A) 9%
B) 10%
C) 11%
D) 13%
Correct Answer: B
Solution :
P = Rs. 3000, A = Rs. 3993, n = 3 yr \[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\] \[\therefore \] \[{{\left( 1+\frac{r}{100} \right)}^{n}}=\frac{A}{P}\] \[{{\left( 1+\frac{r}{100} \right)}^{3}}=\frac{3993}{3000}=\frac{1331}{1000}\] \[{{\left( 1+\frac{r}{100} \right)}^{3}}={{\left( \frac{11}{10} \right)}^{3}}\] \[\Rightarrow \] \[1+\frac{r}{100}=\frac{11}{10}\] \[\Rightarrow \] \[\frac{r}{100}=\frac{11}{10}-1\] \[\Rightarrow \] \[\frac{r}{100}=\frac{11}{10}\] \[\Rightarrow \] \[r=\frac{100}{10}\] R = 10%
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