Question

a sum becomes Rs 1323 in 2 years and Rs 138 9.15 in 3 years when is interest compound annually find the sum?​

Answer 1

$\textbf{Concept used:}$

$\text{Compound interest formula:}$

$\boxed{\textbf{Amount,}\bf\,A=P(1+\frac{r}{100})^n}$

$\text{When n=2 years,}$

$1323=\displaystyle\,P(1+\frac{r}{100})^2$.....(1)

$\text{When n=3 years,}$

$1389.15=\displaystyle\,P(1+\frac{r}{100})^3$.....(2)

$\text{Divide (2) by (1), we get}$

$\displaystyle\frac{P(1+\frac{r}{100})^3}{P(1+\frac{r}{100})^2}=\frac{1389.15}{1323}$

$\implies\displaystyle\,1+\frac{r}{100}=\frac{1389.15}{1323}$

$\implies\displaystyle\,100+r=\frac{138915}{1323}$

$\implies\displaystyle\,r=\frac{138915}{1323}-100$

$\implies\displaystyle\,r=\frac{138915-132300}{1323}$

$\implies\displaystyle\,r=\frac{6615}{1323}$

$\implies\bf\,r=5%$

$\text{Now (1) becomes}$

$\displaystyle\,1323=P(1+\frac{5}{100})^2$

$\displaystyle\,1323=P(\frac{105}{100})^2$

$\displaystyle\,1323=P(\frac{21}{20})^2$

$\displaystyle\,P=\frac{1323{\times}20{\times}20}{21{\times}21}$

$P=\displaystyle\,\frac{63{\times}20{\times}20}{21}$

$\displaystyle\,P=\frac{3{\times}20{\times}20}{1}$

$\implies\displaystyle\textbf{P=Rs.1200}$

$\therefore\textbf{The required sum is Rs. 1200}$

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