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Sagot :
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.
### 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.
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