Karan invested a sum of money ‘P’ for two years in scheme A at 20% p.a. on compound interest. The amount received at end of two years from scheme A, reinvested in scheme B for 4 years that offers 25% p.a. on simple interest. If total interest received from scheme B is Rs 16500 more than P, then find ‘P’?

The correct option is Rs 37500.

Let's break down the problem step-by-step to find the initial principal amount, $P$.

Step 1: Calculate the amount received from Scheme A
In Scheme A, a sum of money $P$ is invested for 2 years at a compound interest rate of 20% per annum (p.a.).
The formula for the total amount $A$ under compound interest is:

$A=P{(1+\frac{r}{100})}^t$

Given:
Rate, $r=20\%$
Time, $t=2$ years

Substituting these values into the formula:

$A=P{(1+\frac{20}{100})}^2$

$A=P{(1+\frac{1}{5})}^2$

$A=P{\left(\frac{6}{5}\right)}^2$

$A=\frac{36}{25}P=1.44P$

Step 2: Calculate the simple interest received from Scheme B
The amount received from Scheme A ($1.44P$) is reinvested in Scheme B as the principal amount.
Scheme B offers simple interest at 25% p.a. for 4 years.

The formula for simple interest (SI) is:

$\text{SI}=\frac{\text{Principal}\times \text{Rate}\times \text{Time}}{100}$

Here:
Principal for Scheme B = $1.44P$
Rate = $25\%$
Time = $4$ years

Substituting these values:

${\text{SI}}_B=\frac{1.44P\times 25\times 4}{100}$

${\text{SI}}_B=\frac{1.44P\times 100}{100}$

${\text{SI}}_B=1.44P$

Step 3: Set up the equation using the given condition
We are given that the total interest received from Scheme B is Rs 16500 more than $P$:

${\text{SI}}_B-P=16500$

Substitute ${\text{SI}}_B=1.44P$ into the equation:

$1.44P-P=16500$

$0.44P=16500$

$P=\frac{16500}{0.44}$

$P=\frac{1650000}{44}$

$P=37500$

Therefore, the sum of money $P$ is Rs 37500.