Question

A sum of money placed at Compound Interest (C.I.) doubles itself in 5 years. It will amount to eight times itself at the same rate of interest in:-
A. 7 years
B. 10 years
C. 15 years
D. 20 years

Answer 1

Hint: To find the number of years in which Compound Interest amounts to eight times given that a sum of money placed at Compound Interest (C.I.) doubles itself in 5 years. We shall take the sum of the money, i.e. the principal amount be ‘X’. Now, according to the question, after 5 years, X becomes 2X, and after next five years, i.e. after 10 years, 2X will become 4X because of the same, and so on. We will use \[A\ \ =\ \ P\ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{n}}\] . Let us take \[A\] as $2X$ , $P$ as $X$ and $n$ as $5$ . We shall put these values in the equation to find the answer. To find the number of years in which the sum that is given to us will become 8 times of itself, take \[A\] as $8X$ , $P$ as $X$ . After substitution and comparing both the equations, we will get the number of years.

Complete step by step answer:
We need to find the number of years in which Compound Interest amounts to eight times.
We know that \[A\ \ =\ \ P\ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{n}}\] is the formula for finding compound interest, where
$A$ is the final amount, $P$ is the initial principal balance, \[R\] is the interest rate, $n$ is the of times interest applied per time period.
Let us take \[A\] as $2X$ , $P$ as $X$ and $n$ as $5$ . We shall put these values in the equation to find the answer.
So, now the equation that we get is:-
\[\begin{align}
  & 2X\ \ =\ \ X\ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{5}} \\
 & \\
\end{align}\]
Cancelling $X$ on both sides, we will get
\[2\ \ =\ \ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{5}}\]
Removing the power of $5$ from RHS, we will get
\[{{2}^{\dfrac{1}{5}}}\ \ =\ \ 1\ \ +\ \ \dfrac{R}{100}\ \,\ \ \ \ .....(a)\]
Now, as mentioned in the question, we have to find the number of years in which the sum that is given to us will become 8 times of itself.
Let us take \[A\] as $8X$ , $P$ as $X$
\[8X\ \ =\ \ X\ \ \times \ \ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{n}}\]
Cancelling $X$ on both sides, we will get
\[8\ \ =\ \ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{n}}\]
Now, using equation (a) that we have found before, we can find the value of ‘n’.
\[8\ \ =\ \ {{\left( {{2}^{\dfrac{1}{5}}} \right)}^{n}}\]
$8$ can be written as
\[{{2}^{3}}\ \ =\ \ {{2}^{\dfrac{n}{5}}}\]
Now, on comparing the exponents of both the sides, we get the equation as follows:-
\[\begin{align}
  & \dfrac{n}{5}\ \ =\ \ 3 \\
 &\Rightarrow n\ \ =\ \ 3\ \ \times \ \ 5 \\
 & \Rightarrow n\ \ =\ \ 15 \\
\end{align}\]
Hence, as the value of ‘n’ is 15 years, the answer of this question is also 15 years as it is the number of years in which the sum amounts to eight times itself at the same rate of interest.

So, the correct answer is “Option C”.

Note: Be careful with the equation of CI \[A\ \ =\ \ P\ {{\left( 1\ \ +\ \ \dfrac{R}{100} \right)}^{n}}\] . There can be a chance of making an error in this equation. We can also use the following method to find the answer to this question.
According to the question, the sum of money doubles itself in five years, therefore, after five years, $X$ will become $2X$ .
$X$ after five years = $2X$
$X$ after next five years ,i.e., after 10 years = $2X\times 2=4X$
$X$ after next five years ,i.e., after 15 years = $4X\times 2=8X$
We need to find the year in which CI amounts to eight times itself. From the above step, we get the answer as 15 years.