In how many years will the compound interest on ₹ 1000 at the rate of 10% p.a. be ₹ 210?

Answer

To solve the problem of finding out in how many years the compound interest on ₹1000 at the rate of 10% per annum will be ₹210, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Principal (P) = ₹1000
- Rate of Interest (R) = 10% per annum
- Compound Interest (CI) = ₹210

2. **Determine the Amount (A):**
The formula for Compound Interest is:
\[ CI = A - P \]
Therefore, we can express the amount (A) as:
\[ A = P + CI \]
Substituting the known values:
\[ A = 1000 + 210 = ₹1210 \]

3. **Use the Compound Interest Formula:**
The formula for the amount in compound interest is:
\[ A = P \left(1 + \frac{R}{100}\right)^N \]
Substituting the values we have:
\[ 1210 = 1000 \left(1 + \frac{10}{100}\right)^N \]
Simplifying the equation:
\[ 1210 = 1000 \left(1 + 0.1\right)^N \]
\[ 1210 = 1000 \left(\frac{11}{10}\right)^N \]

4. **Isolate the Exponential Term:**
Dividing both sides by 1000 gives:
\[ \frac{1210}{1000} = \left(\frac{11}{10}\right)^N \]
\[ 1.21 = \left(\frac{11}{10}\right)^N \]

5. **Take Logarithm to Solve for N:**
To solve for N, we can take the logarithm of both sides:
\[ \log(1.21) = N \cdot \log\left(\frac{11}{10}\right) \]
Therefore,
\[ N = \frac{\log(1.21)}{\log\left(\frac{11}{10}\right)} \]

6. **Calculate the Values:**
Using a calculator:
- \(\log(1.21) \approx 0.083\)
- \(\log\left(\frac{11}{10}\right) \approx 0.041\)

Now substituting these values:
\[ N = \frac{0.083}{0.041} \approx 2.02 \]

7. **Conclusion:**
Since N must be a whole number, we can round it to 2 years. Thus, the time required for the compound interest to be ₹210 is approximately **2 years**.