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Sagot :
Let's break this problem into two parts:
Part 1: Finding the Sum
Given:
- Compound interest (CI) per annum exceeds simple interest (SI) per annum by ₹ 360.
- Rate of interest [tex]\( R = 12.5\% \)[/tex] (which is [tex]\( \frac{25}{2}\% \)[/tex]).
- Time [tex]\( T = 2 \)[/tex] years (since this is a standard problem setup for such comparisons).
### Simple Interest Calculation:
Simple interest for [tex]\( T \)[/tex] years is given by:
[tex]\[ \text{SI} = \frac{P \times R \times T}{100} \][/tex]
Here, [tex]\( P \)[/tex] is the principal amount.
So, for 1 year:
[tex]\[ \text{SI_{1 year}} = \frac{P \times 12.5 \times 1}{100} = \frac{25P}{200} = \frac{P}{8} \][/tex]
### Compound Interest Calculation:
Compound interest formula for an annual compounding rate is:
[tex]\[ A = P \left(1 + \frac{R}{100} \right)^T \][/tex]
So, for 1 year:
[tex]\[ A = P \left(1 + \frac{25}{200} \right) = P \left(1 + 0.125 \right) = 1.125P \][/tex]
Compound interest for 1 year:
[tex]\[ \text{CI_{1 year}} = A - P = 1.125P - P = 0.125P \][/tex]
### Difference Calculation:
The difference between compound interest and simple interest for 1 year is given as ₹ 360.
[tex]\[ \text{CI_{1 year}} - \text{SI_{1 year}} = 0.125P - \frac{P}{8} = ₹ 360 \][/tex]
Substitute [tex]\(\frac{P}{8}\)[/tex]:
[tex]\[ 0.125P - \frac{P}{8} = ₹ 360 \][/tex]
Converting 0.125 to fractions:
[tex]\[ \frac{P}{8} - \frac{P}{8} = ₹ 360 \][/tex]
No difference needs any calculation here, let's now switch to per annum base and calculate the principal difference. The formulation itself involves special [tex]\( n \)[/tex] growth rate logarithms. Therefore, this abstraction relates more writing the comparable differences.currency based on derived rate differentials projected value sums.
Part 2: Finding the Time
Given:
- Principal amount [tex]\( P = ₹ 4096 \)[/tex]
- Amount [tex]\( A = ₹ 4913 \)[/tex]
- Interest rate [tex]\( r = 12.5\% \)[/tex]
We will use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{100} \right)^t \][/tex]
Substitute the values:
[tex]\[ 4913 = 4096 \left(1 + \frac{12.5}{100} \right)^t \][/tex]
Convert percentage to decimal:
[tex]\[ 1 + \frac{12.5}{100} = 1.125 \][/tex]
So,
[tex]\[ 4913 = 4096 \times 1.125^t \][/tex]
Taking the natural logarithm (ln) on both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(4913) = \ln(4096 \times 1.125^t) \][/tex]
Using the properties of logarithms:
[tex]\[ \ln(4913) = \ln(4096) + t \ln(1.125) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(4913) - \ln(4096)}{\ln(1.125)} \][/tex]
Given the calculations, the time [tex]\( t \)[/tex] is approximately:
[tex]\[ t \approx 1.544 \text{ years} \][/tex]
Part 1: Finding the Sum
Given:
- Compound interest (CI) per annum exceeds simple interest (SI) per annum by ₹ 360.
- Rate of interest [tex]\( R = 12.5\% \)[/tex] (which is [tex]\( \frac{25}{2}\% \)[/tex]).
- Time [tex]\( T = 2 \)[/tex] years (since this is a standard problem setup for such comparisons).
### Simple Interest Calculation:
Simple interest for [tex]\( T \)[/tex] years is given by:
[tex]\[ \text{SI} = \frac{P \times R \times T}{100} \][/tex]
Here, [tex]\( P \)[/tex] is the principal amount.
So, for 1 year:
[tex]\[ \text{SI_{1 year}} = \frac{P \times 12.5 \times 1}{100} = \frac{25P}{200} = \frac{P}{8} \][/tex]
### Compound Interest Calculation:
Compound interest formula for an annual compounding rate is:
[tex]\[ A = P \left(1 + \frac{R}{100} \right)^T \][/tex]
So, for 1 year:
[tex]\[ A = P \left(1 + \frac{25}{200} \right) = P \left(1 + 0.125 \right) = 1.125P \][/tex]
Compound interest for 1 year:
[tex]\[ \text{CI_{1 year}} = A - P = 1.125P - P = 0.125P \][/tex]
### Difference Calculation:
The difference between compound interest and simple interest for 1 year is given as ₹ 360.
[tex]\[ \text{CI_{1 year}} - \text{SI_{1 year}} = 0.125P - \frac{P}{8} = ₹ 360 \][/tex]
Substitute [tex]\(\frac{P}{8}\)[/tex]:
[tex]\[ 0.125P - \frac{P}{8} = ₹ 360 \][/tex]
Converting 0.125 to fractions:
[tex]\[ \frac{P}{8} - \frac{P}{8} = ₹ 360 \][/tex]
No difference needs any calculation here, let's now switch to per annum base and calculate the principal difference. The formulation itself involves special [tex]\( n \)[/tex] growth rate logarithms. Therefore, this abstraction relates more writing the comparable differences.currency based on derived rate differentials projected value sums.
Part 2: Finding the Time
Given:
- Principal amount [tex]\( P = ₹ 4096 \)[/tex]
- Amount [tex]\( A = ₹ 4913 \)[/tex]
- Interest rate [tex]\( r = 12.5\% \)[/tex]
We will use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{100} \right)^t \][/tex]
Substitute the values:
[tex]\[ 4913 = 4096 \left(1 + \frac{12.5}{100} \right)^t \][/tex]
Convert percentage to decimal:
[tex]\[ 1 + \frac{12.5}{100} = 1.125 \][/tex]
So,
[tex]\[ 4913 = 4096 \times 1.125^t \][/tex]
Taking the natural logarithm (ln) on both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(4913) = \ln(4096 \times 1.125^t) \][/tex]
Using the properties of logarithms:
[tex]\[ \ln(4913) = \ln(4096) + t \ln(1.125) \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(4913) - \ln(4096)}{\ln(1.125)} \][/tex]
Given the calculations, the time [tex]\( t \)[/tex] is approximately:
[tex]\[ t \approx 1.544 \text{ years} \][/tex]
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