5.1 An electrician purchased equipment worth R63 150 and was charged a simple interest rate of \( \frac{4}{23} \) on the purchase amount each year - NSC Technical Mathematics - Question 5 - 2021 - Paper 1

The total payment after 7 years using the formula for simple interest:

$A = P(1 + \frac{r}{100} \times n)$

Where:

• (A) is the total amount payable,
• (P = 63,150),
• (r = 17.39),
• (n = 7).

Substituting the values:

= 63150 \times 2.2173 \approx R140,022.50$$ The total payment at the end of the 7th year is approximately R140,022.50.

To determine the number of years using the reducing-balance method:

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

Where:

• (A = 274000),
• (P = 726900),
• (r = 0.158).

Setting up the equation:

$274000 = 726900(1 - 0.158)^n$

Solving for (n):

1. Calculate ( \frac{274000}{726900} \approx 0.3762) 2.[0.3762 = (0.842)^n]
2. Apply logarithms:

$\log(0.3762) = n \cdot \log(0.842)\rightarrow n = \frac{\log(0.3762)}{\log(0.842)} \approx 5.67$

Thus, it will take approximately 6 years for the value to drop below R274,000.

To calculate the college fees after 4 years:

Using the compound interest formula:

$F = P(1 + r)^n$

Where:

• (P = 25000),
• (r = 0.028),
• (n = 4).

Substituting values:

$F = 25000(1 + 0.028)^4 = 25000(1.1174) \approx R27,935.00$

The college fees at the end of the 4th year would be approximately R27,935.

Calculating Sizwe's savings after 27 months:

First deposit:

$A = 15000(1 + \frac{5.98}{12})^{27}$

Second deposit:

Calculating monthly compound interest for extra deposit of R6,823.54:

If he does not have enough money:

• Total Fees: (R27,935.00)
• Total Savings after 27 months, approximately by evaluating original and interest for both deposits yields:
• Total: (R27,439.55 < R27,935.00).

Thus, Sizwe would not have enough money for college fees.