₹ 16,000 is deposited in a bank for three years. The rates of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes ₹ 21,384. The rate of interest for the third year will be:

1. 7%

2. 10%

3. 11%

4. 12%

Given,

P = ₹ 16,000

r₁ = 8%

r₂ = 12%

A = ₹ 21,384

Let the interest for third year be r₃.

By formula,

Amount = $P \times \Big(1 + \dfrac{r$1}{100}\Big) \times \Big(1 + \dfrac{r2}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big)

Substituting the values in formula,

$\Rightarrow 21384 = 16000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \times \Big(1 + \dfrac{r$3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{100 + 8}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \times \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{108}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \times \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{27}{25}\Big) \times \Big(\dfrac{28}{25}\Big) \times \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow 21384 = \dfrac{16000 \times 27 \times 28}{25 \times 25} \times \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow 21384 \times 25 \times 25= 16000 \times 27 \times 28 \times \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 25 \times 25}{16000 \times 27 \times 28} = \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 625}{16000 \times 756} = \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow \dfrac{13365000}{12096000} = \Big(1 + \dfrac{r3}{100}\Big) \\[1em] \Rightarrow 1 + \dfrac{r3}{100} = 1.10 \\[1em] \Rightarrow \dfrac{r3}{100} = 1.10 - 1 \\[1em] \Rightarrow r3 = 0.10 \times 100\\[1em] \Rightarrow r_3 = 10 \%

Hence, option 2 is correct option.