Question

The sum of money placed at a compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
$
  (a){\text{ 16}} \\
  (b){\text{ 8}} \\
  (c){\text{ 12}} \\
  (d){\text{ 20}} \\
 $

Answer 1

Hint: Principal amounts the money deposited initially into the back, so it gets doubled in 4 years. Direct use of the formula for compound interest will help establish relation between time and the amount. Then time can easily be calculated for which the principal amount gets 8 times of itself.

Complete step-by-step answer:
As we know the formula for compound interest which is given as
$ \Rightarrow A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$…………………………. (1)
Where A = total amount received after the compound interest.
             P = Principle amount.
             r = rate of interest.
             t = time in years.
Now it is given that a sum of money placed at compound interest doubles itself in 4 years.
Therefore A = 2P and t = 4 years.
So substitute this values in above equation we have,
 $ \Rightarrow 2P = P{\left( {1 + \dfrac{r}{{100}}} \right)^4}$
Now simplify it we have,
$ \Rightarrow {\left( {1 + \dfrac{r}{{100}}} \right)^4} = 2$
$ \Rightarrow \left( {1 + \dfrac{r}{{100}}} \right) = {\left( 2 \right)^{\dfrac{1}{4}}}$……………………………. (2)
Now we have to calculate in how many years it will amount to eight times itself.
i.e. A = 8P and we have to calculate the value of no. of years.
So let the number of years be n.
So from equation (1)
$ \Rightarrow 8P = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$
$ \Rightarrow {\left( {1 + \dfrac{r}{{100}}} \right)^n} = 8$
Now from equation (2) substitute the value in above equation we have,
$ \Rightarrow {\left( 2 \right)^{\dfrac{1}{4} \times n}} = 8$
Now simplify the above equation we have,
$ \Rightarrow {\left( 2 \right)^n} = {8^4} = {2^{\left( {3 \times 4} \right)}}$
So on comparing we have,
$ \Rightarrow n = 3 \times 4 = 12$ Years.
So in 12 years the amount becomes 8 times to itself.
Hence option (C) is correct.

Note: Whenever we face such types of problems the key concept is simply to have a good grasp over the basic formula involving compound interest, or sometimes even simple interest direct formulas also help in getting on the right track to teach the answer. Calculation in advance problems of this kind involves time to be in years and not in simple months so use a unitary method for time conversion in years for that type of question.