 0,007=0,084$, i.e. 8,4% p.a. compounded monthly ✓

b) $1+i_{\text{eff}}={\left(1+\frac{0,084}{12}\right)}^{12}=(1,007{)}^{12}=1,087311$ ✓
$i_{\text{eff}}=0,0873$, so the effective rate is 8,73% p.a. ✓

c) Straight line: $0=560000(1-0,125n)$ ✓
$0,125n=1$, so $n=8$. The book value is zero after 8 years ✓

d) Deposits are made on 1 April 2024, 1 May 2024, …, 1 March 2032: $9+7\times 12+3=96$ deposits ✓ (one month after the purchase until the replacement date), with $i=\frac{0,09}{12}=0,0075$.
The fund must be worth R820 000 directly after the 96th deposit:
$820000=\frac{x\left[(1,0075{)}^{96}-1\right]}{0,0075}$ ✓✓
$x=\frac{820000\times 0,0075}{(1,0075{)}^{96}-1}=$ R5 863,17 per month ✓

e) $i=\frac{0,078}{12}=0,0065$. The withdrawals form an ordinary annuity whose present value is R850 000:
$850000=\frac{9200\left[1-(1,0065{)}^{-n}\right]}{0,0065}$ ✓
$1-(1,0065{)}^{-n}=0,600543$, so $(1,0065{)}^{-n}=0,399457$ ✓
$-n\log 1,0065=\log 0,399457$ ✓
$n=-\frac{\log 0,399457}{\log 1,0065}=141,64$ ✓
$n$ is NOT a whole number: after 141 full withdrawals the balance left in the account is less than R9 200, so the 142nd withdrawal would only be a partial one. She can make 141 full withdrawals ✓

Modelled on Nov 2023 Paper 1, Question 6 — same skills, new scenario.

a) $A=60000{\left(1+\frac{0,078}{4}\right)}^{64}$ ✓ ($n=16\times 4=64$ quarters) ✓
$=60000(1,0195{)}^{64}=$ R206 506,62 ✓

b) $19250=38500(1-8i)$ ✓ so $8i=0,5$, giving $i=0,0625=6,25$% p.a. ✓

c) Total repaid $=60\times 6232,18=$ R373 930,80 ✓
Interest $=373930,80-285000=$ R88 930,80 ✓

d) Outstanding balance after 24 instalments:
$B=285000(1,009375{)}^{24}-\frac{6232,18\left[(1,009375{)}^{24}-1\right]}{0,009375}$ ✓✓
$=$ R189 674,67 ✓
After the extra payment: $189674,67-40000=149674,67$ ✓
Now $149674,67=\frac{6232,18\left[1-(1,009375{)}^{-n}\right]}{0,009375}$ ✓
$(1,009375{)}^{-n}=0,774846$, so $n=-\frac{\log 0,774846}{\log 1,009375}=27,34$ ✓
So 28 more instalments are needed (the 28th is smaller than R6 232,18).
Timeline: $24+28=52$ months instead of 60, so the loan is settled 8 months earlier. ✓

Modelled on Nov 2024 Paper 1, Question 7 — same skills, new scenario.

a) Price $=28500(1,062{)}^4$ ✓ $=$ R36 252,91 ✓

b) Deposits run from 1 April 2026 to 1 October 2029, one at the START of every quarter: $3+4+4+4=15$ deposits ✓, with $i=\frac{0,08}{4}=0,02$.
Value on 1 October 2029, directly after the 15th deposit:
$F=\frac{8000\left[(1,02{)}^{15}-1\right]}{0,02}$ ✓ $=$ R138 347,34 ✓
One further quarter of growth to 1 January 2030: $138347,34\times 1,02=$ R141 114,29 ✓

c) Debt at the end of the THREE payment-free months, with $i=\frac{0,114}{12}=0,0095$:
$480000(1,0095{)}^3=$ R493 810,37 ✓
The instalments then form an ordinary annuity, so
$493810,37=\frac{11500\left[1-(1,0095{)}^{-n}\right]}{0,0095}$ ✓
$(1,0095{)}^{-n}=0,592070$ ✓
$n=-\frac{\log 0,592070}{\log 1,0095}=55,43$ ✓
So 55 full instalments plus a smaller 56th payment are needed. The final payment falls at the end of month $3+56=59$, i.e. 59 months after the loan was granted. ✓

d) Balance directly after the 55th full instalment (end of month 58):
$B=493810,37(1,0095{)}^{55}-\frac{11500\left[(1,0095{)}^{55}-1\right]}{0,0095}$ ✓✓
$=$ R4 949,61 ✓
One month later this balance has grown to $4949,61\times 1,0095=$ R4 996,63, the final payment ✓ (indeed less than R11 500).

Modelled on Nov 2025 Paper 1, Question 7 — same skills, new scenario.