| A sum of money placed at compound interest doubles itself in 4 yr. In how many years will it amount to four times itself? [SSC (CGL) 2011] |
A) 12 yr
B) 13 yr
C) 8 yr
D) 16 yr
Correct Answer: C
Solution :
| Let \[A=2x,\]Then, \[P=x.\] So, \[\frac{A}{P}=2\] |
| We know that, |
| \[A=P{{\left( 1+\frac{r}{100} \right)}^{t}}\] |
| \[\Rightarrow \] \[\frac{A}{P}={{\left( 1+\frac{r}{100} \right)}^{4}}\] |
| \[\Rightarrow \] \[2={{\left( 1+\frac{r}{100} \right)}^{4}}\] |
| On squaring both sides, we get |
| \[{{2}^{2}}={{\left( 1+\frac{r}{100} \right)}^{8}}\] |
| \[\Rightarrow \] \[4={{\left( 1+\frac{r}{100} \right)}^{8}}\] |
| \[\therefore \]It will become 4 times itself in 8 yr. |
| Alternate Method |
| If a certain sum, at compound interest becomes x time in \[{{t}_{1}}\,\,yr\]and \[y\]times in \[{{t}_{2}}\,\,yr.\]Then, |
| \[{{x}^{\frac{1}{{{t}_{1}}}}}={{y}^{\frac{1}{{{t}_{2}}}}}\] |
| Given, \[{{t}_{1}}=4\,\,yr,\]\[x=2,\]\[{{t}_{2}}=?\] |
| and \[y=4\] |
| \[\Rightarrow \]\[{{(2)}^{1/4}}={{(4)}^{1/{{t}_{2}}}}\] |
| \[\Rightarrow \]\[{{(2)}^{1/4}}={{(2)}^{2/{{t}_{2}}}}\] |
| On comparing both sides, we get |
| \[\frac{2}{{{t}_{2}}}=\frac{1}{4}\] |
| \[\therefore \] \[{{t}_{2}}=8\,\,yr\] |
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