At what rate of compound interest p.a. will ₹ 20,000 amount to ₹ 26,620 in 3 years?

1. 4%

2. 6%

3. 8%

4. 10%

Given,

A = ₹ 26,620

P = ₹ 20,000

n = 3 years

Let rate of interest be r.

$A = P\Big(1 + \dfrac{r}{100}\Big)^{n}$

$\Rightarrow 26620 = 20000 \Big(1 + \dfrac{R}{100}\Big)^{3} \\[1em] \Rightarrow \dfrac{26620}{20000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = \Big(1 + \dfrac{R}{100}\Big) \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{11-10}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{100}{10} = R \\[1em] \Rightarrow R = 10\%$

Hence, option 4 is correct option.