Grade 12 Maths · Financial Maths · Use the present value formula

Present Value Annuity Grade 12

Use the present value formula for loans paid back in equal instalments, for example find the monthly payment on a R300 000 home loan.

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Present Value Grade 12 | IntroductionKevinmathscience

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Question 7.2 · May/June 2025 · Paper 1 · National (DBE) · 5 marks

7.2 Tino invests

R 500000

in an account earning interest at the rate of

6%

p.a., compounded quarterly. Tino decides to withdraw

R 11250

at the end of every

3

months. Tino will continue making these regular withdrawals until there is no money in the account. How many withdrawals of

R 11250

will Tino be able to make?(5)

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

More past paper questions: use the present value formula

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3. Question 7.2.2 · September 2024 · Paper 1 · Provincial · 4 marks
7.2 Micaela buys furniture to the value of R10 000. She borrows the money on 1 February 2023 from a financial institution that charges interest at a rate of 9,5% p.a. compounded monthly. Micaela agrees to pay monthly instalments of R450. The loan agreement allows Micaela to start paying equal monthly instalments from 01 August 2023.
7.2.2 How many months will it take Micaela to pay back the loan?(4)
Show the memo (4 ticks)Memo
1. ✓
  Substitute into Present Value formula $10402, 15 = \frac{450 [ 1 - ( 1 + i ) ^{- n} ]}{i}$
  1 mark · Use the present value formula
2. ✓
  Isolate the bracket $1 - (1, 0079)^{- n} = 0, 183$
  1 mark · Solve a linear equation
3. ✓
  Correct use of logs $0, 8169 = (1, 0079)^{- n}$ $- n = lo g_{1, 0079} 0, 8169$ $n = 25, 63...$
  1 mark · Find an exponent using logs
4. ✓
  Answer $n = 26$ months
  1 mark · Find an exponent using logs
4. 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
5. Question 7.3.2 · November 2024 · Paper 1 · National (DBE) · 7 marks
7.3 Tshepo was granted a loan of R100 000 on 1 March 2022 at an interest rate of 13,5% p.a., compounded monthly. Tshepo agreed to repay the loan over 5 years in monthly instalments of R2 300,98, starting on 1 April 2022.
7.3.2 Tshepo paid R22 300,98 (his monthly instalment and an additional R20 000) on 1 March 2024 into the loan account. He continues to pay the original monthly instalment thereafter. How many months earlier will Tshepo repay the loan?(7)
Show the memo (6 ticks)Memo
1. ✓
  Calculate Balance after 2 years (24 months) $P = \frac{x [ 1 - ( 1 + i ) ^{- n} ]}{i}$
  Remaining period $n = 60 - 24 = 36$
  Balance $= \frac{2300 , 98 [ 1 - ( 1 + \frac{13 , 5}{1200} ) ^{- 36} ]}{\frac{13 , 5}{1200}}$
  2 marks · Calculate the outstanding loan balance
2. ✓
  Determine outstanding balance valueBalance $= R 67804, 94$
  1 mark · Calculate the outstanding loan balance
3. ✓
  Adjust balance for lump sum paymentNew Balance $= 67804, 94 - 20000 = R 47804, 94$
  1 mark · Calculate the outstanding loan balance
4. ✓
  Calculate new repayment period $47804, 94 = \frac{2300 , 98 [ 1 - ( 1 + \frac{13 , 5}{1200} ) ^{- n} ]}{\frac{13 , 5}{1200}}$
  $0, 766... = (1, 01125)^{- n}$
  1 mark · Use the present value formula
5. ✓
  Solve for n using logs $- n = lo g_{1, 01125} (0, 766...)$
  $n = 23, 83$
  1 mark · Find an exponent using logs
6. ✓
  Calculate time savedOriginal remaining: 36 months
  Time saved $= 36 - 23, 83 \approx 12, 17$
  Answer: 12 months earlier
  1 mark · Find an exponent using logs

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