| On what sum does the difference between the compound interest and the simple interest for 3 yr at 10% is Rs. 31? [SSC (CGL) 2011] |
A) Rs. 1500
B) Rs. 1200
C) Rs. 1100
D) Rs. 1000
Correct Answer: D
Solution :
| Let the sum be Rs. \[x.\] |
| \[r=10\]%and \[t=3\,\,yr\] |
| \[SI=\frac{x\times r\times t}{100}\] |
| \[SI=\frac{x\times 10\times 3}{100}=\frac{3}{10}x\] |
| \[CI=\left[ {{\left( 1+\frac{r}{100} \right)}^{t}}-1 \right]x=\left[ {{\left( 1+\frac{10}{100} \right)}^{3}}-1 \right]x\] |
| \[=\left[ {{\left( \frac{11}{10} \right)}^{3}}-1 \right]x=\left( \frac{1331}{1000}-1 \right)x=\frac{331}{1000}x\] |
| According to the question, |
| \[CI-SI=31\] |
| \[\Rightarrow \]\[\frac{331}{1000}x-\frac{3}{10}x=31\] |
| \[\Rightarrow \]\[\frac{(331-300)}{1000}x=31\] |
| \[\Rightarrow \] \[\frac{31}{1000}x=31\] |
| \[\therefore \] \[x=1000\] |
| \[\therefore \] \[\text{Sum}=\text{Rs}\text{. 1000}\] |
| Alternate Method |
| When difference between the CI and SI on a certain sum of money for 3 yr at r % rate is Rs. x, then |
| Difference between SI and CI |
| \[=\frac{{{\Pr }^{2}}(300+r)}{{{(100)}^{3}}}\] |
| \[\Rightarrow \] \[31=\frac{P\times {{(10)}^{2}}(300+10)}{1000000}\] |
| \[\Rightarrow \] \[31=\frac{P\times 100\times 310}{1000000}\] |
| \[\Rightarrow \] \[31=\frac{31P}{1000}\] |
| \[\Rightarrow \] \[P=1000.\] |
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