A cellphone bought in 2022 cost R8 000 - NSC Technical Mathematics - Question 5 - 2022 - Paper 1

At the start of the survey in 2011, there were 10,621 white rhinos.

Graph g represents the reducing-balance method.

Let the initial population be $P = 10,621$ and the final population be $3459$. The depreciation rate is $r = 0.128$. The formula for the reducing-balance method is:

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

Setting up the equation:

$3459 = 10621(1 - 0.128)^n$

We can simplify this to:

$\frac{3459}{10621} = (0.872)^n$

Taking logarithms on both sides gives:

$n = \frac{\log(\frac{3459}{10621})}{\log(0.872)}$

Calculating:

$n \approx 8.1905 \text{ years (rounded to nearest year)}$

To determine if Samuel will have enough for the boat cruise costing R35,000, we first calculate the value of his investment at the end of 5 years.

1. Calculate the value of the investment after 2 years:

   $A = P(1 + i)^n = 20000 \left(1 + \frac{0.06}{12}\right)^{2 \times 12}$

   Calculate:

   $A_2 = 20000 \left(1 + 0.005\right)^{24} \approx 20000 imes 1.12749 \approx R22,549.62$

2. Add the additional deposit of R5,000:

   $Total = 22549.62 + 5000 = R27,549.62$

3. Calculate the final amount after 3 years at the new rate of 5% per annum compounded half-yearly:

   $A = P(1 + i)^n = 27549.62 \left(1 + \frac{0.05}{2}\right)^{3 \times 2}$

   Calculate:

   $A_5 = 27549.62 \left(1 + 0.025\right)^{6}\approx 27549.62 \cdot 1.159274 \approx R31,941.66$

Since R31,941.66 < R35,000, he will NOT have enough money.