A) Rs.12100
B) Rs.10720
C) Rs.11500
D) Rs.13215
E) Rs.9210
Correct Answer: C
Solution :
Let the principal be Rs. x. Then, \[Cl=P{{\left( 1+\frac{r}{100} \right)}^{n}}-P\] \[=x\left\{ {{\left( 1+\frac{20}{100} \right)}^{2}}-1 \right\}=x\left\{ {{\left( \frac{6}{5} \right)}^{2}}-1 \right\}\] \[=x\left\{ \frac{36}{25}-1 \right\}=x\left\{ \frac{36-25}{25} \right\}\] \[=x\left\{ \frac{11}{25} \right\}\] Now, according to the question, \[5060=\frac{11x}{25}\] or, \[x=\frac{5060\times 25}{11}=460\times 25=Rs.\,11500\] Method II. 20% per annum compounded annually Step I. \[20+20+\frac{20\times 20}{100}=44%\] Now, 44% of x = 5060 \[\therefore \]\[x=\frac{5060\times 100}{44}=Rs.\,11500\]
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