A) ` 9025
B) ` 9152
C) ` 9215
D) ` 9251
Correct Answer: B
Solution :
| Sol. [b] |
| When the Principal P Amounts A in n years using compound interest then |
| \[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\] |
| \[\therefore \,\,\,P{{\left( 1+\frac{r}{100} \right)}^{2}}\] . (i) |
| and \[10684\,=\,P{{\left( 1+\frac{r}{100} \right)}^{3}}\] .. (ii) |
| divide (ii) by (i), we get |
| \[\frac{10648}{9680}\,=\left( 1+\frac{r}{100} \right)\] |
| \[\frac{10648}{9680}\,-1=\frac{r}{100}\] |
| \[\frac{10648-9680}{9680}=\frac{r}{100}\,\] |
| \[\frac{968}{9680}\,=\,\frac{r}{100}\Rightarrow r=10%\] |
| Now by using eq. (i) |
| \[9680\,=p{{\left( 1+\frac{10}{100} \right)}^{2}}\] |
| \[p=\,9680\,\times \,\frac{100}{121}=\]` 8000 |
| Now P= 8000, Time= \[1\frac{2}{5}\] yr, Rate = 10% |
| \[A=8000{{\left( 1+\frac{10}{100} \right)}^{1}}\,\times \left( 1+\frac{10}{100}\times \frac{2}{5} \right)\] |
| \[A=8000\times \frac{110}{100}\times \frac{26}{25}=\] `9152 |
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