Answer

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Hint: Based on the concept of linear equations in two variables.

Let the starting salary be Rs. X and the annual increment is R.

Now, the problem says the salary will have an yearly increment of R.

According to the question, his salary after 4 years is Rs 1500.

$ \Rightarrow {\text{X + 4R = 1500 - (1)}}$

And his salary is Rs 1800 after 10 years.

$ \Rightarrow {\text{X + 10R = 1800 - (2)}}$

Subtracting equation (1) from (2), we get

$

\Rightarrow {\text{X + 10R - X - 4R = 1800 - 1500}} \\

\Rightarrow 6{\text{R = 300}} \\

\Rightarrow {\text{R = 50}} \\

$

Putting the value of R in equation (2), we get

$

\Rightarrow {\text{X + 10(50) = 1800}} \\

\Rightarrow {\text{X = 1800 - 500}} \\

\Rightarrow {\text{X = 1300}} \\

$

So, the starting salary is Rs. 1300 and the yearly increment is Rs. 50.

Note:- There are two unknowns. To find two unknowns we require minimum two equations. And the equations can be solved by either the substitution method or the elimination method.

Let the starting salary be Rs. X and the annual increment is R.

Now, the problem says the salary will have an yearly increment of R.

According to the question, his salary after 4 years is Rs 1500.

$ \Rightarrow {\text{X + 4R = 1500 - (1)}}$

And his salary is Rs 1800 after 10 years.

$ \Rightarrow {\text{X + 10R = 1800 - (2)}}$

Subtracting equation (1) from (2), we get

$

\Rightarrow {\text{X + 10R - X - 4R = 1800 - 1500}} \\

\Rightarrow 6{\text{R = 300}} \\

\Rightarrow {\text{R = 50}} \\

$

Putting the value of R in equation (2), we get

$

\Rightarrow {\text{X + 10(50) = 1800}} \\

\Rightarrow {\text{X = 1800 - 500}} \\

\Rightarrow {\text{X = 1300}} \\

$

So, the starting salary is Rs. 1300 and the yearly increment is Rs. 50.

Note:- There are two unknowns. To find two unknowns we require minimum two equations. And the equations can be solved by either the substitution method or the elimination method.

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