Certainly! Let's solve this problem step-by-step.
### Given Data:
- Principal (P): Rs 55,000
- Compound Interest (CI): Rs 4,488
- Rate of interest per rupee per year: 4 paisa
### Conversion of interest rate:
- 4 paisa per rupee means:
[tex]\( \text{Rate} (r) = \frac{4}{100} = 0.04 \)[/tex] per year
### Required:
- Number of years (t) to pay a compound interest of Rs 4,488.
### Compound Interest Formula:
The compound interest formula is:
[tex]\[ A = P(1 + r)^t \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount
- [tex]\( P \)[/tex] is the principal amount
- [tex]\( r \)[/tex] is the annual interest rate
- [tex]\( t \)[/tex] is the time period in years
### Step 1: Find the final amount (A).
The compound interest (CI) is given by:
[tex]\[ \text{CI} = A - P \][/tex]
Rearranging to solve for [tex]\( A \)[/tex]:
[tex]\[ A = \text{CI} + P \][/tex]
Substituting the values:
[tex]\[ A = 4,488 + 55,000 \][/tex]
[tex]\[ A = 59,488 \][/tex]
### Step 2: Plug values into the compound interest formula.
[tex]\[ 59,488 = 55,000 (1 + 0.04)^t \][/tex]
### Step 3: Solve for [tex]\( t \)[/tex].
1. Divide both sides by the principal (P):
[tex]\[ \frac{59,488}{55,000} = (1 + 0.04)^t \][/tex]
2. Simplify the left side:
[tex]\[ \frac{59,488}{55,000} \approx 1.0816 \][/tex]
3. Convert the equation to logarithmic form to solve for [tex]\( t \)[/tex]:
[tex]\[ \log(1.0816) = t \log(1.04) \][/tex]
4. Rearrange to isolate [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\log(1.0816)}{\log(1.04)} \][/tex]
5. Calculate the values:
- [tex]\( \log(1.0816) \approx 0.0348 \)[/tex]
- [tex]\( \log(1.04) \approx 0.0170 \)[/tex]
6. Plug these calculations back in:
[tex]\[ t \approx \frac{0.0348}{0.0170} \][/tex]
After calculating:
[tex]\[ t \approx 2 \][/tex]
### Answer:
The number of years Ashok takes to pay the compound interest of Rs 4,488 at the rate of 4 paisa per rupee per year is approximately 2 years.
