Question
The difference between CI and SI on a certain sum of money for 3 years at 5% p.a. is Rs. 122. Find the sum invested = ?
Answer: Option C
$$\eqalign{
& {\text{Rate of interest = 5}}\% {\text{ p}}{\text{.c}}{\text{.p}}{\text{.a}}{\text{.}} \cr
& {\text{If time 3 years than CI}} - {\text{SI}} \cr
& {\text{ = }}P\left[ {{{\left( {\frac{R}{{100}}} \right)}^3} + 3{{\left( {\frac{R}{{100}}} \right)}^2}} \right] \cr
& \Rightarrow 122 = P\left[ {{{\left( {\frac{5}{{100}}} \right)}^3} + 3{{\left( {\frac{5}{{100}}} \right)}^2}} \right] \cr
& \Rightarrow 122 = P\left( {\frac{{125}}{{1000000}} + \frac{{75}}{{10000}}} \right) \cr
& \Rightarrow 122 = P\left[ {\frac{{125 + 7500}}{{1000000}}} \right] \cr
& \Rightarrow 122 = P\left[ {\frac{{7525}}{{1000000}}} \right] \cr
& \Rightarrow P = \frac{{122 \times 1000000}}{{7625}} \cr
& \Rightarrow P = {\text{Rs}}{\text{. 16000}} \cr} $$
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$$\eqalign{
& {\text{Rate of interest = 5}}\% {\text{ p}}{\text{.c}}{\text{.p}}{\text{.a}}{\text{.}} \cr
& {\text{If time 3 years than CI}} - {\text{SI}} \cr
& {\text{ = }}P\left[ {{{\left( {\frac{R}{{100}}} \right)}^3} + 3{{\left( {\frac{R}{{100}}} \right)}^2}} \right] \cr
& \Rightarrow 122 = P\left[ {{{\left( {\frac{5}{{100}}} \right)}^3} + 3{{\left( {\frac{5}{{100}}} \right)}^2}} \right] \cr
& \Rightarrow 122 = P\left( {\frac{{125}}{{1000000}} + \frac{{75}}{{10000}}} \right) \cr
& \Rightarrow 122 = P\left[ {\frac{{125 + 7500}}{{1000000}}} \right] \cr
& \Rightarrow 122 = P\left[ {\frac{{7525}}{{1000000}}} \right] \cr
& \Rightarrow P = \frac{{122 \times 1000000}}{{7625}} \cr
& \Rightarrow P = {\text{Rs}}{\text{. 16000}} \cr} $$
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