Question

Swathi starts her job with certain monthly salary and earns a fixed increment every year. If her salary was Rs. 22500 per month after 6 years of service and Rs. 30000 per month after 11 years of service. Find her salary after 8 years of service (in Rs.). 

A) 25500 

B) 29500 

C) 25000 

D) 22343

Answer 1

Correct option is (A) 25500

Let Swathi's first salary is Rs x & fixed increment is Rs y.

\(\therefore\) Swathi's salary after 6 years = Rs (x + 6y)

Swathi's salary after 11 years = Rs (x + 11y)

Swathi's salary after 8 years = Rs (x + 8y)

\(\therefore\) x + 6y = 22500     __________(1)

& x + 11y = 30000   __________(2)

From (1) & (2), we get

22500 - 6y + 11y = 30000

\(\Rightarrow\) 5y = 30000 - 22500

= 7500

\(\Rightarrow\) y = \(\frac{7500}5\) = 1500

\(\therefore\) From (1), we get

x = 22500 - 6y

= 22500 - 6 \(\times\) 1500

= 22500 - 9000

= 13500

\(\therefore\) Swathi's salary after 8 years of service = Rs (x + 8y)

= Rs (13500 + 8 \(\times\) 1500)

= Rs (13500 + 12000)

= Rs 25500

Answer 2

Correct option is A) 25500