NBT Mathematics: Finance, Growth and Decay Flashcards

R15 000 is invested at 8% per annum compounded annually for 3 years. What is the value of the investment at the end of the period?

Answer: R18 895.68

Using A = P(1 + i)ⁿ: A = 15 000(1.08)³ = 15 000 × 1.259712 = R18 895.68. Note that compounded interest grows faster than simple interest because each year's interest also earns interest.

A car originally worth R200 000 depreciates at 15% per year on a reducing-balance basis. What is its value after 4 years?

Answer: R104 401.25

Reducing-balance depreciation: A = P(1 − i)ⁿ = 200 000(0.85)⁴ = 200 000 × 0.52200625 ≈ R104 401.25. The value reduces by 15% of the current (not original) value each year.

A bank offers a savings account at 12% per annum compounded monthly. What is the effective annual interest rate?

Answer: 12.68%

Effective annual rate = (1 + i_nominal/n)ⁿ − 1 = (1 + 0.12/12)¹² − 1 = (1.01)¹² − 1 ≈ 0.12683 = 12.68%. More frequent compounding always produces an effective rate above the nominal rate.

R500 is deposited at the end of each month into an account earning 6% per annum compounded monthly. What is the approximate future value of these payments after 3 years?

Answer: R19 668.00

Future value of an annuity: FV = PMT × [(1 + i)ⁿ − 1] / i, where PMT = 500, i = 0.06/12 = 0.005, n = 36. FV = 500 × [(1.005)³⁶ − 1] / 0.005 = 500 × [1.19668 − 1] / 0.005 ≈ 500 × 39.336 ≈ R19 668.

A loan of R50 000 is taken at 9% per annum compounded monthly and repaid over 2 years with equal monthly payments. What is the approximate monthly repayment?

Answer: R2 284.59

Monthly rate i = 0.09/12 = 0.0075, n = 24 months. PMT = PV × i / [1 − (1 + i)⁻ⁿ] = 50 000 × 0.0075 / [1 − (1.0075)⁻²⁴] = 375 / [1 − 0.83583] = 375 / 0.16417 ≈ R2 284.59.

R8 000 is invested at 10% per annum simple interest. How many years will it take for the investment to double in value?

Answer: 10 years

Simple interest: A = P(1 + rn). We need A = 16 000. So 16 000 = 8 000(1 + 0.10 × n) → 2 = 1 + 0.1n → 1 = 0.1n → n = 10 years.