If at the same rate of interest, in 2 years, the simple interest is Rs. 40 and the compound interest is Rs. 44, then what is the principal (in Rs.)?

Correct option is A

Given:

Simple Interest (SI) for 2 years = ₹40

Compound Interest (CI) for 2 years = ₹44

Rate of interest (r) is the same for both.

Formula Used:

SI =$$\frac{P \times r \times t}{100}$$​​

CI = P$$\left(1 + \frac{r}{100}\right)^t$$​- P

Solution:

( t = 2) years

SI = 40

40 =$$\frac{P \times r \times 2}{100}$$​​

40 =$$\frac{2Pr}{100}$$​​

40 =$$\frac{Pr}{50}$$​​

P r = 40$$\times$$​ 50

Pr = 2000 _____(Equation 1)

( t = 2) years

CI = 44

44 = P$$\left(1 + \frac{r}{100}\right)^2$$​ - P

44 = P$$\left[\left(1 + \frac{r}{100}\right)^2 - 1\right]$$​​

Pr = 2000

P =$$\frac{2000}{r}$$​​

44 =$$\frac{2000}{r} \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]$$
Let x = $$\frac {r}{100}$$​​

44 =$$\frac{2000}{100x} \left[(1 + x)^2 - 1\right]$$​​

44 =$$\frac{20}{x} \left[(1 + x)^2 - 1\right]$$​​

44 =$$\frac{20}{x} \left[1 + 2x + x^2 - 1\right]$$​​

44 =$$\frac{20}{x} \left[2x + x^2\right]$$​​

44 = 20$$\left[2 + x\right]$$​​

44 = 40 + 20x

44 - 40 = 20x

4 = 20x

x =$$\frac{4}{20} = \frac{1}{5}$$​​

​$$\frac{r}{100} = \frac{1}{5} \implies r = 20$$​​​

Pr = 2000
P $$\times 20 = 2000$$​​

P = 100

Alternate method:  

​​r =$$\frac{4}{20}$$$$\times100$$ =20%   

P=  $$\frac {20 \times 100}{20} = 100$$