Answer: Option D. -> Rs.12,320
Answer: (d)Using Rule 3,A = P$(1 + r_1/100)(1 + r_2/100)$= 10000$(1 + 10/100)(1 + 12/100)$= 10000 × $11/10 × 28/25$ = Rs.12320

Answer: Option D. -> 1$1/2$ years
Answer: (d)Using Rule 1 and 2,Interest is compounded half yearly.Rate of interest = 5%Time = $n/2$ years (let)or n half-yearsA = P$(1 + R/100)^T$9261 = 8000$(1 + 5/100)^n$$9261/8000 = (21/20)^n$$(21/20)^3 = (21/20)^n$n = 3 half years= $3/2$ years = $1{1}/2$ years

Answer: Option B. -> Rs.142.40
Answer: (b)Using Rule 3,Amount = $2000(1 + 4/100)(1 + 3/100)$= 2000 ×1.04 ×1.03 = Rs.2142.40CI = Rs.(2142.40 - 2000) = Rs.142.40

Answer: Option A. -> Rs.3750
Answer: (a)Difference of CI and SI for two years= Rs.(954 - 900) = Rs.54Sum= Difference in CI and SI × $(100/{Rate})^2$Rate = ${2 × \text"Difference" × 100}/\text"Simple interest"$= ${2 × 5400}/900 = 12%$Sum = 54 × $(100/12)^2$= $54 × 25/3 × 25/3$ = Rs.3750Using Rule 10,C.I. = Rs.954, S.I.=Rs.900, P=?C.I.= S.I.$(1 + R/200)$954 = 900$(1 + R/200)$$954/900 = 1 + R/200$$954/900 - 1 = R/200$${954 - 900}/900 = R/200$$54/9 = R/2$R = 12%Now S.I. = ${P × R × T}/100$900 = ${P × 12 × 2}/100$P = Rs.3750

Answer: Option A. -> Rs.100000
Answer: (a)Let the principal be Rs.P. For 4 years,S.I. = $\text"Principal × Time × Rate"/100$= ${P × 4 × 5}/100$ = Rs.$P/5$C.I. = P$[(1 + R/100)^T - 1]$= P$[(1 + 10/100)^4 - 1]$= P$[(11/10)^4 - 1]$= P$(14641/10000 - 1) = {4641P}/10000$According to the question,${4641P}/10000 - P/5$ = 26410${4641P - 2000P}/10000$ = 2641${2641P}/10000$ = 2641P = Rs.10000

Answer: Option B. -> Rs.2,400
Answer: (b)C.I. = P$[(1 + R/100)^T - 1]$2544 = P$[(1 + 12/100)^2 - 1]$2544 = P$[(28/25)^2 - 1]$2544 = P$(784/625 - 1)$2544 = P$({784 - 625}/625)$2544 = ${P × 159}/625$P = ${2544 × 625}/159$ = Rs.10000S.I. = ${P × R × T}/100$= ${10000 × 2 × 12}/100$ = Rs.2400Using Rule 10,Here, C.I. = Rs.2544, R = 12%, S.I. = ?C.I. = S.I.$(1 + R/200)$2544 = S.I.$(1 + 12/200)$2544 = S.I.$(212/200)$S.I. = ${2544 × 200}/212$ = Rs.2400

Answer: Option D. -> Rs.2,400
Answer: (d)C.I.= P$(1 + r/100)^t$ - P2448 = P$[(1 + r/100)^t - 1]$or 2448 = P$[(1 + 4/100)^2 - 1]$2448 = P$[676/625 - 1]$2448 = P$[51/625]$P = ${2448 × 625}/51$ = Rs.30,000S.I. = ${30000 × 4 × 2}/100$ = Rs.2400Using Rule 10,Here, C.I. = Rs.2448, R = 4%, S.I. = ?C.I.= S.I.$(1 + R/200)$2448 = S.I.$(1 + 4/200)$2448 = S.I.$(1 + 1/50)$2448 = S.I.$(51/50)$S.I. = ${2448 × 50}/51$ = Rs.2400

Answer: Option B. -> Rs.400
Answer: (b)Compound interest = P $[(1 + R/100)^T - 1]$410 = P$[(1 + 5/100)^2 - 1]$410 = P$[(1 + 1/20)^2 - 1]$410 = P$[(21/20)^2 - 1]$410 = P$(441/400 - 1)$410 = P$(41/400)$P = ${410 × 400}/41$ = Rs.4000S.I. = $\text"Principal × Time × Rate"/100$= ${4000 × 2 × 5}/100$ = Rs.400Using Rule 10,Here, C.I. = Rs.410, R = 5%, S.I. = ?C.I. = S.I.$(1 + R/200)$410 = S.I.$(1 + 5/200)$410 = S.I.$(205/200)$S.I. = ${410 × 200}/205$ = Rs.400

Answer: Option D. -> Rs.1,00,000
Answer: (d)Using Rule 1,Sum borrowed = Rs.xSimple interest after 4 years= ${x × 4 × 5}/100$ = Rs.$x/5$Amount lent of on compound interest = Rs.$x/2$C.I. = P$[(1 + R/100)^T - 1]$= $x/2[(1 + 10/100)^4 - 1]$= $x/2[(1.1)^4 - 1]$= $x/2$ (1.4641 - 1) = Rs.${0.4641x}/2$${0.4641x}/2 - x/5$ = 3205${2.3205x - 2x}/10$ = 32050.3205x = 32050$x = 32050/{0.3205}$ = Rs.100000

Answer: Option C. -> 2000
Answer: (c)S.I. for 2 years= $2/3$ × 540 = Rs.360C.I. - S.I.= 376.20 - 360 = Rs.16.20Rate of interest= ${16.20}/180$ × 100 = 9% per annumPrincipal = $\text"S.I. × 100"/\text"Time × Rate"$= ${180 × 100}/{1 × 9}$ = Rs.2000