A) 9%
B) 10%
C) 11%
D) 13%
Correct Answer: B
Solution :
| P Rs. 3000, A = Rs. 3993, n = 3 yr |
| \[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\] |
| \[\therefore \] \[{{\left( 1+\frac{r}{100} \right)}^{n}}=\frac{A}{P}\] |
| \[{{\left( 1+\frac{r}{100} \right)}^{3}}=\frac{3993}{3000}=\frac{1331}{1000}\] |
| \[{{\left( 1+\frac{r}{100} \right)}^{3}}={{\left( \frac{11}{10} \right)}^{3}}\] |
| \[\Rightarrow \] \[1+\frac{r}{100}=\frac{11}{10}\] |
| \[\Rightarrow \] \[\frac{r}{100}=\frac{11}{10}-1\]\[\Rightarrow \]\[\frac{r}{100}=\frac{1}{100}\] |
| \[\Rightarrow \] \[r=\frac{100}{10}\]\[\Rightarrow \]\[r=10%\] |
You need to login to perform this action.
You will be redirected in
3 sec
