A) 10
B) 8
C) 5
D) 8
Correct Answer: A
Solution :
| [a] r = x%, P = Rs. 3000, A = Rs. 3993, t = 3 years \[\therefore \] \[A=P{{\left( 1+\frac{r}{100} \right)}^{t}}\] \[\Rightarrow \] \[3993=3000{{\left( 1+\frac{x}{100} \right)}^{3}}\] \[\Rightarrow \] \[\frac{3993}{3000}={{\left( 1+\frac{x}{100} \right)}^{3}}\] \[\Rightarrow \] \[{{\left( 1+\frac{x}{100} \right)}^{3}}=\frac{1331}{1000}={{\left( \frac{11}{10} \right)}^{3}}\] \[\Rightarrow \] \[1+\frac{x}{100}=\frac{11}{10}\] \[\Rightarrow \] \[\frac{x}{100}=\frac{11}{10}-1\] \[\Rightarrow \] \[\frac{x}{100}=\frac{1}{10}\] \[\Rightarrow \] \[x=\frac{100}{10}\] \[\Rightarrow \] \[x=10\] |
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