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In what time will ₹64,000 amount to ₹88,360 at [tex]17 \frac{1}{2} \%[/tex] per annum, interest compounded yearly?

Sagot :

To determine the time it will take for ₹64,000 to amount to ₹88,360 at an annual interest rate of [tex]\(17.5\%\)[/tex] (or [tex]\(0.175\)[/tex] when expressed as a decimal), with interest compounded yearly, we can follow these steps:

### Step-by-Step Solution

1. Identify the given values:
- Principal (P): ₹64,000
- Amount (A): ₹88,360
- Annual interest rate (r): [tex]\(17.5\%\)[/tex] or [tex]\(0.175\)[/tex]

2. Use the compound interest formula:
The compound interest formula for yearly compounding is given by:
[tex]\[ A = P(1 + r)^t \][/tex]
Where:
- [tex]\(A\)[/tex] is the amount after time [tex]\(t\)[/tex],
- [tex]\(P\)[/tex] is the principal amount,
- [tex]\(r\)[/tex] is the annual interest rate,
- [tex]\(t\)[/tex] is the time in years.

3. Rearrange the formula to solve for [tex]\(t\)[/tex]:
[tex]\[ (1 + r)^t = \frac{A}{P} \][/tex]
Taking the natural logarithm (log) of both sides, we get:
[tex]\[ \log((1 + r)^t) = \log\left(\frac{A}{P}\right) \][/tex]
By using the properties of logarithms (specifically, [tex]\(\log(a^b) = b \log(a)\)[/tex]), we can simplify this to:
[tex]\[ t \cdot \log(1 + r) = \log\left(\frac{A}{P}\right) \][/tex]
Solving for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\log\left(\frac{A}{P}\right)}{\log(1 + r)} \][/tex]

4. Substitute the known values into the equation:
[tex]\[ t = \frac{\log\left(\frac{88,360}{64,000}\right)}{\log(1 + 0.175)} \][/tex]
[tex]\[ t = \frac{\log(1.38)}{\log(1.175)} \][/tex]

5. Calculate the values of the logarithms and the final value of [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\log(1.38)}{\log(1.175)} \][/tex]

Thus, the time [tex]\(t\)[/tex] required for ₹64,000 to amount to ₹88,360 at an annual interest rate of [tex]\(17.5\%\)[/tex], compounded yearly, is approximately:

[tex]\(\boxed{2 \text{ years}}\)[/tex]

This shows that it will take very close to 2 years for the principal to grow to the given amount with the given annual compound interest rate.