NBT Mathematics: Finance, Growth and Decay

Question 1

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?

Accepted Answer

R18 895.68

Answer

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.

Question 2

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

Accepted Answer

R104 401.25

Answer

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.

Question 3

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

Accepted Answer

12.68%

Answer

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.

Question 4

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?

Accepted Answer

R19 668.00

Answer

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.

Question 5

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?

Accepted Answer

R2 284.59

Answer

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.

(NBT) National Benchmark Test Practice Test

(NBT) National Benchmark Test Practice Test

NBT Mathematics: Finance, Growth and Decay