Answer: Option B. -> Rs. 4000
Let the principal = 1000, r = 10%, T = 3Year
Rate r = 10% = $$\frac{1}{10}$$
Amount = 300 + 30 + 1 + 1000(Principal) = 1331
⇒1331 unit → 5324
⇒ 1unit → $$\frac{5324}{1331}$$
∴ 1000unit = $$1000 \times \frac{5324}{1331}$$     = 4000

Answer: Option C. -> Rs. 25200
$$\eqalign{
& {\text{Rate of interest = 5}}\% {\text{ per annum}} \cr
& {\text{Time = 2 year}} \cr
& {\text{Accroding to question,}} \cr
& \Rightarrow P\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} - 1} \right] - \frac{{P \times r \times t}}{{100}}{\text{ = 63}} \cr
& \Rightarrow P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{P \times 5 \times 2}}{{100}}{\text{ = 63}} \cr
& \Rightarrow P\left[ {{{\left( {1 + \frac{5}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr
& \Rightarrow P\left[ {{{\left( {\frac{{105}}{{100}}} \right)}^2} - 1} \right] - \frac{{10P}}{{100}}{\text{ = 63}} \cr
& \Rightarrow P\left( {\frac{{11025 - 10000}}{{10000}}} \right) - \frac{{10P}}{{100}} = 63 \cr
& \Rightarrow \frac{{1025P}}{{10000}} - \frac{{10P}}{{100}} = 63 \cr
& \Rightarrow \frac{{1025P - 1000P}}{{10000}} = 63 \cr
& \Rightarrow 25P = Rs.630000 \cr
& \Rightarrow P = \frac{{630000}}{{25}} \cr
& \Rightarrow P = Rs. 25200 \cr
& {\text{Hence}},\,{\text{sum Rs}}{\text{. 25200}} \cr} $$

Answer: Option D. -> Rs. 81.60
$$\eqalign{
& {\text{Rate }}\% {\text{ = 4}}\% \cr
& {\text{Time (}}{{\text{t}}_1}) = 2\,{\text{years}} \cr
& {\text{SI for 2 years}} \cr
& {\text{ = 4}} \times {\text{2 = 8}}\% \cr
& {\text{CI for 2 years}} \cr
& {\text{ = 4 + 4 + }}\frac{{4 \times 4}}{{100}} \cr
& = 8.16\% \cr
& \operatorname{Required} \,CI \cr
& = \frac{{80}}{8} \times 8.16 \cr
& = Rs.\,81.60 \cr} $$

Answer: Option C. -> 5.5%
$$\eqalign{
& {\text{Let the rate be R}}\% {\text{ p}}{\text{.a}}{\text{.}} \cr
& {\text{Then,}} \cr
& {\text{2000}}{\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} = 2226.05 \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} = \frac{{222605}}{{200000}} \cr
& \, \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} = \frac{{44521}}{{40000}} \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{100}}} \right)^2} = {\left( {\frac{{221}}{{200}}} \right)^2} \cr
& \Rightarrow 1 + \frac{{\text{R}}}{{100}} = \frac{{211}}{{200}} \cr
& \Rightarrow \frac{{\text{R}}}{{100}} = \frac{{11}}{{200}} \cr
& \Rightarrow {\text{R}} = \frac{{11}}{2}\% \cr
& \Rightarrow {\text{R}} = 5.5\% \cr} $$

Answer: Option A. -> Rs. 16000
$$\eqalign{
& \Rightarrow P\left[ {{{\left( {\frac{{21}}{{20}}} \right)}^3} - 1 - \frac{3}{{20}}} \right] = 122 \cr
& \Rightarrow P\left[ {\frac{{{{21}^3} - {{20}^3} - 3 \times {{20}^2}}}{{{{20}^3}}}} \right] = 122 \cr
& \Rightarrow P\left[ {\frac{{9261 - 8000 - 1200}}{{8000}}} \right] = 122 \cr
& \Rightarrow P \times \frac{{61}}{{8000}} = 122 \cr
& \Rightarrow P = \frac{{8000 \times 122}}{{61}} \cr
& \Rightarrow P = {\text{Rs}}{\text{.}}\,16000 \cr} $$

Answer: Option D. -> Rs. 432
$$\eqalign{
& {\text{Effective rate of CI for 2 years}} \cr
& {\text{= 5 + 5 + }}\frac{{5 \times 5}}{{100}} \cr
& = 10.25\% \cr
& {\text{Effective rate of SI for 3 years}} \cr
& {\text{= 6}} \times {\text{3 = 18% }} \cr
& {\text{According to question}} \cr
& {\text{Required SI}} \cr
& {\text{= }}\frac{{246}}{{10.25}} \times 18 \cr
& = {\text{Rs. 432}} \cr} $$

Answer: Option A. -> 2 years
$$\eqalign{
& {\text{Amount = Rs}}{\text{. }}\left( {30000 - 4347} \right) \cr
& {\text{Amount = Rs}}{\text{. }} 34347 \cr
& {\text{Let the time be }}n{\text{ years}} \cr
& {\text{Then,}} \cr
& {\text{30000}}{\left( {1 + \frac{7}{{100}}} \right)^n} = 34347 \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{34347}}{{30000}} \cr
& \Leftrightarrow {\left( {\frac{{107}}{{100}}} \right)^n} = \frac{{11449}}{{10000}} = {\left( {\frac{{107}}{{100}}} \right)^2} \cr
& \therefore n = {\text{ 2 years}} \cr} $$

Answer: Option D. -> 10%
$$\eqalign{
& {\text{Amount after three years}} \cr
& {\text{ = Rs. 2662}} \cr
& {\text{Amount after two years}} \cr
& {\text{ = Rs. 2420}} \cr
& {\text{Net interest earned in the }}{{\text{3}}^{{\text{rd}}}}{\text{ year}} \cr
& {\text{ = }}\,{\text{2662}} - {\text{2420}} \cr
& {\text{ = Rs}}{\text{. 242}} \cr
& {\text{Rate of interest (r)}} \cr
& {\text{ = }}\frac{{242}}{{2420}} \times {\text{100 = 10% }} \cr} $$
(∴ 2nd year's amount is principal for 3rd year)

Answer: Option D. -> 20%
$$\eqalign{
& {\text{Let the rate be R% p}}{\text{.a}}{\text{. }} \cr
& {\text{Then,}} \cr
& {\text{10000}}{\left( {1 + \frac{{\text{R}}}{{2 \times 100}}} \right)^4} = 14641 \cr
& \Rightarrow {\left( {1 + \frac{{\text{R}}}{{200}}} \right)^4} = \frac{{14641}}{{10000}} = {\left( {\frac{{11}}{{10}}} \right)^4} \cr
& \Rightarrow 1 + \frac{{\text{R}}}{{200}} = \frac{{11}}{{10}} \cr
& \Rightarrow \frac{{\text{R}}}{{200}} = \frac{1}{{10}} \cr
& \Rightarrow {\text{R}} = {\text{20% }} \cr} $$

Answer: Option C. -> Rs. 4050
$$\eqalign{
& {\text{Rate of interest}} \cr
& {\text{r}} = {\text{12}}\frac{1}{2}\% = \frac{1}{8} \cr} $$
    Year    
    Principal    
    Installment    
 
⇒ I
8×9 →
9×9
......(i)
⇒ II
64 →
81
......(ii)
Since, installment is equal hence multiply equation (i) by 9
⇒ Total principal = 72 + 64 = 136 units
136 units → 6800
1 units → 50
81 units → 4050
⇒ Each installment = Rs. 4050