A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs 1500 after 4 year of service and Rs 1800 after 10 years of service, what was his starting salary and what is the annual increment?
Answer
367.8k+ views
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.
Last updated date: 01st Oct 2023
•
Total views: 367.8k
•
Views today: 3.67k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Who had given the title of Mahatma to Gandhi Ji A Bal class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE
