Rahul borrowed some money at the rate of 6% p.a. for the first three years , 9% p.a. for the next years five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is Rs. 8160 , the money borrowed by him ( in Rs. ) was

Answer

To solve the problem step by step, we will break down the calculation of the total simple interest paid by Rahul over the 11 years based on the different interest rates for different time periods.

### Step 1: Define the variables
Let the amount of money borrowed by Rahul be \( x \) (in Rs).

### Step 2: Calculate the interest for the first 3 years
The interest rate for the first 3 years is 6% per annum.

The formula for simple interest is:
\[ \text{Simple Interest} = \frac{P \times R \times T}{100} \]
Where:
- \( P \) = Principal amount (the money borrowed, \( x \))
- \( R \) = Rate of interest (6% for the first 3 years)
- \( T \) = Time (3 years)

Thus, the interest for the first 3 years is:
\[ \text{SI}_1 = \frac{x \times 6 \times 3}{100} = \frac{18x}{100} = 0.18x \]

### Step 3: Calculate the interest for the next 5 years
The interest rate for the next 5 years is 9% per annum.

Using the same formula:
\[ \text{SI}_2 = \frac{x \times 9 \times 5}{100} = \frac{45x}{100} = 0.45x \]

### Step 4: Calculate the interest for the last 3 years
The interest rate for the period beyond 8 years (which is 3 years) is 13% per annum.

Thus, the interest for these 3 years is:
\[ \text{SI}_3 = \frac{x \times 13 \times 3}{100} = \frac{39x}{100} = 0.39x \]

### Step 5: Calculate the total interest
Now, we can add all the simple interests calculated:
\[ \text{Total Interest} = \text{SI}_1 + \text{SI}_2 + \text{SI}_3 \]
\[ \text{Total Interest} = 0.18x + 0.45x + 0.39x = (0.18 + 0.45 + 0.39)x = 1.02x \]

### Step 6: Set up the equation
According to the problem, the total interest paid by Rahul at the end of 11 years is Rs. 8160. Therefore, we can set up the equation:
\[ 1.02x = 8160 \]

### Step 7: Solve for \( x \)
To find \( x \), we divide both sides by 1.02:
\[ x = \frac{8160}{1.02} \]
Calculating this gives:
\[ x = 8000 \]

### Conclusion
The amount of money borrowed by Rahul is Rs. 8000.

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