Answer: Option B. -> Rs. 5039
$$\eqalign{
& {\text{Le the rate be R }}\% {\text{ p}}{\text{.a}}{\text{.}} \cr
& {\text{Then,}} \cr
& {\text{4000}}\left( {1 + \frac{{{\text{R }}}}{{100}}} \right) = 4320 \cr
& \Rightarrow 1 + \frac{{{\text{R }}}}{{100}} = \frac{{4320}}{{4000}} = \frac{{108}}{{100}} \cr
& \Rightarrow \frac{{{\text{R }}}}{{100}} = \frac{8}{{100}} \cr
& \Rightarrow {\text{R }} = 8 \cr
& \therefore {\text{Amount after 3 yeras}} \cr
& {\text{ = Rs}}{\text{. }}\left[ {4000 + {{\left( {1 + \frac{8}{{100}}} \right)}^3}} \right] \cr
& {\text{ = Rs}}{\text{. }}\left( {4000 \times \frac{{27}}{{25}} \times \frac{{27}}{{25}} \times \frac{{27}}{{25}}} \right) \cr
& {\text{ = Rs}}{\text{. }}\left( {\frac{{629856}}{{125}}} \right) \cr
& {\text{ = Rs}}{\text{. }}5038.848 \approx 5039 \cr} $$

Answer: Option B. -> Rs. 7569
$$\eqalign{
& {\text{Rate of interest (r)}} \cr
& {\text{ = 8}}\frac{3}{4}\% = \frac{7}{{80}} = \frac{{87 \to {\text{ Installment}}}}{{80 \to {\text{Principal}}}} \cr} $$
⇒  
  I  
  80×87  
  →  
  87×87  
......(i)
⇒  
  II  
  6400  
  →  
  7569  
......(ii)
Since, installment is equal hence multiply equation (i) by 87
⇒ Total principal = 6960 + 6400 = 13360
⇒ 13360 units = Rs. 13360
⇒ 1 units = Rs. 1
⇒ 7569 units = Rs. 7569
∴ Each installment = Rs. 7569

Answer: Option C. -> Rs. 1750
$$\eqalign{
& {\text{C}}{\text{.I}}{\text{.}} \cr
& {\text{ = Rs}}{\text{.}}\left[ {4000 \times {{\left( {1 + \frac{{10}}{{100}}} \right)}^2} - 4000} \right] \cr
& = {\text{ Rs}}{\text{.}}\left( {4000 \times \frac{{11}}{{10}} \times \frac{{11}}{{10}} - 4000} \right) \cr
& = {\text{Rs}}{\text{. 840}} \cr
& \therefore {\text{Sum = Rs}}{\text{.}}\left( {\frac{{420 \times 100}}{{3 \times 8}}} \right) \cr
& = {\text{Rs}}{\text{. }}1750 \cr} $$

Answer: Option D. -> Rs. 6615
$$\eqalign{
& {\text{Amount = 6000}}{\left( {1 + \frac{5}{{100}}} \right)^2} \cr
& \Rightarrow {\text{Amount = 6000}} \times \frac{{21}}{{20}} \times \frac{{21}}{{20}} \cr
& \Rightarrow {\text{Amount = Rs. 6615}} \cr} $$

Answer: Option B. -> $$1\frac{1}{2}$$ years
$$\eqalign{
& {\text{Let the time be }}n{\text{ year}} \cr
& {\text{Then,}} \cr
& {\text{800}} \times {\left( {1 + \frac{5}{{100}}} \right)^{2n}} = 926.10 \cr
& \Leftrightarrow {\left( {1 + \frac{5}{{100}}} \right)^{2n}} = \frac{{9261}}{{8000}} \cr
& \Leftrightarrow {\left( {\frac{{21}}{{20}}} \right)^{2n}} = {\left( {\frac{{21}}{{20}}} \right)^3} \cr
& \Leftrightarrow 2n = 3 \cr
& \Leftrightarrow n = \frac{3}{2} \cr
& \therefore n = 1\frac{1}{2}{\text{years}} \cr} $$

Answer: Option B. -> Rs. 6615
$$5\% = \frac{1}{{20}} = \frac{{21 \to {\text{ Installment}}}}{{20 \to {\text{ Principal}}}}$$
Year    
    Principal    
    Installment
 
⇒ I
20×21 →
21×21
......(i)
⇒ II
400→
441
.....(ii)
Since, installment is equal, hence multiply equation (i) by 21
⇒ Total principal = 420 + 400 = 820
⇒ 820 units = Rs. 12300
⇒ 1 units = Rs. 15
⇒ 441 units = Rs. 6615
∴ Each installment = Rs. 6615

Answer: Option C. -> Rs. 3972
$$\eqalign{
& {\text{Let P}} = {\text{Rs}}.100 \cr
& {\text{Then,}} \cr
& {\text{S}}{\text{.I}}{\text{. = Rs}}.60{\text{ and}} \cr
& {\text{T = 6 years}} \cr
& {\text{R = }}\frac{{100 \times 60}}{{100 \times 6}}{\text{ = 10% p}}{\text{.a}}{\text{.}} \cr
& {\text{Now}} \cr
& {\text{P = Rs 12000,}} \cr
& {\text{T = 3 years and}} \cr
& {\text{R = 10% p}}{\text{.a}}{\text{.}} \cr
& \therefore {\text{C}}{\text{.I}}{\text{. = Rs}}{\text{.}}\left[ {12000 \times \left\{ {{{\left( {1 + \frac{{10}}{{100}}} \right)}^3} - 1} \right\}} \right] \cr
& = {\text{Rs}}{\text{.}}\left( {12000 \times \frac{{331}}{{1000}}} \right) \cr
& = {\text{Rs}}{\text{. }}3972 \cr} $$

Answer: Option A. -> 100%
$$\eqalign{
& {\text{Principal}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Amount}} \cr
& \,\,\,\,\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,{\text{4}} \cr
& \Rightarrow 4 = 1{\left( {1 + \frac{r}{{100}}} \right)^2} \cr
& \Rightarrow 4 = {\left( {1 + \frac{r}{{100}}} \right)^2} \cr
& \Rightarrow r = 100\% \cr
& \cr
& {\text{Alternate}} \cr
& {\text{Principal}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{Amount}} \cr
& \,\,\,\,\,\,\,\,\,\root 2 \of 1 \,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\root 2 \of 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,2 \cr
& \Rightarrow {\text{Rate of interest}} \cr
& {\text{ = }}\frac{{\left( {2 - 1} \right)}}{1} \times 100 = 100\% \cr} $$

Answer: Option C. -> Rs. 8750
When the money is compounded half yearly the effective rate of interest for 6 months = $$\frac{{16}}{2}$$ = 8% = $$\frac{{2}}{25}$$
Let principal = (25)2 = 625
⇒ 4 units → 56
⇒ 1 unit → 14
⇒ Principal = 14 × 625 = Rs. 8750

Answer: Option B. -> 19%
Let the R% p.a.
Then,
$$\left[ {26000 \times {{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 26000} \right] - $$       $$\left( {\frac{{26000 \times {\text{R}} \times 3}}{{100}}} \right) = $$     $$2994.134$$
$$ \Rightarrow 26000\left[ {{{\left( {1 + \frac{{\text{R}}}{{100}}} \right)}^3} - 1 - \frac{{3{\text{R}}}}{{100}}} \right] = $$        $$2994.134$$
$$ \Rightarrow 26000$$ $$\left[ {\frac{{{{\left( {100 - {\text{R}}} \right)}^3} - 1000000 - 30000{\text{R}}}}{{1000000}}} \right] = $$        $$2994.134$$
$$ \Rightarrow 26\left[ {\left\{ {1000000 + {{\text{R}}^3} + 300{\text{R}}\left( {100 + {\text{R}}} \right) - 1000000 - 30000{\text{R}}} \right\}} \right] = 2994134$$
$$ \Rightarrow {{\text{R}}^3} + 300{{\text{R}}^2} = \frac{{2994134}}{{26}} = $$      $$115159$$
$$ \Rightarrow {{\text{R}}^2}\left( {{\text{R}} + 300} \right) = 115159$$
$$ \Rightarrow {\text{R = 19}}\% $$