Answer: Option D. -> Rs.2522
Answer: (d)Principal = $\text"S.I. × 100"/\text"Time × Rate"$= ${1600 × 100}/{5 × 2}$ = Rs.16000C.I. = P$[(1 + R/100)^T - 1]$= 16000$[(1 + 5/100)^3 –1]$= 16000$[(21/20)^3 - 1]$= $16000(9261/8000 - 1)$= ${16000 × 1261}/8000$ = Rs.2522

Answer: Option B. -> 5%
Answer: (b)S.I. on Rs.2400 for 1 year= Rs.(2, 520 - 2, 400) = Rs.120Rate = $\text"S.I. × 100"/ \text"Principal × Time"$%= ${120 × 100}/{2400 × 1}$ = 5%Using Rule 7(i),Here, b - a = 4 - 3 = 1B = Rs.2520, A = Rs.2400R% = $(B/A - 1)$ × 100%= $[2520/2400 - 1] × 100%$= $[{2520 - 2400}/2400] × 100%$= $120/2400 × 100%$ = 5%

Answer: Option D. -> Rs.7
Answer: (d)Principal = $\text"S.I. × 100"/ \text"Time × Rate"$= ${350 × 100}/{2 × 4}$ = Rs.4375C.I. = P$[(1 + R/100)^T - 1]$= 4375$[(1 + 4/100)^2 - 1]$= 4375$[(1 + 1/25)^T - 1]$= 4375$[(26/25)^2 - 1]$= 4375$(676/625 - 1)$= ${4375 × 51}/625$ = Rs.357Required difference= Rs.(357 - 350) = Rs.7

Answer: Option C. -> Rs.3000
Answer: (c)P$(1 + r/100)^2$ = 4500 ...(i)P$(1 + r/100)^4$ = 6750 .....(ii)On dividing equation (ii) by equation (i), we get$(1 + r/100)^2 = 6750/4500$From equation (i),P × $6750/4500$ = 4500P = ${4500 × 4500}/6750$ = Rs.3,000Using Rule 7(ii),Here, b - a = 4 - 2 = 2B = Rs.6750, A = Rs.4500R% = $[(B/A)^{1/2} - 1]$ × 100%= $[(6750/4500)^{1/2} - 1] × 100%$= $[(3/2)^{1/2} - 1] ×$ 100%$(3/2)^{1/2} = 1 + R/100$$3/2 = (1 + R/100)^2$A = P$(1 + R/100)^2$4500 = P × $3/2$ ⇒ P = Rs.3000

Answer: Option D. -> Rs.30,000
Answer: (d)Let the principal be Rs. x.When the interest is compounded annually,C.I. = P$[(1 + R/100)^T - 1]$= P$[(1 + 20/100)^2 - 1]$= P$[(6/5)^2 - 1]$= P$(36/25 - 1)$ = Rs.${11P}/25$When the interest is compounded half–yearly,C.I. = P$[(1 + 10/100)^4 - 1]$= P$[(11/10)^4 - 1]$= P$(14641/10000 - 1)$= Rs.${4641P}/10000$${4641P}/10000 - {11P}/25$ = 723${4641P - 4400P}/10000$ = 723${241P}/10000$ = 723P = ${723 × 10000}/241$ = Rs.30000

Answer: Option B. -> 2.5%
Answer: (b)A = P$(1 + R/100)^T$3840 = P$(1 + R/100)^4$ ....(i)3936 = P$(1 + R/100)^5$ ...(ii)Dividing equation (ii) by equation (i),$3936/3840 = 1 + R/100$$R/100 = 3936/3840$ - 1= ${3936 - 3840}/3840 = 96/3840$R = $96/3840 × 100$ = 2.5%Using Rule 7(i),Here, b - a = 5 - 4 = 1B = Rs.3,936, A = Rs.3,840R%= $(B/A - 1)$ × 100%= $(3936/3840 - 1)$ × 100%= $({3936 - 3840}/3840) × 100%$= $96/3840 × 100% = 10/4%$ = 2.5%

Answer: Option D. -> 8%
Answer: (d)Using Rule 6,Difference of 2 years= ${p × r^2}/10000$32 = ${5000 × r^2}/10000$$r^2 = {32 × 10000}/5000$ = 64r = $√{64}$ = 8%

Answer: Option D. -> 72,000
Answer: (d)If the interest is compounded half yearly,C.I. = P$[(1 + R/100)^T - 1]$= P$[(1 + 5/100)^2 - 1]$= P$[(21/20)^2 - 1] = {41P}/400$S.I. = ${P × R × T}/100 = {P × 10}/100 = P/10$${41P}/400 - P/10$ = 180${41P - 40P}/400$ = 180$P/400 = 180$P = Rs.72000Using Rule 6,Here, C.I. - S.I. = Rs.180Interest is compounded half yearlyR = $10/5$ = 5%, T = 2 yearsC.I. - S.I. = P$(R/100)^2$180 = P$(5/100)^2$P = 180 × 20 × 20 ⇒ P = Rs.72000

Answer: Option A. -> 1000
Answer: (a)Let the sum be xr = 10%, n = 3 yearsS.I. = ${x × r × n}/100$S.I.= ${x × 10 × 3}/100 = 3/10x$C.I.= $[(1 + r/100)^n - 1]x$= $[(1 + 10/100)^3 - 1]x$= $[(11/10)^3 - 1]x$$(1331/1000 - 1)x = 331/1000x$$331/1000x - 3/10x$ = 31or $({331 - 300})/1000x = 31$or $31/1000x$ = 31or x = 1000Sum = Rs.1000Using Rule 6,Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?C.I. - S.I. = P × $(R/100)^2 × (3 + R/100)$31 = P × $(10/100)^2(3 + 10/100)$31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000

Answer: Option C. -> 1,20,000
Answer: (c)Using Rule 6,Simple ApproachSum = (CI - SI)$(100/r)^2$= 768 × $(100/8)^2$ = Rs.1,20,000