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 Find the principal which yields a simple interest of Rs. 20 and compound interest of Rs. 21 in two years, at the same percent rate per annum?

 Find the principal which yields a simple interest of Rs. 20 and compound interest of Rs. 21 in two years, at the same percent rate per annum?

A). Rs. 190

B). Rs. 200

C). Rs. 220

D). Rs. 240

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This Question has 1 answers.

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Let the principal be **P** and the rate of interest per annum be **R%**.

### Step 1: Using Simple Interest Formula
Simple Interest (SI) is given by:
$ SI = \frac{P \times R \times T}{100} $

Given that **SI for 2 years** is Rs. 20:
$ \frac{P \times R \times 2}{100} = 20 $
$ P \times R = 1000 $ ----(Equation 1)

### Step 2: Using Compound Interest Formula
Compound Interest (CI) for 2 years is given by:
$ CI = P \left(1 + \frac{R}{100} \right)^2 - P $

Given that **CI for 2 years** is Rs. 21:
$ P \left(1 + \frac{R}{100} \right)^2 - P = 21 $

Expanding using the binomial approximation:
$ P \left( 1 + \frac{2R}{100} + \frac{R^2}{10000} \right) - P = 21 $
$ P \times \left( \frac{2R}{100} + \frac{R^2}{10000} \right) = 21 $
$ P \times \frac{2R}{100} + P \times \frac{R^2}{10000} = 21 $

Using Equation 1:
$ \frac{1000 \times 2}{100} + P \times \frac{R^2}{10000} = 21 $
$ 20 + P \times \frac{R^2}{10000} = 21 $
$ P \times \frac{R^2}{10000} = 1 $
$ P \times \frac{(1000/P)^2}{10000} = 1 $
$ \frac{1000000}{10000P} = 1 $
$ \frac{100}{P} = 1 $
$ P = 200 $

Thus, the correct answer is:

$\boxed{200}$ **Rs.**

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