Question

Swati 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. $24000$
• B. $25500$
• C. $26000$
• D. $24500$

Answer 1

The correct option is B

$25500$

Step 1: Use the given information

It is given that Swati starts her job with a certain monthly salary and earns a fixed increment every year.

Let Swati's joining salary be Rs. $x$ and her increment is Rs. $y$.

So, the total money which she earns is $\left(x+y\right)$.

Her salary was $\text{Rs}.22,500$ per month after 6 years of service.

$\Rightarrow x+6y=22500...\left(1\right)$

Also, her salary was $\text{Rs}.30,000$ per month after 11 years of service.

$\Rightarrow x+11y=30000...\left(2\right)$

Step 2: Solve the system of equations by the elimination method.

Subtract equation (2) from (1) to get:

$\begin{array}{rcl}& \Rightarrow & \left(x+6y\right)-\left(x+11y\right)=22500-30000\\ & \Rightarrow & -5y=-7500\\ & \Rightarrow & y=1500\end{array}$

Solve for $x$:

$\begin{array}{rcl}x& =& 22500-6y\\ & =& 22500-6\left(1500\right)\\ & =& 13500\end{array}$

Therefore, her current salary is $\left(x+y\right)=15000$.

Her salary after 8 years is $\left(x+8y\right)=13500+8\left(1500\right)=25500$.

Thus, Swati's salary after 8 years is $\text{Rs.}25500$.

Hence, option (b) is correct.