Answer :
Sure! Let's break down the solution step-by-step.
Given:
- Principal amount, [tex]\( P = ₹4096 \)[/tex]
- Amount after interest, [tex]\( A = ₹4913 \)[/tex]
- Annual interest rate, [tex]\( r = 12\frac{1}{2}\% \)[/tex] which can be written as [tex]\( 12.5\% \)[/tex] or [tex]\( 0.125 \)[/tex] in decimal form.
The formula for compound interest when the interest is compounded annually is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Since the interest is compounded annually, [tex]\( n = 1 \)[/tex]. Therefore, the formula simplifies to:
[tex]\[ A = P (1 + r)^t \][/tex]
Rearranging this formula to solve for the time, [tex]\( t \)[/tex], we have:
[tex]\[ (1 + r)^t = \frac{A}{P} \][/tex]
Taking the natural logarithm on both sides to isolate [tex]\( t \)[/tex]:
[tex]\[ t \cdot \ln(1 + r) = \ln\left(\frac{A}{P}\right) \][/tex]
Finally, solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{A}{P}\right)}{\ln(1 + r)} \][/tex]
Substitute the given values into the formula:
- [tex]\( A = 4913 \)[/tex]
- [tex]\( P = 4096 \)[/tex]
- [tex]\( r = 0.125 \)[/tex]
So,
[tex]\[ t = \frac{\ln\left(\frac{4913}{4096}\right)}{\ln(1 + 0.125)} \][/tex]
[tex]\[ t = \frac{\ln(4913 / 4096)}{\ln(1.125)} \][/tex]
By solving the above expression, we find:
[tex]\[ t ≈ 1.544 \][/tex]
Thus, the time it will take for ₹4096 to amount to ₹4913 at an annual interest rate of [tex]\( 12\frac{1}{2}\% \)[/tex] is approximately 1.544 years.
Given:
- Principal amount, [tex]\( P = ₹4096 \)[/tex]
- Amount after interest, [tex]\( A = ₹4913 \)[/tex]
- Annual interest rate, [tex]\( r = 12\frac{1}{2}\% \)[/tex] which can be written as [tex]\( 12.5\% \)[/tex] or [tex]\( 0.125 \)[/tex] in decimal form.
The formula for compound interest when the interest is compounded annually is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Since the interest is compounded annually, [tex]\( n = 1 \)[/tex]. Therefore, the formula simplifies to:
[tex]\[ A = P (1 + r)^t \][/tex]
Rearranging this formula to solve for the time, [tex]\( t \)[/tex], we have:
[tex]\[ (1 + r)^t = \frac{A}{P} \][/tex]
Taking the natural logarithm on both sides to isolate [tex]\( t \)[/tex]:
[tex]\[ t \cdot \ln(1 + r) = \ln\left(\frac{A}{P}\right) \][/tex]
Finally, solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{A}{P}\right)}{\ln(1 + r)} \][/tex]
Substitute the given values into the formula:
- [tex]\( A = 4913 \)[/tex]
- [tex]\( P = 4096 \)[/tex]
- [tex]\( r = 0.125 \)[/tex]
So,
[tex]\[ t = \frac{\ln\left(\frac{4913}{4096}\right)}{\ln(1 + 0.125)} \][/tex]
[tex]\[ t = \frac{\ln(4913 / 4096)}{\ln(1.125)} \][/tex]
By solving the above expression, we find:
[tex]\[ t ≈ 1.544 \][/tex]
Thus, the time it will take for ₹4096 to amount to ₹4913 at an annual interest rate of [tex]\( 12\frac{1}{2}\% \)[/tex] is approximately 1.544 years.
