Calculating the Period of an Investment (Grade 12 NSC Matric Mathematics): Revision Notes

Worked Example: Basic Investment Period Calculation

Question: Thembile invests R 3500 into a savings account which pays 7.5% per annum compounded yearly. After an unknown period of time his account is worth R 4044.69. For how long did Thembile invest his money?

Solution:

Step 1: Write down the compound interest formula and identify known values

• $A = P(1 + i)^n$
• $A = 4044.69$
• $P = 3500$
• $i = 0.075$ (7.5% as a decimal)
• $n = ?$ (what we need to find)

Step 2: Substitute values and solve for n

• $4044.69 = 3500(1 + 0.075)^n$
• $4044.69 = 3500(1.075)^n$
• $\frac{4044.69}{3500} = (1.075)^n$
• $1.1556... = (1.075)^n$

Step 3: Apply logarithms

• $n = \log_{1.075}\left(\frac{4044.69}{3500}\right)$
• $n = \frac{\log(4044.69/3500)}{\log(1.075)}$
• $n = \frac{\log(1.1556)}{\log(1.075)}$
• $n = 2.00...$

Answer: The R 3500 was invested for 2 years.

Worked Example: Planning Investment Duration

Question: Margo has R 12 000 to invest and needs the money to grow to at least R 30 000 to pay for her daughter's studies. If it is invested at a compound interest rate of 9% per annum, determine how long (in full years) her money must be invested.

Solution:

Step 1: Write down the formula and known values

• $A = P(1 + i)^n$
• $A = 30 000$
• $P = 12 000$
• $i = 0.09$ (9% as a decimal)

Step 2: Substitute and solve for n

• $30 000 = 12 000(1 + 0.09)^n$
• $30 000 = 12 000(1.09)^n$
• $\frac{30 000}{12 000} = (1.09)^n$
• $2.5 = (1.09)^n$

Step 3: Apply logarithms

• $n = \log_{1.09}(2.5)$
• $n = \frac{\log(2.5)}{\log(1.09)}$
• $n = 10.632...$

Step 4: Consider practical rounding

Since we need "at least" R 30 000, and 10 years will not deliver the required amount, the money must be invested for at least 11 years.