Question

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was $\u20b91500$ after $4$ year of service and $\u20b91800$ after $10$ years of service, what was his starting salary and what is the annual increment?

Open in App

Solution

Let $\u20b9x$ be the starting salary of a man and $\u20b9y$ be the fixed annual increment.

Then,

Salary after $4$ years of service $=\u20b9(x+4y)$

Salary after $10$ years of service $=\u20b9(x+10y)$

As per the question, given that,

$(x+4y)=\u20b91500$…………eq.$1$

$(x+10y)=\u20b91800$………..eq.$2$

On subtracting equation ($1$) from equation ($2$), we get,

$6y=300$

$y=\frac{300}{6}=50$

Substituting $y=50$ in equation ($1$), we get,

$x+4\times 50=1500$

$x+200=1500$

$x=1500-200$

$x=1300$

**Therefore, the starting salary of a man was **$\u20b91300$** and the annual increment is **$\u20b950$**.**

0

View More