Timelines (Grade 11 NSC Matric Mathematics): Revision Notes

Worked Example: Mixed Interest Types

Problem: R 5500 is invested for 4 years. For the first year, it earns 11% p.a. simple interest, then 12.5% p.a. compounded quarterly for the remaining 3 years. Find the final value.

Solution:

Step 1: Draw timeline and identify variables

The timeline shows $T_0$ (start), $T_1$ (end of year 1), and $T_4$ (end of year 4).

Step 2: Calculate value at $T_1$ using simple interest

Using the simple interest formula: $A = P(1 + in)$

$A = 5500(1 + 0.11 \times 1) = 5500(1.11) = \text{R}5105$

Step 3: Calculate compound interest from $T_1$ to $T_4$

Since interest compounds quarterly for 3 years:

• $n = 3 \times 4 = 12$ periods
• $i = \frac{0.125}{4} = 0.03125$ per quarter

The accumulated amount from $T_1$ becomes the new principal:

$A = P(1 + i)^n = 6105(1 + 0.03125)^{12} = \text{R}8831.88$

Step 4: Final answer

The investment value after 4 years is R 8831.88.

Worked Example: Multiple Deposits with Changing Rates

Problem: R 150 000 is invested for 6 years at 12% p.a. compounded half-yearly for the first 4 years, then 8.5% p.a. compounded yearly. Additional deposits of R 8000 (after 1 year) and R 2000 (after 5 years) are made.

Solution:

Step 1: Set up timeline

Timeline shows: $T_0$ (start), $T_1$ (+R 8000), $T_4$ (rate change), $T_5$ (+R 2000), $T_6$ (end).

Step 2: Calculate initial deposit growth

From $T_0$ to $T_4$ (4 years, half-yearly compounding):

• $n_1 = 4 \times 2 = 8$ periods
• $i_1 = \frac{0.12}{2} = 0.06$

From $T_4$ to $T_6$ (2 years, yearly compounding):

• $n_2 = 2$ periods
• $i_2 = 0.085$

Total growth: $A = P(1 + i_1)^{n_1}(1 + i_2)^{n_2}$

$A = 150000(1 + 0.06)^8(1 + 0.085)^2$

Step 3: Calculate R 8000 deposit growth

This deposit grows for 5 years total:

• 3 years at 12% half-yearly: $n_3 = 6$, $i_3 = 0.06$
• 2 years at 8.5% yearly: $n_4 = 2$, $i_4 = 0.085$

$A = 8000(1 + 0.06)^6(1 + 0.085)^2$

Step 4: Calculate R 2000 deposit growth

This deposit grows for only 1 year at 8.5%:

$A = 2000(1 + 0.085)^1$

Step 5: Final calculation

Total value = Initial deposit growth + R 8000 deposit growth + R 2000 deposit growth

$A = 150000(1 + 0.06)^8(1 + 0.085)^2 + 8000(1 + 0.06)^6(1 + 0.085)^2 + 2000(1 + 0.085)^1$

$A = \text{R}296977.00$

Worked Example: Handling Withdrawals

Problem: R 60 000 is invested at 7% p.a. compounded quarterly for 18 months, then 5% p.a. compounded monthly. After 3 years, R 5000 is withdrawn. What remains after 5 years?

Solution:

Step 1: Set up timeline

Timeline shows rate change at 18 months ($T_{1.5}$) and withdrawal at 3 years ($T_3$).

Step 2: Calculate initial growth

First 1.5 years (quarterly): $n_1 = 1.5 \times 4 = 6$, $i_1 = \frac{0.07}{4}$

Next 3.5 years (monthly): $n_2 = 3.5 \times 12 = 42$, $i_2 = \frac{0.05}{12}$

Growth without withdrawal: $A = 60000\left(1 + \frac{0.07}{4}\right)^6\left(1 + \frac{0.05}{12}\right)^{42}$

Step 3: Account for withdrawal

We must subtract both the R 5000 withdrawn and the interest it would have earned:

Interest the R 5000 would have earned over final 2 years: $A = 5000\left(1 + \frac{0.05}{12}\right)^{24}$

Step 4: Final calculation

Final amount = Growth without withdrawal - R 5000 - Interest on withdrawn amount

$A = 60000\left(1 + \frac{0.07}{4}\right)^6\left(1 + \frac{0.05}{12}\right)^{42} - 5000\left(1 + \frac{0.05}{12}\right)^{24}$

$A = \text{R}73762.19$