A company bought a photocopier for R150 000 on 1 July 2022 - NSC Mathematics - Question 6 - 2023 - Paper 1

Using the straight-line depreciation formula:

$\text{Depreciation} = \frac{(P - S)}{n}$

Where:

• $P = 150,000$, the initial cost.
• $S$ = salvage value (assumed to be $R0$ for simplicity).
• $n = 5$, the lifespan in years.

In this case, the annual depreciation is:

$\text{Depreciation} = \frac{150,000 - 0}{5} = R30,000$

After 5 years, total depreciation is:

$\text{Total Depreciation} = 5 \times 30,000 = R150,000$

Thus, the trade-in value will be:

$\text{Trade-in Value} = P - \text{Total Depreciation} = 150,000 - 150,000 = R0$

To find the monthly deposit required for the sinking fund, we use the future value of an ordinary annuity formula:

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

Where:

• $F$ is the future value needed to buy the new photocopier (R205,761.18 from part 6.1.1).
• $i = \frac{7.85\, \%}{12} = 0.00654167$ (monthly interest rate).
• $n = 5 \times 12 = 60$ months.

Rearranging to find the monthly deposit:</br>

$P = \frac{F \cdot i}{(1 + i)^n - 1}$

Substituting the known values:

$P = \frac{205,761.18 \cdot 0.00654167}{(1 + 0.00654167)^{60} - 1}$

Calculating this yields:

After performing the calculations, find the monthly deposit amount.

First, we will use the loan payment formula, which can be rearranged to solve for $n$:

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

Where:

• $P = 200,000$, the principal amount.
• $A = 6,000$, the periodic payment.
• $i = \frac{5.25\%}{4} = 0.013125$ (quarterly interest).

Rearranging to find $n$:

$n = -\frac{\log(1 - \frac{P \cdot i}{A})}{\log(1 + i)}$

Substituting:

1. Find $i$: $i = 0.013125$.
2. Calculate: $\log(1 - \frac{200,000 \cdot 0.013125}{6,000})$
3. Find $n$.
4. Divide the result by 4 to convert to years: $n \approx ext{(result of calculation)}$.