Answer: Option D. -> Rs.4,900
Answer: (d)A = P$(1 + R/100)^T$7000 = P$(1 + R/100)^4$....(i)10000 = P$(1 + R/100)^8$.......(ii)Dividing equation (ii) by (i)$10000/7000 = (1 + R/100)^4$$10/7 = (1 + R/100)^4$From equation (i),7000 = P × $10/7$ ⇒ P = Rs.4900Using Rule 7(iii),Here, b - a = 8 - 4 = 4B = Rs.10,000, A = Rs.7000R% = $((B/A)^{1/n} - 1)$ × 100%R% = $[(10000/7000)^{1/4} - 1]$= $[(10/7)^{1/4} - 1]$$1 + R/100 = (10/7)^{1/4}$$(1 + R/100)^4 = 10/7$7000 = $P × 10/7$Since, Amount = P$(1 + R/100)^4$P = Rs.4900

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 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 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. -> 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. -> 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 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 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 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$[(1 + r/100)^n - 1]$