Sanjay starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was ₹ 31,000 after four years of service and ₹ 40,000 after 10 years, find his initial salary and annual increment.

Answer

To solve the problem, we can set up a system of linear equations based on the information provided.

### Step 1: Define Variables
Let:
- \( x \) = initial monthly salary
- \( y \) = annual increment

### Step 2: Set Up Equations
From the problem, we know:
1. After 4 years, Sanjay's salary is ₹31,000.
- This can be expressed as:
\[ x + 4y = 31,000 \]

2. After 10 years, Sanjay's salary is ₹40,000.
- This can be expressed as:
\[ x + 10y = 40,000 \]

### Step 3: Solve the Equations
We have the following system of equations:
1. \( x + 4y = 31,000 \) (Equation 1)
2. \( x + 10y = 40,000 \) (Equation 2)

To eliminate \( x \), we can subtract Equation 1 from Equation 2:
\[ (x + 10y) - (x + 4y) = 40,000 - 31,000 \]
This simplifies to:
\[ 6y = 9,000 \]

### Step 4: Calculate the Annual Increment
Now, divide both sides by 6 to find \( y \):
\[ y = \frac{9,000}{6} = 1,500 \]
So, the annual increment \( y \) is ₹1,500.

### Step 5: Substitute Back to Find Initial Salary
Now, substitute \( y \) back into Equation 1 to find \( x \):
\[ x + 4(1,500) = 31,000 \]
This simplifies to:
\[ x + 6,000 = 31,000 \]
Now, subtract 6,000 from both sides:
\[ x = 31,000 - 6,000 = 25,000 \]

### Conclusion
Thus, Sanjay's initial monthly salary is ₹25,000 and his annual increment is ₹1,500.

### Summary of the Solution
- Initial monthly salary \( x = ₹25,000 \)
- Annual increment \( y = ₹1,500 \)