Annuities & Loan Repayments — Grade 12 Maths (CAPS)

Future value for savings and present value for loans, working out a monthly instalment, and finding the outstanding balance part-way through a loan.

1The two formulae

Future value (saving up)
F = (x[(1 + i)ⁿ − 1])/i

Present value (paying off)
P = (x[1 − (1 + i)⁻ⁿ])/i

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Choose by asking when the lump sum exists. Saving towards an amount in the future ⇒ F. Borrowing money you have now ⇒ P. The only difference in the formula is the sign of the exponent.

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i is the rate per payment period, not per year. For 9% p.a. compounded monthly, i = 0,09/12 = 0,0075 and n counts months.

2Future value: saving

Worked ExampleR1 500 is deposited monthly for 20 years at 9% p.a. compounded monthly. Determine the future value.

1
i = 0,09/12 = 0,0075 and n = 20 × 12 = 240.
Convert to the payment period first, always.
2
F = (1 500[(1,0075)²⁴⁰ − 1])/0,0075.
3
= R1 001 830,30.
Total deposited was only 240 × 1 500 = R360 000 — the rest is interest.

3Present value: a loan instalment

Worked ExampleA bond of R800 000 is repaid monthly over 20 years at 9% p.a. compounded monthly. Determine the instalment.

1
i = 0,0075, n = 240, P = 800 000.
2
800 000 = (x[1 − (1,0075)⁻²⁴⁰])/0,0075.
Present value — the money exists now.
3
x = (800 000 × 0,0075)/(1 − (1,0075)⁻²⁴⁰).
Rearrange for x.
4
= R7 197,81 per month.
Over 240 months that is about R1,73 million repaid on an R800 000 loan.

4Outstanding balance

The balance still owing is the present value of the payments still to come — not the amount already paid subtracted from the loan.

Worked ExampleHow much is still owing on that bond after 5 years?

1
Payments made: 5 × 12 = 60. Payments remaining: 240 − 60 = 180.
2
Balance = (7 197,81[1 − (1,0075)⁻¹⁸⁰])/0,0075.
Present value of the 180 payments still due, using the instalment found above.
3
= R709 656,62.
After a quarter of the term only about R90 000 of the R800 000 has been repaid — the early payments are mostly interest. (Working with the unrounded instalment gives R709 656,39.)

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If a question says payments start one month after the loan, the formula works as it stands. If there is a delay, adjust n and be explicit about it.

5Using logarithms to find n

When the unknown is the number of periods, it sits in the exponent — and the only way down from an exponent is a logarithm.

A = P(1 + i)ⁿ ⇒ n = (log(A ÷ P))/(log(1 + i))

Worked ExampleHow long will R50 000 take to grow to R80 000 at 9% p.a. compounded annually?

1
80 000 = 50 000(1,09)ⁿ.
2
1,6 = (1,09)ⁿ.
Divide both sides by 50 000.
3
n = (log 1,6)/(log 1,09).
Take logs of both sides and use the power law.
4
n = 5,45 years.
Interest is only paid at year end, so in practice you would wait 6 full years.

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Any base works, as long as you use the same base top and bottom. The calculator's log and ln keys both give the same n.

Worked ExampleSecond example — rounding n: how long will it take R15 000 to grow to R25 000 at 9% p.a. compounded annually?

1
25 000 = 15 000(1,09)ⁿ
A = P(1 + i)ⁿ
2
(1,09)ⁿ = (25 000)/(15 000) = 1,6667
Isolate the power first — always.
3
n = (log 1,6667)/(log 1,09)
Take logs of both sides: n log 1,09 = log 1,6667.
4
n = 5,93 years
5
So it takes 6 years, since interest is only added at the end of each year.
Round UP in a compound-interest context — at 5,93 years the money has not yet arrived.

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Rounding 5,93 down to 5 years, or leaving it as 5,93 when the interest compounds annually. Read the compounding period: with annual compounding the balance only changes at whole years, so the answer is 6.

6Sinking funds

A sinking fund is a future-value annuity with a purpose: a business saves monthly so that it can replace equipment when the old one is scrapped. The exam version almost always has three parts.

Worked ExampleA machine costing R800 000 depreciates at 12% p.a. on a reducing balance and is replaced in 5 years. A new machine inflates at 8% p.a. A sinking fund earns 10% p.a. compounded monthly. Find the monthly payment.

1
Scrap value: 800 000(1 − 0,12)⁵ = R422 185,53
Reducing-balance depreciation, A = P(1 − i)ⁿ.
2
Cost of the new machine: 800 000(1 + 0,08)⁵ = R1 175 462,46
Inflation is compound growth.
3
Amount the fund must produce: 1 175 462,46 − 422 185,53 = R753 276,93
The old machine is traded in — this is the step that is missed.
4
753 276,93 = (x[(1 + 0,10/12)⁶⁰ − 1])/(0,10/12)
Future value of an annuity, 60 monthly payments.
5
x = R9 727,60 per month

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The order is always the same: depreciate the old, inflate the new, subtract, then annuity. Write those four steps down before touching the calculator.

7Comparing investment and loan options

The ATP asks you to critically analyse options and make an informed decision. The marks are for the comparison and the sentence at the end, not only for the arithmetic.

Convert every option to the same basis before comparing — usually the effective annual rate, or the total amount after the same number of years.
A lower nominal rate can still be worse if it compounds less often. 10% compounded monthly beats 10,2% compounded annually.
For loans, compare the total repaid, not the instalment. A longer term gives a smaller monthly payment and a much larger total.
Finish with a sentence: which option, and why. That is where the final mark sits.

Worked ExampleOption A: 11% p.a. compounded monthly. Option B: 11,4% p.a. compounded annually. Which is the better investment?

1
Option A effective: (1 + 0,11/12)¹² − 1 = 0,11572 → 11,57%
Convert the nominal rate to an effective annual rate.
2
Option B is already annual: 11,4%
3
11,57% > 11,4%, so Option A is better.
Compare like with like.
4
Option A is the better investment because its effective annual rate of 11,57% exceeds Option B's 11,4%.
The concluding sentence earns its own mark.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
R1 500 is deposited monthly for 20 years at 9% p.a. compounded monthly. Determine the future value.
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i = 0,0075, n = 240 → F = R1 001 830,30
2
How much was actually deposited in total in that annuity?
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240 × 1 500 = R360 000
3
A bond of R800 000 is repaid monthly over 20 years at 9% p.a. compounded monthly. Determine the monthly instalment.
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x = (800 000 × 0,0075)/(1 − (1,0075)⁻²⁴⁰) = R7 197,81
4
Determine the outstanding balance on that bond after 5 years.
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240 − 60 = 180 payments remain ✓; Balance = (7 197,81[1 − (1,0075)⁻¹⁸⁰])/0,0075 ✓✓ = R709 656,62 ✓ (R709 656,39 with the unrounded instalment)
5
Write down the value of i for 9% p.a. compounded monthly.
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0,09/12 = 0,0075
6
Which formula do you use when saving towards a future amount?
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Future value, F = (x[(1 + i)ⁿ − 1])/i
7
Which formula do you use for a loan taken out today?
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Present value, P = (x[1 − (1 + i)⁻ⁿ])/i
8
How is the outstanding balance on a loan calculated?
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As the present value of the payments still outstanding.
9
How long will it take R20 000 to double at 8% p.a. compounded annually? Give your answer in whole years.
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40 000 = 20 000(1,08)ⁿ ✓; (1,08)ⁿ = 2 ✓; n = (log 2)/(log 1,08) = 9,006 ✓; so 10 years (interest is only added at year end) ✓
10
A vehicle costing R450 000 depreciates at 15% p.a. on a reducing balance. Calculate its scrap value after 4 years.
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450 000(1 − 0,15)⁴ ✓✓ = R234 902,81 ✓
11
In a sinking-fund question, state the THREE amounts you must calculate before working out the monthly payment.
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The scrap value of the old asset ✓; the inflated cost of the new one ✓; and the difference between them, which is what the fund must produce ✓
12
Option A pays 9,6% p.a. compounded monthly; Option B pays 9,9% p.a. compounded annually. Which is better? Show your reasoning.
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A effective = (1 + 0,096/12)¹² − 1 ✓ = 0,10034 = 10,03% ✓; B = 9,9% ✓; Option A is better because its effective annual rate is higher ✓
13
Why can a loan with a lower monthly instalment cost you more overall?
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A lower instalment usually means a longer term ✓, so interest is charged for more months and the total repaid is larger ✓