A) Rs. 4000
B) Rs. 2500
C) Rs. 3000
D) Rs. 3050
Correct Answer: C
Solution :
| Let the sum be Rs. x and. |
| rate of interest per annum be r. |
| \[\therefore \] \[4500=x{{\left( 1+\frac{r}{100} \right)}^{2}}\] |
| \[6750=x{{\left( 1+\frac{r}{100} \right)}^{4}}\] |
| On dividing Eq. (ii) by Eq. (i), we get |
| \[\frac{6750}{4500}={{\left( 1+\frac{r}{100} \right)}^{2}}\]\[\Rightarrow \]\[{{\left( 1+\frac{r}{100} \right)}^{2}}=\frac{9}{6}=\frac{3}{2}\] |
| Substituting this value in Eq. (i), we get |
| \[x\times \frac{3}{2}=4500\]\[\Rightarrow \]\[x=\frac{4500\times 2}{3}\]= Rs. 3000 |
You need to login to perform this action.
You will be redirected in
3 sec
