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 |
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