A sum of money was lent out at simple interest at [tex]11\%[/tex] for [tex]\frac{7}{2}[/tex] years and [tex]4 \frac{9}{2}[/tex] years respectively. If the interest difference between 9 years is 5500, find the sum.



Answer :

Sure, let's work through this problem step-by-step.

### Step 1: Understand the Problem
We need to determine the sum of money (the principal amount) that was lent out at a simple interest rate of 11% for two different time periods (3.5 years and 8.5 years). The difference in the interest accrued over these two periods is given as Rs 5500.

### Step 2: Set Up the Equations For Simple Interest

The formula for simple interest (SI) is:
[tex]\[ SI = P \times R \times T \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount
- [tex]\( R \)[/tex] is the rate of interest per year (expressed as a decimal)
- [tex]\( T \)[/tex] is the time period in years

Given:
- The rate of interest [tex]\( R = 11\% = \frac{11}{100} = 0.11 \)[/tex]
- The first time period [tex]\( T_1 = \frac{7}{2} = 3.5 \)[/tex] years
- The second time period [tex]\( T_2 = 4 + \frac{9}{2} = 4 + 4.5 = 8.5 \)[/tex] years
- The difference in the interest accrued over the two periods is Rs 5500

### Step 3: Calculate the Interest for Each Period

Calculate the interest accrued over 3.5 years [tex]\( I_1 \)[/tex] and 8.5 years [tex]\( I_2 \)[/tex]:

[tex]\[ I_1 = P \times R \times T_1 \][/tex]
[tex]\[ I_1 = P \times 0.11 \times 3.5 \][/tex]
[tex]\[ I_1 = 0.385 \times P \][/tex]

[tex]\[ I_2 = P \times R \times T_2 \][/tex]
[tex]\[ I_2 = P \times 0.11 \times 8.5 \][/tex]
[tex]\[ I_2 = 0.935 \times P \][/tex]

### Step 4: Set Up the Equation for the Difference in Interest

The difference in interest for the two periods is given as Rs 5500:
[tex]\[ I_2 - I_1 = 5500 \][/tex]

Substitute the expressions for [tex]\( I_1 \)[/tex] and [tex]\( I_2 \)[/tex]:
[tex]\[ 0.935P - 0.385P = 5500 \][/tex]

### Step 5: Solve for [tex]\( P \)[/tex]

Combine like terms:
[tex]\[ (0.935 - 0.385)P = 5500 \][/tex]
[tex]\[ 0.55P = 5500 \][/tex]
[tex]\[ P = \frac{5500}{0.55} \][/tex]
[tex]\[ P = 10000 \][/tex]

Therefore, the principal amount [tex]\( P \)[/tex] is Rs 10000.

### Step 6: Verification (Optional)

To verify, calculate the interest accrued over each period using [tex]\( P = Rs 10000 \)[/tex]:

For 3.5 years:
[tex]\[ I_1 = 0.385 \times 10000 = 3850 \][/tex]

For 8.5 years:
[tex]\[ I_2 = 0.935 \times 10000 = 9350 \][/tex]

Difference in interest:
[tex]\[ I_2 - I_1 = 9350 - 3850 = 5500 \][/tex]

The difference matches the given value, thus verifying our calculations.

### Answer

The sum of money (principal amount) that was lent out is Rs 10000.