Question

A man is appointed in a job with a monthly salary of certain amount and a fixed amount of
annual increment. If his salary was ` 19,800 per month at the end of the first month after 3
years of service and ` 23,400 per month at the end of the first month after 9 years of service,
find his starting salary and his annual increment. (Use matrix inversion method to solve the
problem.)

Answer 1

Answer:

Let he started working with a salary of X and earn fixed increment of Y every year.

Than ,

⟼ Salary + Increment = Salary for next year.

A/q,

After 3 years :-

➪ x + 3y = 19,800 --------- Equation (1) .

After 9 years :-

➪ x + 9y = 23,400 --------- Equation (2) .

subtracting Equation (1) From Equation (2) , we get,

➪ (x + 9y) - (x + 3y) = 23400 - 19800

➪ x - x + 9y - 3y = 3600

➪ 6y = 3600

➪ y = 600 .

Putting value of y in Equation (1) Now,

➪ x + 3*600 = 19800

➪ x = 19800 - 1800

➪ x = 18,000 .

Hence, His starting salary was Rs.18,000 and annual increment was Rs.600.

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Answer 2

Answer:Let x be the salary of a man at the end of first month in the first year and let d be the fixed annual increment.

Salary at the beginning of second year = x + d

Salary at the beginning of third year = x + 2d

Salary at the beginning of fourth year = x + 3d

Given salary after 3 years = ₹ 19,800

Salary after 3 years Means salary at the beginning of fourth year (see the attachment)