A) Rs 1500
B) Rs 1600
C) Rs 1700
D) Rs 1800
Correct Answer: A
Solution :
Let \[x\] be the sum of amount. \[\therefore \] \[SI=\frac{x\times 3\times 10}{100}=\frac{3x}{10}\] and \[CI=x\,\left( 1+\frac{10}{100} \right)-x\] \[=x\times \,{{\left( \frac{11}{10} \right)}^{3}}-x=x\,\left[ {{\left( \frac{11}{10} \right)}^{3}}-1 \right]\] According to the questions, \[x\left[ {{\left( \frac{11}{10} \right)}^{3}}-1 \right]-\frac{3x}{10}=46.5\] \[\Rightarrow \] \[x(1.331-1)-\frac{3x}{10}=46.6\] \[\Rightarrow \] \[x(0.331-\frac{3x}{10}=46.5\] \[\Rightarrow \] \[x\left( 0.331-\frac{3}{10} \right)\,=46.5\] \[\Rightarrow \] \[x(0.331-0.3)=46.5\] \[\Rightarrow \] \[x\times 0.031=46.5\] \[\therefore \] \[x=\frac{46.5}{0.031}=\frac{46500}{31}=1500\] Hence, the sum was Rs.1500.You need to login to perform this action.
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