A) \[1\frac{1}{2}\,\,yr\]
B) \[2\,\,yr\]
C) \[2\frac{1}{2}\,\,yr\]
D) \[3\,\,yr\]
Correct Answer: A
Solution :
The rate of interest is compounded half yearly, |
\[\therefore \] r = 10% per half year |
Let time \[=\frac{T}{2}\]year = half year |
According to the question, |
Amount = Principal \[{{\left( 1+\frac{\text{Rate}}{100} \right)}^{\text{Time}}}\] |
\[13310=10000{{\left( 1+\frac{10}{100} \right)}^{T}}\Rightarrow \frac{13310}{10000}={{\left( \frac{11}{10} \right)}^{T}}\] |
\[\Rightarrow \] \[{{\left( \frac{11}{10} \right)}^{T}}=\frac{1331}{1000}\Rightarrow {{\left( \frac{11}{10} \right)}^{T}}={{\left( \frac{11}{10} \right)}^{3}}\] |
\[\therefore \] T = 3 half yr\[=1\frac{1}{2}\]yr |
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