Answer
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Hint: To solve this question, first of all we will assume variables for principal amount and rate of interest. We will use the formula for the future value at compound interest. The formula for future value is FV = $ {{\left( 1+\dfrac{R}{100} \right)}^{n}}\times P $ Since the time period for compounding is not given, we assume that the principal amount is compounded annually. Taking the future value twice that of principal amount, we will try to find a relation between principal amount and rate of interest. Then, we will make the future value 8 times that of the principal amount and use the derived relation to find the number of years required for it.
Complete step-by-step answer:
Let P be the principal amount which compounds at a rate of interest R.
It is given that it takes 15 years for the money to get doubled.
Thus, after 15 years, future value = 2(Principal amount)
$ \Rightarrow $ FV = 2P
So, we will substitute FV = 2P and n = 15 in the formula for future value.
$ \Rightarrow $ 2P = $ {{\left( 1+\dfrac{R}{100} \right)}^{15}}\times $ P
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{15}} $ = 2……(1)
Now, we need to find the number of years required for the future value to be 8 times the principal amount.
Let the number of years be n.
We will substitute future value = 8(principal amount)
Thus, FV = 8(P)
$ \Rightarrow $ 8P = $ {{\left( 1+\dfrac{R}{100} \right)}^{n}}\times $ P
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}} $ = 8
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{2}^{3}} $
From (1), we know that $ {{\left( 1+\dfrac{R}{100} \right)}^{15}} $ = 2
$ \begin{align}
& \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{\left[ {{\left( 1+\dfrac{R}{100} \right)}^{15}} \right]}^{3}} \\
& \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{\left( 1+\dfrac{R}{100} \right)}^{45}} \\
\end{align} $
$ \Rightarrow $ n = 45
Therefore, it takes 45 years for the principal amount to become 8 times.
So, the correct answer is “Option C”.
Note: There is a shortcut logical way to solve this question. If it takes 15 years for the money to get doubled. It will take more than 15 years to double the doubled money. i.e. it will take 30 years for the amount to be 4 times the initial amount. Now, it takes another 15 years to double the already 4 times increased amount. Therefore, it takes 45 years to get 8 times the initial amount.
Complete step-by-step answer:
Let P be the principal amount which compounds at a rate of interest R.
It is given that it takes 15 years for the money to get doubled.
Thus, after 15 years, future value = 2(Principal amount)
$ \Rightarrow $ FV = 2P
So, we will substitute FV = 2P and n = 15 in the formula for future value.
$ \Rightarrow $ 2P = $ {{\left( 1+\dfrac{R}{100} \right)}^{15}}\times $ P
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{15}} $ = 2……(1)
Now, we need to find the number of years required for the future value to be 8 times the principal amount.
Let the number of years be n.
We will substitute future value = 8(principal amount)
Thus, FV = 8(P)
$ \Rightarrow $ 8P = $ {{\left( 1+\dfrac{R}{100} \right)}^{n}}\times $ P
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}} $ = 8
$ \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{2}^{3}} $
From (1), we know that $ {{\left( 1+\dfrac{R}{100} \right)}^{15}} $ = 2
$ \begin{align}
& \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{\left[ {{\left( 1+\dfrac{R}{100} \right)}^{15}} \right]}^{3}} \\
& \Rightarrow {{\left( 1+\dfrac{R}{100} \right)}^{n}}={{\left( 1+\dfrac{R}{100} \right)}^{45}} \\
\end{align} $
$ \Rightarrow $ n = 45
Therefore, it takes 45 years for the principal amount to become 8 times.
So, the correct answer is “Option C”.
Note: There is a shortcut logical way to solve this question. If it takes 15 years for the money to get doubled. It will take more than 15 years to double the doubled money. i.e. it will take 30 years for the amount to be 4 times the initial amount. Now, it takes another 15 years to double the already 4 times increased amount. Therefore, it takes 45 years to get 8 times the initial amount.
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