Let Principal =P, Rate % =R % Show
t=4 years \( \Large \therefore \) Amount =2P, \( \Large \textbf{Case1:} 2P=P \left(1+\frac{R}{100}\right)^{4}\) \( \Large 2= \left(1+\frac{R}{100}\right)^{4}\)..............1 \( \Large \textbf{Case2:} \) Let after t years it will be 8 times \( \Large 8P=P \left(1+\frac{R}{100}\right)^{t}\) \( \Large\left(2\right)^{3} = \left(1+\frac{R}{100}\right)^{t}\)...............2 By using eq. 1 and 2 \( \Large\left(1+\frac{R}{100}\right)^{12}= \left(1+\frac{R}{100}\right)^{t} \) By comparing both sides, t= 12 years A. 9 years B. 8 years C. 27 years D. 12 years Solution(By Examveda Team)$$\eqalign{ & {\text{Let}}, \cr & {\text{Principal}} = Rs.\,100 \cr & {\text{Amount}} = Rs.\,200 \cr & {\text{Rate}} = r\% \cr & {\text{Time}} = 4\,{\text{years}} \cr & {\text{Now}}, \cr & A = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & 200 = 100 \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} \cr & 2 = {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^4} - - - - - - \left( i \right) \cr & {\text{If}}\,{\text{sum}}\,{\text{become}}\,{\text{8}}\,{\text{times}}\,{\text{in}}\,{\text{the}}\,{\text{time}}\,n\,{\text{years}} \cr & {\text{then,}} \cr & 8 = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {2^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} - - - - - - \left( {ii} \right) \cr & {\text{Using}}\,{\text{eqn}}\,\left( i \right)in\left( {ii} \right),\,{\text{we}}\,{\text{get}} \cr & {\left( {{{\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]}^4}} \right)^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^{12}} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\text{Thus}},\,n = 12\,{\text{years}}. \cr} $$ Home A sum of money placed at compound interest Doubles itself in 4 years in how many years will it amount to 8 times ? Open in App Solution Let, Principal = Rs. 100. Amount = Rs. 200. Rate = r% Time = 4 years. Now, A = P*[1+ (r/100)]^n; 200 = 100*[1+(r/100)]^4; 2 = [1+(r/100)]^4; ........... (i) If sum become 8 times in the time n years, then, 8 = (1+(r/100))^n; 2^3 = (1+(r/100))^n; ........ (ii) Using eqn (i) in (ii), we get; ([1+(r/100)]^4)^3 = (1+(r/100))^n; [1+(r/100)]^12 = (1+(r/100))^n; Thus, n = 12 years.Suggest Corrections 3 Similar questions Q. A sum of money at compound interest (compounded annually) doubles itself in 4 years. In how many years will it amount to eight times of itself ? Q. A sum of money placed at compound interest doubles itself in 4 years. It will amount to eight times itself at the same rate in Q. A sum of money under compound interest doubles itself in 4 years. In how many years will it become 16 times itself? Q. A sum of money doubles itself in 3 years at compound interest, when the interest is compounded annually.In how many years will the sum amount to 16 times of itself? Q. A sum of money doubles itself in $3$ years at CI, when the interest is compounded annually. In how many years will it amount to 16 times of itself? View More
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At what rate will a sum of money doubles itself in 4 years?∴ The rate of interest is 25%
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At what rate a sum becomes double of itself in 8 years?5% Was this answer helpful?
How long will it take for an amount to become double of itself at 4% per annum simple interest?=(x×4100×x)years = 25 years.
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