Answer: Option A. -> 10%
Answer: (a)Let the rate of interest = R% per annum.We know thatA = P$(1 + R/100)^T$2420 = P$(1 + R/100)^2$ ....(i)2662 = P$(1 + R/100)^3$ ...(ii)Dividing equation (ii) by (i),$1 + R/100 = 2662/2420$$R/100 = 2662/2420$ - 1$R/100 = {2662 - 2420}/2420$= $242/2420 = 1/10$R = $1/10 × $100 = 10%Using Rule 7(i),Here, b - a = 3 - 2 = 1B = Rs.2,662, A= Rs.2,420R% = $(B/A - 1)$ × 100%= $(2662/2420 -1)$ × 100%= $[{2662 - 2420}/2420]$ × 100%= $242/2420 × 100%$ = 10%

Answer: Option A. -> 20
Answer: (a)If the principal be Rs.P, thenA = P$(1 + R/100)^T$1440 = P$(1 + R/100)^2$ ...(i)and 1728 = P$(1 + R/100)^3$ ...(ii)On dividing equation (ii) by (i),$1728/1440 = 1 + r/100$$r/100 = 1728/1440$ - 1= ${1728 - 1440}/1440 = 288/1440$r = ${288 × 100}/1440$r = 20% per annumUsing Rule 7(i),Here, b - a = 3 - 2 = 1B = Rs.1728, A = Rs.1440R% = $(B/A - 1)$ × 100%= $(1728/1440 - 1) × 100%$= $({1728 - 1440}/1440) × 100%$= $[288/1440] × 100%$ = 20%

Answer: Option C. -> 625
Answer: (c)Using Rule 6,The difference between compound interest and simple interest for two years= ${\text"Principal" × (Rate)^2}/{100 × 100}$1 = ${\text"Principal" × (4)^2}/10000$Principal = $10000/16$ = Rs.625

Answer: Option D. -> 2400
Answer: (d)Using 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 C. -> 32,000
Answer: (c)Using Rule 6,Time = $3/2 × 2 = 3$ half yearsRate = $10/2$ = 5% per half year[Since, when r → $r/2$, then t → 2t]Difference = P$(r^3/1000000 + {3r^2}/10000)$244 = P$(125/1000000 + 75/10000)$244 = P$(7625/1000000)$P = ${244 × 1000000}/7625$ = Rs.32000

Answer: Option B. -> 2051.20
Answer: (b)Using Rule 1,S.I. = $\text"Principal × Time × Rate"/100$= ${32000 × 4 × 10}/100$ = Rs.12800C.I. = P$[(1 + R/100)^4 - 1]$= 32000$[(1 + 10/100)^4 - 1]$= 32000 $[(1.1)^4$ - 1]= 32000 (1.4641 - 1)= 32000 × 0.4641= Rs.14851.2Required difference= 14851.2 - 12800 = Rs.2051.2

Answer: Option C. -> 43.41
Answer: (c)Using 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 C. -> 8750
Answer: (c)Using Rule 6,Rate of interest = 8% per halfyearTime = 2 half yearsDifference of interests = ${PR^2}/100$56 = $P × (8)^2/(100)^2$P = ${56 × 10000}/64$ = 8750

Answer: Option D. -> 625
Answer: (d)Using Rule 6,When difference between the CI and SI on a certain sum of money for 2 years at r % rate is x, thenSum = x × $(100/r)^2$= 1 × $(100/4)^2$ = Rs.625

Answer: Option C. -> 330.75
Answer: (c)Using Rule 1,If A = Amount, P = Principal, r = Rate of Compound Interest (C.I.), n = no. of years then,A=P$(1 + r/100)^n$, C.I. = A - PC.I. = P$[(1 + r/100)^n - 1]$