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 $(P) = R s . 100$
then Amount $(A) = R s . 200$
Period $(t) = 3$ years.
Let $R$ be the rate $%$ .
Since, we have $A = P (1 + \frac{R}{100})^{3}$
Substitute given values, we get
$\Rightarrow 200 = 100 (1 + \frac{R}{100})^{3}$
$\Rightarrow \frac{200}{100} = (1 + \frac{R}{100})^{3}$
$\Rightarrow \sqrt[3]{2} = (1 + \frac{R}{100})$
$\Rightarrow 1.2599 = 1 + \frac{R}{100}$
$\Rightarrow \frac{R}{100} = 1.2599 - 1$
$\Rightarrow R = 0.2599 \times 100$
$∴ R = 25.99 %$ per annum.

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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(t)=3 years. Let R be the rate %.Since, we have A=P(1+R100)3Substitute given values, we get ⇒200=100(1+R100)3⇒200100=(1+R100)3⇒3√2=(1+R100)⇒1.2599=1+R100⇒R100=1.2599−1⇒R=0.2599×100∴R=25.99% per annum.

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