Summary (Grade 11 NSC Matric Mathematics): Revision Notes

Worked Example 1: Simple vs Compound Interest

Problem: Calculate the accumulated amount after 3 years if R5000 is invested at 8% per annum using both simple and compound interest.

Simple interest solution:

• $A = P(1 + in)$
• $A = 5000(1 + 0.08 \times 3)$
• $A = 5000(1 + 0.24)$
• $A = 5000 \times 1.24 = \text{R6 240}$

Compound interest solution:

• $A = P(1 + i)^n$
• $A = 5000(1 + 0.08)^3$
• $A = 5000(1.08)^3$
• $A = 5000 \times 1.259712 = \text{R6 298.56}$

Result: The compound interest generates R58.56 more due to earning interest on previously earned interest.

Worked Example 2: Effective Annual Interest Rate

Problem: Calculate the effective annual interest rate equivalent to a nominal interest rate of 8.75% per annum compounded monthly.

Solution:

• $1 + i = \left(1 + \frac{i(m)}{m}\right)^m$
• $1 + i = \left(1 + \frac{0.0875}{12}\right)^{12}$
• $1 + i = (1 + 0.007292)^{12}$
• $1 + i = (1.007292)^{12}$
• $1 + i = 1.0912$
• $i = 0.0912 = 9.12\%$

Result: The effective annual rate (9.12%) is higher than the nominal rate (8.75%) because of monthly compounding.

Worked Example 3: Comparing Investment Options

Problem: Determine which option provides better returns for paying back a student loan:

• Option A: 9.1% per annum compounded quarterly
• Option B: 9% per annum compounded monthly
• Option C: 9.3% per annum compounded half-yearly

Solution: Calculate the effective annual rate for each option:

Option A:

• $1 + i = \left(1 + \frac{0.091}{4}\right)^4 = (1.02275)^4 = 1.0940$
• Effective rate = 9.40%

Option B:

• $1 + i = \left(1 + \frac{0.09}{12}\right)^{12} = (1.0075)^{12} = 1.0938$
• Effective rate = 9.38%

Option C:

• $1 + i = \left(1 + \frac{0.093}{2}\right)^2 = (1.0465)^2 = 1.0953$
• Effective rate = 9.53%

Answer: Option B (9% compounded monthly) is the best choice as it has the lowest effective annual interest rate.