Question

A sum of money doubles itself in 4 years at compound interest. How many years it will become eight times at the same rate of interest.
(a) 12 years
(b) 18 years
(c) 24 years
(d) 16 years

Answer 1

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.
\[\Rightarrow 8={{\left( 1+\dfrac{\left( {{2}^{\dfrac{1}{4N}}}-1 \right)}{N} \right)}^{NT}}\]
\[\Rightarrow 8={{\left( 1+{{2}^{\dfrac{1}{4N}}}-1 \right)}^{NT}}\]
\[\Rightarrow 8={{\left( {{2}^{\dfrac{1}{4N}}} \right)}^{NT}}\]
\[\Rightarrow 8={{2}^{\dfrac{T}{4}}}\]
\[\Rightarrow {{2}^{3}}={{2}^{\dfrac{T}{4}}}\]
\[\Rightarrow \dfrac{T}{4}=3\]
\[\Rightarrow T=12\text{ years}\]