The problem describes a new electricity charging system with an installation fee of K15. The first 200 units are charged at 50 toea per unit, and units exceeding 200 are charged at 60 toea per unit. We need to: (i) Construct a formula for this charging function. (ii) Calculate the charge for 200 kilowatt hours. (iii) Calculate the number of units consumed if the total charge is K120.

1. Problem Description

The problem describes a new electricity charging system with an installation fee of K

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5. The first 200 units are charged at 50 toea per unit, and units exceeding 200 are charged at 60 toea per unit. We need to:

(i) Construct a formula for this charging function.

(ii) Calculate the charge for 200 kilowatt hours.

(iii) Calculate the number of units consumed if the total charge is K

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2. Solution Steps

(i) Constructing the charging function:

Let $C$ be the total charge in Kina, and $x$ be the number of units (kilowatt hours).

The installation fee is K

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5. The cost for the first 200 units is $200 \times 0.50 = 100$ Kina (since 50 toea = K0.50).

If $x \le 200$, the total cost is the installation fee plus the cost of the units used, so $C = 15 + 0.50x$.

If $x > 200$, the total cost is the installation fee plus the cost of the first 200 units plus the cost of the units exceeding

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0. The number of units exceeding 200 is $x - 200$. The cost of these units is $0.60(x - 200)$.

So, if $x > 200$, $C = 15 + 100 + 0.60(x - 200) = 115 + 0.60(x - 200) = 115 + 0.60x - 120 = 0.60x - 5$.

Therefore, the charging function is:

$C(x) = \begin{cases} 15 + 0.50x, & \text{if } x \le 200 \\ 0.60x - 5, & \text{if } x > 200 \end{cases}$

(ii) Charge for 200 kilowatt hours:

Since $x = 200$, we use the first part of the function:

$C(200) = 15 + 0.50(200) = 15 + 100 = 115$.

The charge for 200 kilowatt hours is K

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5.

(iii) Units consumed for a charge of K120:

We need to find $x$ such that $C(x) = 120$.

First, let's assume $x \le 200$.

Then $15 + 0.50x = 120$.

$0.50x = 120 - 15 = 105$.

$x = \frac{105}{0.50} = 210$.

However, this contradicts our assumption that $x \le 200$. So, $x > 200$.

Now, let's assume $x > 200$.

Then $0.60x - 5 = 120$.

$0.60x = 120 + 5 = 125$.

$x = \frac{125}{0.60} = \frac{1250}{6} = \frac{625}{3} \approx 208.33$.

Since $x > 200$, this result is valid.

Therefore, the corresponding units consumed are approximately 208.33 kilowatt hours.

3. Final Answer

(i) $C(x) = \begin{cases} 15 + 0.50x, & \text{if } x \le 200 \\ 0.60x - 5, & \text{if } x > 200 \end{cases}$

(ii) K115

(iii) 208.33 units