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

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 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. -> 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 C. -> 10%
Answer: (c)Using Rule 1,Let the rate of interest per annum be r%According to the question,${5000 × 2 × r}/100 + {3000 × 4 × r}/100 = 2200$100r + 120r = 2200220 r = 2200r = $2200/220$ = 10%

Answer: Option D. -> Rs.600
Answer: (d)Using Rule 1,Simple interest gained from Rs.500= ${500 × 12 × 4}/100$ = Rs.240Let the other Principal be x.S.I. gained = Rs.(480 - 240) = Rs.240${x × 10 × 4}/100$ = 240$x = {240 × 100}/40$ = Rs.600

Answer: Option D. -> 7%
Answer: (d)Using Rule 1,Principal (P) = Rs.1600T = 2 years 3 months= $(2 + 3/12) yrs. = (2 + 1/4) yrs. = 9/4 yrs.$S.I = Rs.252R = % rate of interest per annumR = ${100 × S.I.}/{P × t}$= ${100 × 252}/{1600 × 9/4}$Rate of interest = 7% per annum.

Answer: Option C. -> 13%
Answer: (c)S.I. for 1$1/2$ years= Rs.(873 - 756) = Rs.117S.I. for 2 years= Rs.$(117 × 2/3 × 2)$ = Rs.156Principal = 756 - 156 = Rs.600Now, P = 600, T = 2, S.I. = 156R = ${100 × S.I.}/{P × T}$= ${100 × 156}/{600 × 2}$ = 13%Using Rule 12,Rate of interest= $({A_1 - A_2}/{A_2T_1 - A_1T_2})$ × 100= $({756 - 873}/{873 × 2 - 756 × 7/2}) × 100$= $({- 117}/{1746 - 2646}) × 100$= $({- 117}/{- 900})$ × 100 = 13%