Certain sum is lent at 10% compound interest per annum. If the interest accrued during the year 2024 was ₹1,331 then the interest accrued during the year 2022, was:

1. ₹ 1,210

2. ₹ 1,100

3. ₹ 1,464.10

4. ₹ 1,610.51

Given, Rate of interest = 10% per annum

Interest accrued during the year 2024 = ₹1,331

Let ₹ P be the principal amount at the beginning of 2022.

$\text{Interest for the year 2022} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100} \\[1em] = \dfrac{\text{P} \times 10 \times 1}{100} \\[1em] = \dfrac{\text{P}}{10}$

Amount at end of first year, 2022 = P + I

= P + $\dfrac{\text{P}}{10} = \dfrac{\text{10P + P}}{10} = \dfrac{\text{11P}}{10}$

$\text{Interest for the year 2023 }= \dfrac{\dfrac{\text{11P}}{10} \times 10 \times 1}{100}\\[1em] = \dfrac{\text{11P}}{100}$

Amount at end of second year, 2023 = P + I

= $\dfrac{\text{11P}}{10} + \dfrac{\text{11P}}{100} = \dfrac{\text{110P + 11P}}{100} = \dfrac{\text{121P}}{100}$

$\text{Interest for the year 2024 }= \dfrac{\dfrac{\text{121P}}{100} \times 10 \times 1}{100}\\[1em] = \dfrac{\text{121P}}{1000}$

It is given, the interest accrued in 2024 = ₹1,331

$\Rightarrow \dfrac{\text{121P}}{1000} = 1331\\[1em] \Rightarrow \text{P} = \dfrac{1000 \times 1331}{121}\\[1em] \Rightarrow \text{P} = 11,000$

Interest for year 2022 = $\dfrac{P}{10} = \dfrac{11,000}{10}$ = 1,100.

Hence, option 2 is correct option.