Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

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Solution

Let Principal (P) = Rs. 100
then Amount (A) = Rs. 200
Period (n) = 3 years.
Let R be the rate $%$ p.a.
Then $\frac{A}{P} = {(1 + \frac{R}{100})}^{n}$
$\Rightarrow \frac{200}{100} = {(1 + \frac{R}{100})}^{3} \Rightarrow {(1 + \frac{R}{100})}^{3} = \frac{2}{1}$
$\Rightarrow (1 + \frac{R}{100}) = \sqrt[3]{2} \Rightarrow 1 + \frac{R}{100} = 1.2599$
$\frac{R}{100} = 1.2599 - 1.0000 = 0.2599$
$∴ R = 0.2599 \times 100 = 25.99$
$∴$ Rate $= 25.99 %$ p.a

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Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

Let Principal (P) = Rs. 100 then Amount (A) = Rs. 200 Period (n) = 3 years. Let R be the rate % p.a. Then AP=(1+R100)n ⇒200100=(1+R100)3⇒(1+R100)3=21 ⇒(1+R100)=3√2⇒1+R100=1.2599 R100=1.2599−1.0000=0.2599 ∴R=0.2599×100=25.99 ∴ Rate =25.99% p.a