A) 13.75%
B) 14%
C) 14.5%
D) 13.5%
E) None of these
Correct Answer: B
Solution :
\[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\] \[18194.4=14000{{\left( 1+\frac{r}{100} \right)}^{2}}\] \[{{\left( 1+\frac{r}{100} \right)}^{2}}=\frac{18194.4}{14000}=\frac{181944}{140000}\] \[=\frac{12996}{10000}={{\left( \frac{114}{100} \right)}^{2}}\] \[1+\frac{r}{100}=\frac{114}{100}\] \[\frac{r}{100}=\frac{14}{100}\]\[\therefore r=14%\]per annumYou need to login to perform this action.
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