Answer: Option C. -> 6,000
Answer: (c)Let the sum Rs.x. Then,C.I. = $x(1 + 5/100)^2 - x$= ${441x}/400 - x = {441x - 400x}/400$= $41/400$xNow, S.I. = ${x × 5 × 2}/100 = x/10$(C.I.) - (S.I.)= ${41x}/400 - x/10$= ${41x - 40x}/400 = x/400$$x/400 = 15$x = 15 × 400 = 6000Hence, the sum is Rs.6000Using Rule 6,The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will beC.I - S.I. = P$(R/100)^2 = {S.I. × R}/200$ For 3 years, C.I. - S.I. = P$(R/100)^2 × (3 + R/100)$

Answer: Option A. -> 4
Answer: (a)S.I. = Rs.${2500 × 2 × 4}/100$ = Rs.200C.I. = Rs.2500$[(1 + 4/100)^2 - 1]$= Rs.2500$[(26/25)^2 - 1]$= Rs.${(676 - 625)}/625$ × 2500= Rs.$51/625 × 2500$ = Rs.204The required difference= C.I. - S.I. = Rs.(204 - 200) = Rs.4Using Rule 6,Here, C.I. - S.I.= ?, P = Rs.2500, R = 4%, T = 2C.I. - S.I.= P$(R/100)^2$= 2500$(4/100)^2$= 2500 × $1/25 × 1/25$C.I.–S.I. = Rs.4

Answer: Option D. -> 50000
Answer: (d)Using Rule 6,When difference between the compound interest and simple interest on a certain sum of money for 2 years at r % rate is x, then the sum is given by$x(100/r)^2$ Here x = Rs.80, r = 40%Required sum = 80$(100/4)^2$= 80 × 25 × 25 = Rs.50000

Answer: Option A. -> 6500
Answer: (a)Let the sum be x. Then,C.I. = $x(1 + 10/100)^2 - x = {21x}/100$S.I. = ${x × 10 × 2}/100 = x/5$C.I. - S.I. = ${21x}/100 - x/5 = x/100$Given that, $x/100$ = 65x = 6500Hence, the sum is Rs.6500.Using Rule 6,Here, C.I. - S.I. = Rs.65, R = 10%, T = 2 years, P = ?C.I. - S.I. = P$(R/100)^2$65 = P$(10/100)^2$ ⇒ P = Rs.6500

Answer: Option B. -> 2,000
Answer: (b)Using Rule 6,Difference between C.I. and S.I for 3 years= $\text"PR"^2/(100)^2(R/100 + 3)$15.25 = ${P × 25}/10000(5/100 + 3)$15.25 = ${P × 305}/{400 × 100}$P = ${15.25 × 400 × 100}/305$ = Rs.2000

Answer: Option A. -> 4800
Answer: (a)Simple ApproachDifference of SI and CI for 3 years= ${PR(300 + R)}/{100^3}$Since, ${P × 25 × 305}/{100 × 100 × 100}$ = 36.60P = ${36.60 × 100 × 100 × 100}/{25 × 305}$ = Rs.4800Using Rule 6,C.I.–S.I. = Rs.36.60, R = 5%, P =?, T = 3yrs.C.I. - S.I.= P$(R/100)^2 × (3 + R/100)$36.60 = P$(5/100)^2 × (3 + 5/100)$36.60 = P × $25/{100^2} × 305/100$P = ${36.60 × 100 × 100 × 100}/{25 × 305}$P = $36600000/{25 × 305}$ = Rs.4800

Answer: Option B. -> 2500
Answer: (b)Using Rule 6,Sum = Difference$(100/r)^2$= 4 × $(100/4)^2$ = Rs.2500

Answer: Option A. -> 625
$$\eqalign{
& {\text{Let}}\,{\text{the}}\,{\text{sum}}\,{\text{be}}\,Rs.\,x.\,{\text{Then}}, \cr
& {\text{C}}{\text{.I}}{\text{.}} = {x{{\left( {1 + \frac{4}{{100}}} \right)}^2} - x} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{{676}}{{625}}x - x} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{51}}{{625}}x \cr
& {\text{S}}{\text{.I}}{\text{.}} = {\frac{{x \times 4 \times 2}}{{100}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2x}}{{25}} \cr
& \therefore \frac{{51x}}{{625}} - \frac{{2x}}{{25}} = 1 \cr
& \Rightarrow x = 625 \cr} $$

Answer: Option B. -> Rs. 121
$$\eqalign{
& {\text{Amount}} \cr
& = {1600 \times {{\left( {1 + \frac{5}{{2 \times 100}}} \right)}^2} + 1600 \times \left( {1 + \frac{5}{{2 \times 100}}} \right)} \cr
& = {1600 \times \frac{{41}}{{40}} \times \frac{{41}}{{40}} + 1600 \times \frac{{41}}{{40}}} \cr
& = {1600 \times \frac{{41}}{{40}}\left( {\frac{{41}}{{40}} + 1} \right)} \cr
& = {\frac{{1600 \times 41 \times 81}}{{40 \times 40}}} \cr
& = Rs.\,3321 \cr
& \therefore C.I. = Rs.\,\left( {3321 - 3200} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = Rs.\,121 \cr} $$

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