Answer: Option B. -> Rs. 2400
$$\eqalign{
& {\text{S}}{\text{.I}}{\text{.}} = {\text{Rs}}{\text{. 210}} \cr
& {\text{R}} = 3\frac{3}{4}\% = \frac{{15}}{4}\% \cr
& {\text{T}} = {\text{2}}\frac{{\text{1}}}{{\text{3}}}{\text{years}} = \frac{7}{3}{\text{years}} \cr
& \therefore {\text{Sum}} = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 210}}{{\frac{{15}}{4} \times \frac{7}{3}}}} \right) \cr
& = {\text{Rs}}{\text{.}}\left( {\frac{{100 \times 210 \times 4 \times 3}}{{15 \times 7}}} \right) \cr
& = {\text{Rs}}{\text{. }}2400 \cr} $$

Answer: Option A. -> 6 : 3 : 2
Let the principal in each case = 100 units
According to the question,
     
     
  1st part  
  2nd part  
  3rd part  
  Principal  
  →  
  100x6  
  100x3  
  100x2  
  Rate %  
  →  
  10  
  12  
  15  
  Time  
  →  
  6  
  10  
  12  
  Interest  
  →  
  60x6  
  120x3  
  180x2  
Interest Interest is same in each, so equal the interest.
Hence required ratio
= 600 : 300 : 200 of sum
= 6 : 3 : 2

Answer: Option A. -> 20 years
$$\eqalign{
& {\text{Let principal = 10P}} \cr
& {\text{Interest = 10P}} \times \frac{{30}}{{100}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 3P}} \cr
& {\text{According to the question,}} \cr
& {\text{Case (I)}} \cr
& \Rightarrow 3{\text{P = }}\frac{{{\text{10P}} \times {\text{R}} \times {\text{6}}}}{{100}} \cr
& \Rightarrow {\text{R = 5}}\% \cr
& {\text{Case (II)}} \cr
& {\text{Interest = Principal = 10P}} \cr
& \Rightarrow {\text{10P = }}\frac{{{\text{10P}} \times {\text{5}} \times {\text{t}}}}{{100}} \cr
& \Rightarrow {\text{t = 20 years}} \cr} $$

Answer: Option D. -> $$16\frac{2}{3}$$ years
$$\eqalign{
& {\text{Principal = Rs}}{\text{. 1000 }} \cr
& {\text{Rate = 5}}\% \cr
& {\text{Interest for first 10 years}} \cr
& = \frac{{1000 \times 5 \times 10}}{{100}} \cr
& = {\text{Rs}}{\text{. 500}} \cr
& {\text{After 10 years principal}} \cr
& = {\text{(1000}} + {\text{500)}} \cr
& {\text{ = Rs}}{\text{. 1500}} \cr
& {\text{Remaining interest}} \cr
& {\text{ = Rs}}{\text{. (2000}} - {\text{1500)}} \cr
& {\text{ = Rs}}{\text{. 500}} \cr
& {\text{Required time }} \cr
& {\text{ = }}\frac{{500}}{{1500}} \times \frac{{100}}{5} \cr
& = \frac{{100}}{15} \cr
& = \frac{{20}}{3} \cr
& = 6\frac{2}{3}{\text{ years}} \cr
& {\text{Total time}} \cr
& = \left( {10 + 6\frac{2}{3}} \right){\text{years}} \cr
& {\text{ = 16}}\frac{2}{3}{\text{ years}} \cr} $$

Answer: Option D. -> Rs. 1785
Let the money added by Rahul be Rs. x
Then,
$$ \Rightarrow \frac{{\left( {1150 + x} \right) \times 9 \times 3}}{{100}} - $$     $$\frac{{1150 \times 6 \times 3}}{{100}} = $$    $$274.95$$
⇒ 1150 × 27 + 27x - 1150 × 18 = 27495
⇒ 27x + 1150 × (27 - 18) = 27495
⇒ 27x = 27495 - 10350
⇒ 27x = 17145
⇒ x = 635
So, sum lent by Rahul to Sachin
= Rs. ( 1150 + 635 )
= Rs. 1785

Answer: Option B. -> Rs. 500
Let the amount invested at 12% be Rs. x and that invested at 10% be Rs. y
$$\eqalign{
& \text{Then,} \cr
& \to 12\% \,{\text{of }}x + 10\% \,{\text{of }}y = 130 \cr
& \Rightarrow 12x + 10y = 13000 \cr
& \Rightarrow 6x + 5y = 6500......{\text{(i)}} \cr
& {\text{And,}} \cr
& \to 10\% \,{\text{of }}x + 12\% \,{\text{of }}y = 134 \cr
& \Rightarrow 10x + 12y = 13400 \cr
& \Rightarrow 5x + 6y = 6700......{\text{(ii)}} \cr
& {\text{Adding (i) and (ii), we get:}} \cr
& 11\left( {x + y} \right) = 13200 \cr
& \Rightarrow x + y = 1200.......({\text{iii}}) \cr
& {\text{Subtracting (i) from (ii),}} \cr
& {\text{we get: }} - x + y = 200.......({\text{iv}}) \cr
& {\text{Adding (iii) and (iv), }} \cr
& {\text{we get}}:2y = 1400\,or\,y = 700 \cr
& {\text{Hence,}} \cr
& {\text{Amount invested at 12%}} \cr
& = \left( {1200 - 700} \right) \cr
& = {\text{Rs}}{\text{. 500}} \cr} $$

Answer: Option A. -> Rs. 210
$$\eqalign{
& {\text{According to the question,}} \cr
& {\text{Principal}} = {\text{Rs}}{\text{. }}2100 \cr
& {\text{Amount}} = {\text{Rs}}{\text{. }}2352 \cr
& {\text{SI}} = {\text{A}} - {\text{P}} \cr
& \,\,\,\,\,\,\, = 2352 - 2100 \cr
& \,\,\,\,\,\,\, = {\text{Rs}}{\text{. }}252 \cr
& {\text{Time = 2 years,}} \cr
& {\text{Let rate = R% }} \cr
& {\text{R = }}\frac{{252}}{{2100}} \times \frac{{100}}{2}{\text{ = 6% }} \cr
& {\text{New rate of interest}} \cr
& {\text{ = (6}} - {\text{1)}} \cr
& {\text{ = 5% }} \cr
& {\text{New interest}} \cr
& {\text{ = }}\frac{{2100 \times 5 \times 2}}{{100}} \cr
& {\text{ = Rs}}{\text{. 210}} \cr
& {\text{Hence required interest}} \cr
& {\text{ = Rs}}{\text{. 210}} \cr} $$

Answer: Option D. -> Rs. 5000
Difference between their rates he gained from both boys
$$\eqalign{
& \Rightarrow (15 \times 5)\% - (12 \times 4)\% \cr
& \Rightarrow 75\% - 48\% \cr
& \Rightarrow 27\% = 1350{\text{ }}({\text{given)}} \cr
& \Rightarrow 100\% = {\text{Rs}}{\text{. 5000}} \cr} $$

Answer: Option B. -> Rs. 600
According to the question,
Principal + SI for 2 year = Rs. 720 ......(i)
Principal + SI for 7 year = Rs. 1020 ......(ii)
Subtracting equation (i) from (ii)
⇒ SI for 5 years = (1020 - 720) = Rs. 300
⇒ SI for 1 years = Rs. 60
⇒ SI for 2 years = 60 × 2 = Rs. 120
⇒ Principal amount = (Amount after 2 years - 2 years SI) = (720 - 120)
⇒ Principal amount = Rs. 600

Answer: Option A. -> Rs. 2000, 3.5 years and 4 years
$$\eqalign{
& {\text{Let each sum}} = {\text{Rs}}{\text{. }}x. \cr
& {\text{Let the first sum be invested for}} \cr
& \left( {T - \frac{1}{2}} \right){\text{years and}} \cr
& {\text{the second sum for }}T{\text{ years}}{\text{.}} \cr
& {\text{Then,}} \cr
& x + \frac{{x \times 8 \times \left( {T - \frac{1}{2}} \right)}}{{100}} = 2560 \cr
& \Rightarrow 100x + 8xT - 4x = 256000 \cr
& \Rightarrow 96x + 8xT = 256000....(i) \cr
& {\text{And,}} \cr
& x + \frac{{x \times 7 \times T}}{{100}} = 2560 \cr
& \Rightarrow 100x + 7xT = 256000....(ii) \cr
& {\text{From(i) and (ii), we get:}} \cr
& 96x + 8xT = 100x + 7xT \cr
& \Rightarrow 4x = xT \cr
& \Rightarrow T = 4 \cr
& {\text{Putting }}T = {\text{4 in (i),we get:}} \cr
& 96x + 32x = 256000 \cr
& \Rightarrow 128x = 256000 \cr
& \Rightarrow x = 2000 \cr
& {\text{Hence,}} \cr
& {\text{each sum}} = {\text{Rs}}{\text{. 2000}} \cr
& {\text{time periods}} = \cr
& {\text{4 years and }}3\frac{1}{2}{\text{years}} \cr} $$