What is the principal amount which earns ₹ 132 as compound interest for the second year at 10% per annum?

1. ₹ 800

2. ₹ 1,050

3. ₹ 1,200

4. Data inadequate

Let the principal amount be ₹ P.

For 1 year :

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100}$

Substituting values we get :

$\Rightarrow I = \dfrac{P \times 10\times 1}{100} = \dfrac{P}{10}$

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

For 2nd year :

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

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100}$

$I = \dfrac{\dfrac{11P}{10} \times 10 \times 1}{100} \\[1em] = \dfrac{11P \times 10}{1000} \\[1em] = \dfrac{11P}{100}.$

Given,

Interest for 2nd year = ₹ 132

$\therefore \dfrac{11P}{100} = 132 \\[1em] \Rightarrow P = \dfrac{132 \times 100}{11} \\[1em] \Rightarrow P = 1,200.$

Hence, Option 3 is correct option.