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Hint: To solve the given question, we will make use of the formula for calculating the compound interest which is given below.
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
where A is the final amount, P is the initial principal balance, r is the interest rate, n is the number of times interest applied and t is the total time elapsed. We will use this formula to find the relation between r and n when the interest is calculated after four years. Similarly, we will find the relation of
r and n after t years when the amount has become 8 times the principal. After this, we will find the value of t with these two relations.Complete step-by-step answer:
We are given the question that a sum of money doubles itself in 4 years at compound interest. Let us assume that the principal is P, then the amount will be 2P. We know that according to the formula of compound interest, we have,
\[A=P{{\left( 1+\dfrac{r}{n} \right)}^{nt}}\]
Here, A is the amount, P is the initial
principal, r is the rate, n is the number of times the interest is applied and t is the total time elapsed. In our case, A = 2P, P = P, r = R, t = 4 and we assume the value of n as 1, i.e. it is compounded annually. Thus, we will get,
\[2P=P{{\left( 1+\dfrac{R}{1} \right)}^{1\times 4}}\]
\[\Rightarrow 2P=P{{\left( 1+R \right)}^{4}}\]
\[\Rightarrow 2={{\left( 1+R \right)}^{4}}........\left( i \right)\]
Now, we have to find the total time after which the amount will become 8 times the
principal. Thus, we have,
\[8P=P{{\left( 1+R \right)}^{1\times T}}\]
\[\Rightarrow 8={{\left( 1+R \right)}^{T}}......\left( ii \right)\]
Here, T is the time assumed after which it will become 8 times. Now, we will cube the equation (i) on both sides. Thus, we will get,
\[\Rightarrow {{\left( 2 \right)}^{3}}={{\left[ {{\left( 1+R \right)}^{4}} \right]}^{3}}\]
\[\Rightarrow 8={{\left( 1+R \right)}^{4\times 3}}\]
\[\Rightarrow 8={{\left( 1+R \right)}^{12}}....\left( iii
\right)\]
From (ii) and (iii), we have,
\[{{\left( 1+R \right)}^{12}}={{\left( 1+R \right)}^{T}}\]
Therefore, T = 12 years.
Hence, option (a) is the right answer.
Note: We have assumed n = 1 while solving. Let us take the value of n as N and solve it. Thus, we will get,
\[2P=P{{\left( 1+\dfrac{R}{N} \right)}^{4N}}\]
\[\Rightarrow 2={{\left( 1+\dfrac{R}{N} \right)}^{4N}}\]
\[\Rightarrow {{2}^{\dfrac{1}{4N}}}=1+\dfrac{R}{N}\]
\[\Rightarrow R=\left( {{2}^{\dfrac{1}{4N}}}-1
\right)N....\left( i \right)\]
Now, another information given is that the amount has become 8 times the principal after T years. Thus,
\[8P=P{{\left( 1+\dfrac{R}{N} \right)}^{NT}}\]
\[\Rightarrow 8={{\left( 1+\dfrac{R}{N} \right)}^{NT}}\]
Now, we will put the value of R from (i) to the above equation.
\[\Rightarrow 8={{\left( 1+\dfrac{\left( {{2}^{\dfrac{1}{4N}}}-1 \right)}{N} \right)}^{NT}}\]
\[\Rightarrow 8={{\left( 1+{{2}^{\dfrac{1}{4N}}}-1 \right)}^{NT}}\]
\[\Rightarrow
8={{\left( {{2}^{\dfrac{1}{4N}}} \right)}^{NT}}\]
\[\Rightarrow 8={{2}^{\dfrac{T}{4}}}\]
\[\Rightarrow {{2}^{3}}={{2}^{\dfrac{T}{4}}}\]
\[\Rightarrow \dfrac{T}{4}=3\]
\[\Rightarrow T=12\text{ years}\]
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⇒ 4 = [21/5]T⇒ 22 = 2T/5⇒ T/5 = 2∴ T = 10 yrs.