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 Rs19, 800 per month at the end of the first month after 3 years of service and Rs23, 400 per month at the end of the first month after 9 years of service, find his starting salary and his annual increment?

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

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.

Answer 2

Solution:

Let initial salary be 'a'

Increment be 'b'

After 3 years his salary was Rs 19,800

→ a+ 3b = 19800

After 9 years his salary was Rs 23,400

→a+9b = 23,400

by substitution method,

→a+3b = 19800......(1)

→a+9b = 23400.....(2)

☞a = 19800-3b.....(3)

Now put the value of a in equation (2),

☞19800-3b=23400

☞ 3b = 23400-19800

☞ 3b = 3600

☞ b = 3600/3

☞ b = 600

Now put the value of be in equation (3),

☞a = 19800-3(600)

☞a = 19800-1800

☞a = 18000

Hence the intial salary = 1800Rs and increment=600Rs