4.1 Wayde invest R500 000 at 7,2% per annum compounded monthly - NSC Mathematics - Question 4 - 2017 - Paper 1

To determine the value of the investment after 5 years:

Substituting $n = 5$ into the expression derived previously:

$A = 500000 \left(1 + \frac{0.072}{12}\right)^{12 \times 5}$

Calculating this:

$A = 500000 \left(1 + 0.006\right)^{60}$

This simplifies to:

$A = 500000 \times (1.006)^{60}$

Calculating $(1.006)^{60} \approx 1.48985$ gives:

$A \approx 500000 \times 1.48985 \approx 714892.41$

Thus, the value of the investment after 5 years is approximately R715 894,21.

To find out when the investment exceeds R1 million, we set up the equation:

$1000000 = 500000 \left(1 + 0.006\right)^{12n}$

This simplifies to:

$2 = \left(1.006\right)^{12n}$

Taking the logarithm on both sides:

$\log(2) = 12n \cdot \log(1.006)$

Solving for $n$ gives:

$n = \frac{\log(2)}{12 \cdot \log(1.006)}$

Calculating the values:

$n \approx \frac{0.3010}{12 \times 0.0026} \approx 9.66$

Therefore, Wayde's investment will exceed R1 million after approximately 10 years.

To find the deposit needed, we first calculate the total loan amount over the 3 years:

The monthly payment is R10,000 for 36 months (3 years), leading to total payments:

$Total Payments = R10,000 \times 36 = R360,000$

Next, apply the formula for the present value of an annuity:

$P = P_m \left(1 - \left(1 + r \right)^{-n} \right) \div r$ where:

• $P_m = 10000$ (monthly payment)
• $r = \frac{0.15}{12} = 0.0125$ (monthly interest rate)
• $n = 36$ (total periods)

Then:

$P = 10000 \left(1 - (1 + 0.0125)^{-36}\right) \div 0.0125$

Computing this gives:

$P \approx R288,472.60$

Finally, to find the deposit, we subtract the present value from the car's price:

$Deposit = 350000 - R288,472.60 \approx R61,527.40$

Thus, Mr. Jones will need to pay approximately R61,527.40 as a deposit.

In this case, the car's price is R350,000 and Mr. Jones wants to finance it for 5 years.

Using a similar formula for the present value of an annuity:

$P = P_m \left(1 - (1 + r)^{-n} \right) \div r$ & where:

• $P = 350000$ (total amount)
• $r = \frac{0.185}{12} \approx 0.01541667$ (monthly interest rate)
• $n = 60$ (total periods over 5 years)

Rearranging gives:

$P_m = P \cdot \frac{r}{1 - (1 + r)^{-n}}$

Substituting the values:

$P_m = 350000 \cdot \frac{0.01541667}{1 - (1 + 0.01541667)^{-60}}$

Calculating this yields:

$P_m \approx R8,628.49$

Therefore, Mr. Jones's monthly payment will be approximately R8,628.49.