6.1 On 31 January 2020, Tshepo made the first of his monthly deposits of R1 000 into a savings account - NSC Mathematics - Question 6 - 2020 - Paper 1

After the last deposit, Tshepo will leave the money to accumulate interest for 1 year. The future value can again be calculated using the formula:

$A = P(1 + i)^n$

Using the total amount from the previous calculation:

• P = 234,888.53
• i = 0.075/12 = 0.00625
• n = 12 months

Now substituting:

$A = 234,888.53(1 + 0.00625)^{12}$

Calculating:

1. First calculate $(1 + 0.00625)^{12} \approx 1.07734$

Substituting this in: $A \approx 234,888.53 * 1.07734 \approx 253,123.54$

Thus, the amount in the account on 31 January 2033 will be approximately R253,123.54.

To find out when the value of the car will depreciate to R92,537.64, we can use the depreciation formula:

$A = P(1 - r)^t$

Where:

• A = R92,537.64 (future value)
• P = R250,000 (original value)
• r = 0.22 (depreciation rate)
• t = time in years

Rearranging to solve for t:

$t = \frac{\log(A/P)}{\log(1 - r)}$

Substituting the known values: $t = \frac{\log(92,537.64/250,000)}{\log(1 - 0.22)}$

Calculating:

1. Calculate $\log(0.37051) \approx -0.4295$
2. Calculate $\log(0.78) \approx -0.1073$
3. Finally, $t \approx \frac{-0.4295}{-0.1073} \approx 4.00$

Thus, it will take approximately 4 years for the car's book value to depreciate to R92,537.64.

For calculating the value of the loan, we can utilize the loan formula:

$P = \frac{R}{(1 - (1 + i)^{-n})/i}$

Where:

• P = Loan amount
• R = Monthly repayment = R1,500
• i = 11.3% p.a. / 12 = 0.113/12
• n = 6 years * 12 = 72 months

Substituting: $P = \frac{1500}{(1 - (1 + 0.00942)^{-72})/0.00942}$

Calculating this:

1. Calculate the denominator: $1 - (1 + 0.00942)^{-72} \approx 0.39136$
2. Thus: $P \approx \frac{1500}{0.39136/0.00942} \approx 78,173.49$

Thus, the value of the loan is approximately R78,173.49.

To find the total interest paid, we first calculate how much Mpho will pay over 5 years:

Total payment over 5 years = Monthly repayment * number of payments = R1,500 * (5 * 12) = R90,000.

Next, we need to find the loan balance after 5 years using:

$B = P \left(1 + i\right)^{-n}$

Using the previously calculated values and substituting:

• P = Loan amount = 78,173.49
• i = 0.00942
• n = 60 (for 5 years)

Thus: $B = 78,173.49 \left(1 + 0.00942\right)^{-60}$

Calculating the balance:

1. Calculate the balance and determine the paid amount: Total paid - Remaining balance gives us total interest paid: R90,000 - (calculated balance)

Thus, the interest paid will be the difference between payments and remaining loan balance.