| An amount of money at compound interest grows upto Rs. 3840 in 4 yr and up to Rs. 3936 in 5 yr. Find the rate of interest. [SSC (CGL) 2012] |
A) 2.5%
B) 2%
C) 3.5%
D) 2.05%
Correct Answer: A
Solution :
| \[A=P{{\left( 1+\frac{r}{100} \right)}^{t}}\] |
| \[\therefore \] \[3840=P{{\left( 1+\frac{r}{100} \right)}^{4}}\] (i) |
| and \[3936=P{{\left( 1+\frac{r}{100} \right)}^{5}}\] ... (ii) |
| On dividing Eq. (ii) by Eq. (i), we get |
| \[\frac{3936}{3840}=1+\frac{r}{100}\] |
| \[\Rightarrow \]\[\frac{3936}{3840}-1=\frac{r}{100}\]\[\Rightarrow \]\[\frac{3936-3840}{3840}=\frac{r}{100}\] |
| \[\Rightarrow \]\[\frac{96}{3840}=\frac{r}{100}\]\[\Rightarrow \]\[r=\frac{96\times 100}{3840}\] |
| \[\therefore \]\[r=2.5%\]per annum |
| Alternate Method |
| Given, \[{{C}_{1}}=3840\]and \[{{C}_{2}}=3936\] |
| \[\Rightarrow \]Rate of interest \[=\frac{{{C}_{2}}-{{C}_{1}}}{{{C}_{1}}}\times 100%\] |
| \[\Rightarrow \] \[r=\frac{3936-3840}{3840}\times 100%\] |
| \[r=\frac{96}{3840}\times 100%\]\[\Rightarrow \]\[r=\frac{960}{384}\] |
| Rate of interest, \[r=25%\]per annum |
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