| A builder borrows Rs. 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 yr in two equal yearly instalments. How much will each instalment be? |
A) Rs. 1352
B) Rs. 1377
C) Rs. 1275
D) Rs. 1283
Correct Answer: A
Solution :
| A = Rs. 2550, R = Rs. 4% per annum, n = 2yr |
| Let each of the two equal instalments be Rs. x. |
| Present worth \[=\frac{\text{Instalment}}{{{\left( 1+\frac{r}{100} \right)}^{n}}}\] |
| \[{{P}_{1}}=\frac{x}{{{\left( 1+\frac{4}{100} \right)}^{1}}}=\frac{x}{1+\frac{1}{25}}=\frac{x}{\frac{26}{25}}\] |
| \[\Rightarrow \] \[{{P}_{1}}=\left( \frac{25}{26} \right)x\] |
| Similarly, \[{{P}_{2}}={{\left( \frac{25}{26} \right)}^{2}}x=\frac{625}{676}x\] |
| \[{{P}_{1}}+{{P}_{2}}=A\] |
| \[\therefore \] \[\frac{25}{26}x+\frac{625}{676}x=2550\] |
| \[\Rightarrow \] \[\frac{(650+625)x}{676}=2550\]\[\Rightarrow \]\[\frac{1275}{676}x=2550\] |
| \[\Rightarrow \] \[x=2550\times \frac{676}{1275}\]\[\Rightarrow \]\[x=Rs.\,1352\] |
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