A) 10%
B) 11%
C) 12%
D) \[12\,\frac{1}{2}\]
Correct Answer: D
Solution :
| [d] Let the principal be Rs. P and rate of interest by R% per annum. |
| \[=\left[ P\times {{\left( 1+\frac{R}{100} \right)}^{2}}-P \right]-\left( \frac{P\times R\times 2}{100} \right)=\frac{P{{R}^{2}}}{{{10}^{4}}}\] |
| Difference of CI and SI for 2 yr |
| \[=\left[ P\times {{\left( 1+\frac{R}{100} \right)}^{3}}-P \right]-\left( \frac{P\times R\times 3}{100} \right)=\frac{P{{R}^{2}}}{{{10}^{4}}}\left( \frac{300+R}{100} \right)\]\[\therefore \] \[\frac{\frac{P{{R}^{2}}}{{{10}^{4}}}\left( \frac{300+R}{100} \right)}{\frac{P{{R}^{2}}}{{{10}^{4}}}}=\frac{25}{8}\] |
| \[\Rightarrow \] \[\left( \frac{300+R}{100} \right)=\frac{25}{8}\] |
| \[\Rightarrow \] \[R=\frac{100}{8}=12\frac{1}{2}\]% |
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