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.

Step-by-step Explanation:

1. Scheme A (Compound Interest):

Let the principal sum invested in Scheme A be P.

Rate of interest for Scheme A, $r=20\%$ per annum.

Time period, $t=2$ years.

The formula for the total amount A received after compound interest is:

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

Substituting the given values:

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

$A=P{\left(\frac{6}{5}\right)}^2=\frac{36}{25}P=1.44P$

2. Scheme B (Simple Interest):

The amount received from Scheme A, which is $1.44P$, is reinvested as the principal ($P_B$) in Scheme B.

Rate of interest for Scheme B, $r_B=25\%$ per annum.

Time period, $t_B=4$ years.

The formula for Simple Interest ($SI$) is:

$SI=\frac{P_B\times r_B\times t_B}{100}$

Substituting the values into the formula:

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

Since $25\times 4=100$, the expression simplifies to:

$SI=1.44P$

3. Finding the value of 'P':

We are given that the total interest received from Scheme B is Rs 16500 more than P:

$SI-P=16500$

Substituting $SI=1.44P$:

$1.44P-P=16500$

$0.44P=16500$

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

$P=37500$

Thus, the sum of money P is equal to Rs 37500.