A man's initial salary is N540.00 a month and increases after each period of six months by N36.00. Find his salary in the eight month of the third year.

Analysis

To solve this problem, we need to track how many times the salary increase has occurred by the time the man reaches the eighth month of his third year.

Concept

This is an arithmetic progression problem where:

• Initial Salary ($a$): N540.00
• Increase ($d$): N36.00
• Interval: Every 6 months.

We need to determine how many "six-month periods" have passed before the 8th month of the 3rd year.

Step-by-Step Solution

1. Calculate the total number of months elapsed: • Year 1: 12 months • Year 2: 12 months • Year 3 (up to the 8th month): 8 months • Total months: $12 + 12 + 8 = 32$ months.

2. Determine the number of increases: The salary increases every 6 months. We divide the total months by 6 to see how many increases have been applied: • $32 \div 6 = 5$ remainder $2$. • This means the salary has increased 5 times by the time he reaches the 8th month of the 3rd year. (The 6th increase would only happen at the end of the 36th month).

3. Calculate the new salary: • Total increase = $5 \times \text{N}36.00 = \text{N}180.00$ • New Salary = Initial Salary + Total Increase • New Salary = $\text{N}540.00 + \text{N}180.00 = \text{N}720.00$

Conclusion

The man's salary in the eighth month of the third year is N720.00.

Exam Tip

When dealing with "increase after every X period" questions, always calculate the total time elapsed first, then use integer division (ignoring the remainder) to find the number of times the increase has been applied. Don't be tempted to add an extra increase just because you are in a new year!