Rates of interest for two consecutive years are 10% and 12% respectively. The percentage increase during these two years is :

1. 22%

2. 23.2%

3. 123.2%

4. 122%

Let initial principal be ₹ x.

For first year :

P = x

R = 10%

T = 1 year

By formula,

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times 1}{100} = \dfrac{x}{10}$.

Amount = P + I = x + $\dfrac{x}{10} = \dfrac{11x}{10}$

For second year :

P = $\dfrac{11x}{10}$

R = 12%

T = 1 year

By formula,

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{11}{10}x \times 12 \times 1}{100} = \dfrac{132x}{1000}$.

A = P + I = $\dfrac{11x}{10} + \dfrac{132x}{1000} = \dfrac{1100x + 132x}{1000} = \dfrac{1232x}{1000}$.

Compound interest = Final Amount - Initial Principal

= $\dfrac{1232x}{1000} - x = \dfrac{1232x - 1000x}{1000} = \dfrac{232x}{1000}$.

Percentage increase = $\dfrac{\text{Compound interest}}{\text{Initial principal}} \times 100$

$= \dfrac{\dfrac{223x}{1000}}{x} \times 100 = \dfrac{223x}{1000x} \times 100$ = 23.2%

Hence, Option 2 is the correct option.