If at same rate of interest, in 2 years, the simple interest is Rs. 44 and compound interest is Rs. 45, then what is the principal (in Rs. )?

Correct option is C

Given:

Simple Interest (SI) for 2 years = ₹44

Compound Interest (CI) for 2 years = ₹45

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 = 44

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

44 =$$\frac{2Pr}{100}$$​​

44 =$$\frac{Pr}{50}$$​​

P r = 44$$\times$$​ 50

Pr = 2200 _____(Equation 1)

( t = 2) years

CI = 45

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

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

Pr = 2200

P =$$\frac{2200}{r}$$​​

45 =$$\frac{2200}{r} \left[\left(1 + \frac{r}{100}\right)^2 - 1\right]$$
Let x = $$\frac {r}{100}$$​​

45 =$$\frac{2200}{100x} \left[(1 + x)^2 - 1\right]$$​​

45 =$$\frac{22}{x} \left[(1 + x)^2 - 1\right]$$​​

45 =$$\frac{22}{x} \left[1 + 2x + x^2 - 1\right]$$​​

45 =$$\frac{22}{x} \left[2x + x^2\right]$$​​

45 = 22$$\left[2 + x\right]$$​​

45 = 44 + 22x

45 - 44 = 22x

1 = 22x

x =$$\frac{1}{22}$$​​

​$$\frac{r}{100} = \frac{1}{22} \implies r = 4.5454\%$$​​​

Pr = 2200
P $$\times \frac{100}{22} = 2200$$​​

P = 484

Alternate method:  

​r =$$\frac{1}{22}$$$$\times100$$ = $$\frac{100}{22}\%$$​   

P=  $$\frac {22 \times 2200}{100 }= 484$$