| The difference between compound and simple interests on a sum of money at 4% per annum for 2 yr is Rs. 8. The sum is [SSC (CGL) 2014] |
A) Rs. 400
B) Rs. 800
C) Rs. 4000
D) Rs. 5000
Correct Answer: D
Solution :
| Given, \[CI-SI=8\]and \[r=4%\] |
| We know that, \[SI=\frac{P\times r\times t}{100}\] |
| and \[CI=P\left[ {{\left( 1+\frac{r}{100} \right)}^{t}}-1 \right]\] |
| According to the question, |
| \[P\left[ {{\left( 1+\frac{r}{100} \right)}^{t}}-1 \right]-\frac{P\times r\times t}{100}=8\] |
| \[\Rightarrow \] \[P\left[ {{\left( 1+\frac{4}{100} \right)}^{2}}-1 \right]-\frac{P\times 4\times 2}{100}=8\] |
| \[\Rightarrow \] \[P\left[ {{\left( \frac{26}{25} \right)}^{2}}-1 \right]-\frac{8P}{100}=8\] |
| \[\Rightarrow \] \[P\left[ \left( \frac{676}{625} \right)-1 \right]-\frac{8P}{100}=8\] |
| \[\Rightarrow \] \[P\left[ \frac{676-625}{625} \right]-\frac{8P}{100}=8\] |
| \[\Rightarrow \] \[\frac{51P}{625}-\frac{8P}{100}=8\] |
| \[\Rightarrow \] \[5100P-5000P=500000\] |
| \[\Rightarrow \] \[100P=500000\] |
| \[\Rightarrow \] \[P=\frac{500000}{100}=\text{Rs}\text{. 5000}\] |
| Alternate Method |
| When difference between the \[CI\] and \[SI\] on a certain sum of money for 2 yr at r% rate is Rs. x. |
| Difference between\[SI\] and \[CI=\frac{R{{r}^{2}}}{{{(100)}^{2}}}\] |
| \[\Rightarrow \] \[8=\frac{P\times 16}{{{(100)}^{2}}}\]\[\Rightarrow \]\[8=\frac{16P}{10000}\] |
| \[\Rightarrow \] \[16P=80000\]\[\Rightarrow \]\[P=\frac{80000}{16}\] |
| \[\therefore \] \[P=\text{Rs}\text{. 5000}\] |
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