Join the conversation on IDNLearn.com and get the answers you seek from experts. Ask anything and receive prompt, well-informed answers from our community of experienced experts.
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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
