Question 3: (Start on a New Page) 3.1 A household in Midrand paid R 1200 including VAT for electricity in December of 2024. The tariff structure of February 2024 - March 2025 (excluding VAT) is as follows: - First 200 kWh @ $$ \mathrm{R} 1,50 $$ per kWh - Next 400 kWh @ $$ \mathrm{R} 2,50 $$ per kWh - Above 600 kWh © $$ \mathrm{R} 4,50 $$ per KWh Calculate the total number of units ( kWh ) used by the household in December 2024 if VAT is charged at $$ 15 \% $$.

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Let the cost excluding VAT be denoted by $$ C $$. We know that the household paid R 1200 including VAT at a rate of $$ 15\% $$. Thus,

$$
C = \frac{1200}{1.15} = \frac{1200 \times 20}{23} = \frac{24000}{23} \approx 1043.48.
$$

The tariff structure (excluding VAT) is:
- First 200 kWh at $$ \mathrm{R} 1.50 $$ per kWh,
- Next 400 kWh (i.e. from 201 kWh to 600 kWh) at $$ \mathrm{R} 2.50 $$ per kWh,
- Above 600 kWh at $$ \mathrm{R} 4.50 $$ per kWh.

Let $$ k $$ be the total number of kWh used. We consider the following cases.

**Case 1: $$ k \leq 200 $$**

In this range, the cost would be:
$$
\text{Cost} = 1.50k.
$$
Since even at maximum consumption of 200 kWh the cost would be
$$
1.50 \times 200 = 300,
$$
and $$ 300 < 1043.48 $$, this case is not applicable.

**Case 2: $$ 200 < k \leq 600 $$**

Here, the cost is given by:
$$
\text{Cost} = 300 + 2.50(k-200).
$$
We set this equal to $$ C $$:
$$
300 + 2.50(k - 200) = \frac{24000}{23}.
$$
Subtract 300 from both sides:
$$
2.50(k - 200) = \frac{24000}{23} - 300.
$$
Divide by 2.50:
$$
k - 200 = \frac{\frac{24000}{23} - 300}{2.50}.
$$
To simplify, write 300 with denominator 23:
$$
300 = \frac{300 \times 23}{23} = \frac{6900}{23}.
$$
Thus,
$$
k - 200 = \frac{\frac{24000 - 6900}{23}}{2.50} = \frac{\frac{17100}{23}}{2.50} = \frac{17100}{23 \times 2.50}.
$$
Recognize that $$ 2.50 = \frac{5}{2} $$, so:
$$
k - 200 = \frac{17100}{23 \times \frac{5}{2}} = \frac{17100 \times 2}{23 \times 5} = \frac{34200}{115} \approx 297.39.
$$
Finally, add 200 to find $$ k $$:
$$
k \approx 200 + 297.39 = 497.39.
$$

**Case 3: $$ k > 600 $$**

If $$ k $$ exceeded 600, the cost would be:
$$
\text{Cost} = 300 + 1000 + 4.50(k-600) = 1300 + 4.50(k-600).
$$
Since $$ 1300 $$ is already greater than $$ 1043.48 $$, this case does not apply.

Thus, the total number of kWh used is approximately

$$
\boxed{497.39 \text{ kWh}}.
$$
The household used approximately 497.39 kWh in December 2024.

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