Emily wants to buy a house. The purchase price is R1 620 000. Calculate the amount financed.

$$\text{Amount financed} = \text{Purchase price} - \text{Deposit}$$ $$\text{Amount financed} = \text{R}1\,620\,000 - \text{R}162\,000 = \text{R}1\,458\,000$$

The amount financed is $\boxed{\text{R}1\,458\,000}$.

1.2.2 Calculate Emily's monthly bond repayment.
Given:

• Present Value ($P$) = R1 458 000
• Nominal annual interest rate ($i_{nom}$) = 11% = 0.11
• Monthly interest rate ($i$) = $\frac{0.11}{12}$
• Term in months ($n$) = $20 \text{ years} \times 12 \text{ months/year} = 240$

$$P = \frac{x[1 - (1+i)^{-n}]}{i}$$ $$\text{R}1\,458\,000 = \frac{x[1 - (1 + \frac{0.11}{12})^{-240}]}{\frac{0.11}{12}}$$

$$x = \frac{\text{R}1\,458\,000 \times \frac{0.11}{12}}{1 - (1 + \frac{0.11}{12})^{-240}}$$ $$x = \frac{13355}{1 - (1.00916666666)^{-240}}$$ $$x = \frac{13355}{1 - 0.11200696}$$ $$x = \frac{13355}{0.88799304}$$ $$x \approx \text{R}15\,039.50$$

Emily's monthly bond repayment is $\boxed{\text{R}15\,039.50}$.

1.2.3 How much will Emily have paid back in total at the end of the 20 years?

$$\text{Total paid} = \text{Monthly repayment} \times \text{Number of payments}$$ $$\text{Total paid} = \text{R}15\,039.50 \times 240$$ $$\text{Total paid} = \text{R}3\,609\,480.00$$

Emily will have paid back a total of $\boxed{\text{R}3\,609\,480.00}$.

1.2.4 If Emily decides to pay an extra R1 500 at the end of each month, how many years will it take to settle her bond?
New monthly repayment ($x_{new}$) = R15 039.50 + R1 500 = R16 539.50
$P = \text{R}1\,458\,000$
$i = \frac{0.11}{12}$

$$P = \frac{x_{new}[1 - (1+i)^{-n_{new}}]}{i}$$ $$\text{R}1\,458\,000 = \frac{\text{R}16\,539.50[1 - (1 + \frac{0.11}{12})^{-n_{new}}]}{\frac{0.11}{12}}$$

Rearrange to solve for $n_{new}$:

$$(1 + \frac{0.11}{12})^{-n_{new}} = 1 - \frac{\text{R}1\,458\,000 \times \frac{0.11}{12}}{\text{R}16\,539.50}$$ $$(1.00916666666)^{-n_{new}} = 1 - \frac{13355}{16539.50}$$ $$(1.00916666666)^{-n_{new}} \approx 1 - 0.8074099$$ $$(1.00916666666)^{-n_{new}} \approx 0.1925901$$

$$-n_{new} \ln(1.00916666666) = \ln(0.1925901)$$ $$n_{new} = -\frac{\ln(0.1925901)}{\ln(1.00916666666)}$$ $$n_{new} \approx -\frac{-1.64736}{0.009124} \approx 180.55 \text{ months}$$

$$\text{Years} = \frac{180.55}{12} \approx 15.046 \text{ years}$$

It will take approximately $\boxed{15.05 \text{ years}}$ to settle her bond.

1.3 Rushaana is comparing two home loan options.

• Property price: R1 800 000
• Deposit: R180 000
• Amount financed ($P$) = R1 800 000 - R180 000 = R1 620 000

1.3.1 Calculate the monthly repayment for each option.
Formula: $x = \frac{P \cdot i}{1 - (1+i)^{-n}}$

Imrah Bank:

• $P = \text{R}1\,620\,000$
• $i = \frac{0.1075}{12}$
• $n = 20 \text{ years} \times 12 = 240$ months

$$x_{\text{Imrah}} = \frac{\text{R}1\,620\,000 \times \frac{0.1075}{12}}{1 - (1 + \frac{0.1075}{12})^{-240}}$$ $$x_{\text{Imrah}} = \frac{14587.5}{1 - (1.00895833333)^{-240}}$$ $$x_{\text{Imrah}} = \frac{14587.5}{1 - 0.120538}$$ $$x_{\text{Imrah}} = \frac{14587.5}{0.879462} \approx \text{R}16\,586.20$$

Endinako Bank:

• $P = \text{R}1\,620\,000$
• $i = \frac{0.0995}{12}$
• $n = 25 \text{ years} \times 12 = 300$ months

$$x_{\text{Endinako}} = \frac{\text{R}1\,620\,000 \times \frac{0.0995}{12}}{1 - (1 + \frac{0.0995}{12})^{-300}}$$ $$x_{\text{Endinako}} = \frac{13447.5}{1 - (1.00829166666)^{-300}}$$ $$x_{\text{Endinako}} = \frac{13447.5}{1 - 0.082725}$$ $$x_{\text{Endinako}} = \frac{13447.5}{0.917275} \approx \text{R}14\,659.80$$

The monthly repayment for Imrah Bank is $\boxed{\text{R}16\,586.20}$ and for Endinako Bank is $\boxed{\text{R}14\,659.80}$.

1.3.2 Calculate the total amount Rushaana will pay back over the term of each loan.

Imrah Bank:

$$\text{Total paid}_{\text{Imrah}} = \text{Monthly repayment} \times \text{Number of payments}$$ $$\text{Total paid}_{\text{Imrah}} = \text{R}16\,586.20 \times 240 = \text{R}3\,980\,688.00$$

Endinako Bank:

$$\text{Total paid}_{\text{Endinako}} = \text{Monthly repayment} \times \text{Number of payments}$$ $$\text{Total paid}_{\text{Endinako}} = \text{R}14\,659.80 \times 300 = \text{R}4\,397\,940.00$$

The total amount paid for Imrah Bank is $\boxed{\text{R}3\,980\,688.00}$ and for Endinako Bank is $\boxed{\text{R}4\,397\,940.00}$.

1.3.3 Which option is the better choice overall? Motivate your answer fully by comparing both the monthly repayments and the total amount paid.

• Monthly Repayment: Endinako Bank has a lower monthly repayment of R14 659.80 compared to Imrah Bank's R16 586.20. This makes Endinako Bank more affordable on a month-to-month basis.
• Total Amount Paid: Imrah Bank results in a total payment of R3 980 688.00, which is R417 252.00 less than Endinako Bank's total payment of R4 397 940.00. Imrah Bank also has a shorter loan term (20 years vs. 25 years).

Conclusion:
If Rushaana can comfortably afford the higher monthly repayment, Imrah Bank is the better choice overall. This is because it results in a significantly lower total amount paid over the life of the loan (saving R417 252.00) and allows her to pay off the loan 5 years sooner. However, if monthly affordability is a critical concern, Endinako Bank offers a lower monthly payment, albeit at a higher total cost and longer term.