A) Rs. 1200
B) Rs. 1250
C) Rs. 1280
D) Rs. 1296
Correct Answer: B
Solution :
| If the principal be Rs. x, then A = P \[{{\left( 1+\frac{R}{100} \right)}^{T}}\] |
| \[\Rightarrow \] \[1352=x{{\left( 1+\frac{4}{100} \right)}^{2}}\] |
| \[\Rightarrow \] \[1352=x{{\left( \frac{26}{25} \right)}^{2}}\] |
| \[\Rightarrow \] \[x=\frac{1352\times 25\times 25}{26\times 26}\]= Rs. 1250 |
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