?
A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{
& {\text{Let}}, \cr
& {\text{Principal}} = Rs.\,100 \cr
& {\text{Amount}} = Rs.\,200 \cr
& {\text{Rate}} = r\% \cr
& {\text{Time}} = 4\,{\text{years}} \cr
& {\text{Now}}, \cr
& A = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr
& 200 = 100 \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} \cr
& 2 = {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} - - - - - - \left( i \right) \cr
& {\text{If}}\,{\text{sum}}\,{\text{become}}\,{\text{8}}\,{\text{times}}\,{\text{in}}\,{\text{the}}\,{\text{time}}\,n\,{\text{years}} \cr
& {\text{then,}} \cr
& 8 = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr
& {2^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} - - - - - - \left( {ii} \right) \cr
& {\text{Using}}\,{\text{eqn}}\,\left( i \right)in\left( {ii} \right),\,{\text{we}}\,{\text{get}} \cr
& {\left( {{{\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]}^4}} \right)^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr
& {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^{12}} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr
& {\text{Thus}},\,n = 12\,{\text{years}}. \cr} $$
Join the Discussion
Login to post a comment or share your explanation.
LoginThere for 3*4=12 years
8times (2^3)----- 4*3
Ans. 12 years
200------->400 in again 4 years then,
400------->800 in 4 years again, thus
the time becomes= 4+4+4= 12 years.
Simplest way to solve.
Thank me later😉
n=number of times
t=years
n2 = 8 i.e. 2^3
n1 = 2 (doubles)
t2 = x
t1 = 4
n2=n1^(t2/t1)
2^3 = 2^(x/4)
3 = x/4
3*4 = x
12 = x
How to solve it?