Question

A man started his job with a certain monthly salary and earned a fixed increment every year. If his salary was $Rs.4500$ after $5$ years of service and $Rs.5550$ after $12$ years of service, what was his starting salary and what was his annual increment.

• A. Starting salary $=Rs.2550$ and annual increment $=Rs.180$
• B. Starting salary $=Rs.3750$ and annual increment $=Rs.150$
• C. Starting salary $=Rs.3500$ and annual increment $=Rs.150$
• D. Starting salary $=Rs.4050$ and annual increment $=Rs.120$

Answer 1

Answer: (B)

Starting salary $=Rs.3750$ and annual increment $=Rs.150$

Let man's initial monthly salary be Rs.x and annual increment be Rs.y
Then According to the problem
After 5 years salary wil become
x+5y $=$ 4500........(1)
After 12 years salary will be
x+12y $=$ 5550........(2)
Hence we get equations
x+5y $=$ 4500........(1)
x+12y $=$ 5550........(2)
Subtracting eq (1) from eq(2)
7y $=$ 1050
$\Rightarrow y=\frac{1050}{7}=150$
Putting value of y in eq(1) we get x $=$ 3750
Solving these two equations, we get x $=$ Rs.3750, y $=$ Rs.150
Hence starting salary $=$ Rs.3750 and annual increment $=$ Rs.150