Grade 10 Maths · Financial Maths · Paper 1

Interest and Growth Grade 10

Real Interest and Growth past paper questions and answers from 2024 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

past paper questions: 14 past paper questions
skills: 5 skills
videos: 5 videos

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Question 1.4 · May/June 2025 · Paper 2 · National (DBE) · 4 marks

An insurance broker signed contracts with 15 people. The monthly premium (in rands) payable on each contract is given below.

134, 215, 325, 326, 362, 429, 515, 531, 598, 610, 624, 728, 923, 1034, 1200

1.4 The insurance company decided to increase the monthly premiums.

Monthly premiums that were less than R500 increased by

18%

Monthly premiums that were equal to or more than R500 increased by

k %

After these increases were applied to the above data, the new mean monthly premium was R686,44.

Calculate the value of

k

.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

Interest and Growth skills

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More Interest and Growth Grade 10 questions and answers

Try each one on paper before you open the memo.

4. Question 4.2.1 · November 2024 · Paper 1 · Provincial · 6 marks
4.2 A television set costs R 20 000. Belinda buys it on hire purchase. She traded in an old television for R 3 000 and also paid a deposit of R 2 000. She made monthly installments of R 900 for a period of 2 years.
4.2.1 Calculate the total interest paid.(6)
Show the memo (3 ticks)Memo
1. ✓
  Calculate balance owedBalance $= R 20000 - R 3000 - R 2000$
  $= R 15000$
  2 marks · Calculate total paid and interest
2. ✓
  Calculate total installments $24 \times R 900 = R 21600$
  2 marks · Calculate total paid and interest
3. ✓
  Calculate interestInterest $= Total Paid - Balance Owed$
  $= R 21600 - R 15000$
  $= R 6600$
  2 marks · Calculate total paid and interest
5. Question WE1 · Siyavula Grade 10 Mathematics, 9.2 Simple interest, Worked example 1 · 3 marks
WE1 Carine deposits R1 000 into a special bank account which pays a simple interest rate of $7%$ p.a. for $3$ years. How much will be in her account at the end of the investment term?(3)
Show the memo (3 ticks)Memo
1. ✓
  FormulaKnown values: $P = 1000$ ; $i = 0, 07$ ; $n = 3$
  $A = P (1 + in)$
  1 mark · Use the simple interest formula
2. ✓
  Substitution $A = 1000 (1 + 0, 07 \times 3)$
  1 mark · Use the simple interest formula
3. ✓
  Answer $A = 1210$
  At the end of $3$ years, Carine will have R1 210 in her bank account.
  1 mark · Use the simple interest formula
6. Question WE2 · Siyavula Grade 10 Mathematics, 9.2 Simple interest, Worked example 2 · 3 marks
WE2 Sarah borrows R5 000 from her neighbour at an agreed simple interest rate of $12, 5%$ p.a. She will pay back the loan in one lump sum at the end of $2$ years. How much will she have to pay her neighbour?(3)
Show the memo (3 ticks)Memo
1. ✓
  FormulaKnown values: $P = 5000$ ; $i = 0, 125$ ; $n = 2$
  $A = P (1 + in)$
  1 mark · Use the simple interest formula
2. ✓
  Substitution $A = 5000 (1 + 0, 125 \times 2)$
  1 mark · Use the simple interest formula
3. ✓
  Answer $A = 6250$
  At the end of $2$ years, Sarah will pay her neighbour R6 250.
  1 mark · Use the simple interest formula
7. Question WE3 · Siyavula Grade 10 Mathematics, 9.2 Simple interest, Worked example 3 · 4 marks
WE3 Prashant deposits R30 000 into a bank account that pays a simple interest rate of $7, 5%$ p.a. How many years must he invest for to generate R45 000?(4)
Show the memo (4 ticks)Memo
1. ✓
  FormulaKnown values: $A = 45000$ ; $P = 30000$ ; $i = 0, 075$
  $A = P (1 + in)$
  1 mark · Use the simple interest formula
2. ✓
  Substitution $45000 = 30000 (1 + 0, 075 \times n)$
  1 mark · Use the simple interest formula
3. ✓
  Value of n $\frac{45 000}{30 000} = 1 + 0, 075 \times n$
  $\frac{45 000}{30 000} - 1 = 0, 075 \times n$
  $\frac{( \frac{45 000}{30 000} ) - 1}{0 , 075} = n$
  $n = 6 \frac{2}{3}$
  1 mark · Use the simple interest formula
4. ✓
  Answer in years and months $\frac{2}{3}$ of a year $= \frac{2}{3} \times 12 = 8$ months
  It will take $6$ years and $8$ months to make R45 000 from R30 000 at a simple interest rate of $7, 5%$ p.a.
  1 mark · Convert units of measurement
8. Question WE4 · Siyavula Grade 10 Mathematics, 9.2 Simple interest, Worked example 4 · 3 marks
WE4 At what simple interest rate should Fritha invest if she wants to grow R2 500 to R4 000 in $5$ years?(3)
Show the memo (3 ticks)Memo
1. ✓
  FormulaKnown values: $A = 4000$ ; $P = 2500$ ; $n = 5$
  $A = P (1 + in)$
  1 mark · Use the simple interest formula
2. ✓
  Substitution $4000 = 2500 (1 + i \times 5)$
  1 mark · Use the simple interest formula
3. ✓
  Answer $\frac{4 000}{2 500} = 1 + i \times 5$
  $\frac{4 000}{2 500} - 1 = i \times 5$
  $\frac{( \frac{4 000}{2 500} ) - 1}{5} = i$
  $i = 0, 12$
  A simple interest rate of $12%$ p.a. will be needed.
  1 mark · Use the simple interest formula
9. Question a · Siyavula Grade 12 Mathematics, 3.4 Present value annuities, Worked example 10 · 4 marks
For calculations, assume the prime lending rate is $8, 5%$ per annum.
A school sells its old bus and uses the proceeds as a $15%$ deposit for the purchase of the new bus, which costs R330 000. To finance the balance of the purchase, the school takes out a loan that is subject to an interest rate of prime $+ 1%$ compounded monthly. The repayment period of the loan is $3$ years.
a Calculate the monthly repayments.(4)
Show the memo (4 ticks)Memo
1. ✓
  Loan amount $P = R 330000 - (\frac{15}{100} \times R 330000) = R 330000 - R 49500 = R 280500$
  1 mark · Calculate with percentages
2. ✓
  Formula and values $x = \frac{P \times i}{[ 1 - ( 1 + i ) ^{- n} ]}$
  $i = \frac{0 , 095}{12}$ (prime $8, 5% + 1% = 9, 5%$ ); $n = 3 \times 12 = 36$
  1 mark · Use the present value formula
3. ✓
  Substitution $x = \frac{280 500 \times \frac{0 , 095}{12}}{[ 1 - ( 1 + \frac{0 , 095}{12} ) ^{- 36} ]}$
  1 mark · Use the present value formula
4. ✓
  Answer $x = R 8985, 24$
  The school must pay R8 985,24 per month.
  1 mark · Use the present value formula
10. Question b · Siyavula Grade 8 Mathematics, 19 Represent data, Pie charts, Worked example 19.4 · 1 mark
The table below shows the average daily water usage for Mpho's household. $Diagram for this question$
b Mpho's household uses, on average, $450$ litres of water per day. How many litres are used for laundry?(1)
Show the memo (1 tick)Memo
1. ✓
  Laundry $450 \times 20% = 450 \times \frac{20}{100} = 90 litres$
  1 mark · Calculate with percentages

Interest and Growth Grade 10 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

Converting Between Fractions, Decimals, and PercentagesProfessor Dave Explains

Grade 8 Mathematics: PercentagesMrs Townsend Mathematics - South Africa

Financial Maths Grade 10 | Simple Interest IntroductionKevinmathscience

Financial Maths Grade 10 | Compound Interest | IntroductionKevinmathscience

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