a) An employee, who received fixed annual increments had a final salary of K90, 000 after 10 years. If her total salary was K650, 000 over the 10 years. What was her initial salary? (10 Marks)​

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To solve this problem, let's denote the employee's initial salary as \( x \) and the fixed annual increment as \( y \).

After 10 years, the employee's final salary is \( x + 10y \), and the total salary over 10 years is the sum of her salaries each year, which is \( 10x + \frac{10(10-1)}{2}y = 10x + 45y \) using the formula for the sum of an arithmetic series.

Given that the final salary is K90,000 and the total salary over 10 years is K650,000, we can set up the following equations:

1. \( x + 10y = 90,000 \) (final salary)
2. \( 10x + 45y = 650,000 \) (total salary over 10 years)

We can solve this system of equations to find the values of \( x \) and \( y \).

From the first equation, we can express \( x \) in terms of \( y \):
\[ x = 90,000 - 10y \]

Substitute this expression for \( x \) into the second equation:
\[ 10(90,000 - 10y) + 45y = 650,000 \]

Now, solve for \( y \):
\[ 900,000 - 100y + 45y = 650,000 \]
\[ -55y = -250,000 \]
\[ y = \frac{-250,000}{-55} = 4545.45 \]

Now that we have found \( y \), we can substitute it back into the first equation to find \( x \):
\[ x = 90,000 - 10(4545.45) = 90,000 - 45,454.5 = 45454.5 \]

So, the employee's initial salary was K45,454.50.

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