Answer: Option A. -> Rs.1,92,000
Answer: (a)A = P$(1 + R/100)^T$24000 = 12000$(1 + R/100)^5$2 = $(1 + R/100)^5$$2^4 = (1 + R/100)^20$i.e. The sum amounts to Rs.192000 after 20 years. Using Rule 11,A certain sum at C.I. becomes x times in $n_1$ year and y times in $n_2$ years then $x^{1/n_1} = y^{1/n_2}$

Answer: Option B. -> 45 years
Answer: (b)A = P$(1 + R/100)^T$2 = 1$(1 + R/100)^15$Cubing on both sides, we have8 = 1$(1 + R/100)^45$Required time = 45 yearsUsing Rule 5,Here, m = 2, t = 15 yearsIt becomes 8 times = $2^3$ timesin t × n years= 15 × 3 = 45 years

Answer: Option C. -> 10%
Answer: (c)If principal = Rs.1000, amount = Rs.1331A = P$(1 + R/100)^T$$1331/1000 = (1 + R/100)^3$$(11/10)^3 = (1 + R/100)^3$$1 + R/100 = 11/10$$R/100 = 1/10$R = $1/10 × 100$ = 10%Using Rule 8,If a sum becomes 'n' times of itself in 't' years on compound interest, then R% = $[n^{1/t} - 1] × 100%$

Answer: Option B. -> 100%
Answer: (b)A = P$(1 + R/100)^T$4 = $(1 + R/100)^2$1 + $R/100$ = 2$R/100$ = 1 ⇒ R = 100%Using Rule 8,Here, n = 4, t = 2 yearsR% = $(n^{1/t} - 1)$ × 100%= $((4)^{1/2} - 1)$ × 100% = 100%

Answer: Option D. -> 50%
Answer: (d)A = P$(1 + R/100)^T$$27/8x = x(1 + R/100)^3$$(3/2)^3 = (1 + R/100)^3$$1 + R/100 = 3/2$$R/100 = 3/2 - 1 = 1/2$R = $1/2$ × 100 ⇒ R = 50%Using Rule 8,n= $27/8$, t = 3 yearsR% = $(n^{1/t} - 1) × 100%$= $((27/8)^{1/3} - 1) × 100%$= $[(3/2) - 1]$ × 100% ⇒ R% = 50%

Answer: Option C. -> 6 years
Answer: (c)A = P$(1 + R/100)^T$Let P = Rs.1, then A = Rs.33 = 1$(1 + R/100)^3$On squaring both sides,9 = 1$(1 + R/100)^6$Time = 6 yearsUsing Rule 11,Here, $x = 3, n_1 = 3, y = 9, n_2$ = ?Using, $x^{1/n_1} = y^{1/n_2}$$(3)^{1/3} = (9)^{1/n_2}$$3^{1/3} = (3^2)^{1/n_2}$$3^{1/3} = 3^{2/n_2}$$1/3 = 2/n_2 ⇒ n_2$ = 6 years

Answer: Option D. -> 12 years
Answer: (d)A sum of Rs.x becomes Rs.2x in 4 years.Similarly, Rs.2x will become 2 × 2x= Rs.4x in next 4 years and Rs.4x will become2 × 4x = Rs.8x in yet another 4 years.So, the total time = 4 + 4 + 4 = 12 yearsUsing Rule 5,A certain sum becomes 'm' times of itself in 't' years on compound interest then the time it will take to become mn times of itself is t × n years.

Answer: Option B. -> Rs.1352
Answer: (b)A = Rs.2550R = 4% per annumn = 2 yearsLet each of the two equal instalments be xPresent worth = $\text"Instalment"/(1 + r/100)^n$$P_1 = x/(1 + 4/100)^1 = x/{1 + 1/25} = x/{26/25}$or $P_1 = 25/26x$Similarly,$P_2 = (25/26)^2x = 625/676x$$P_1 + P_2$ = A$25/26x + 625/676x$ = 2550${(650 + 625)x}/676 = 2550$$1275/676x = 2550$x = 2550 $× 676/1275$ ⇒ x = Rs.1352Using Rule 9,If a sum 'P' is borrowed at r% annual compound interest which is to be paid in 'n' equal annual installments including interest, then(i) For n = 2, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2$(ii) For n = 3, Each annual installment= $p/{(100/{100 + r}) + (100/{100 + r})^2 + (100/{100 + r})^3$

Answer: Option C. -> Rs.4050
Answer: (c)Let the annual instalment be xA = P$(1 + R/T)^T$$x = P_1(1 + 25/200)$$x = P_1 × 9/8$$P_1 = 8/9x$Similarly, $P_2 = 64/81x$$P_1 + P_2$ = 6800$8/9x + 64/81x$ = 6800${72x + 64x}/81 = 6800$${136x}/81 = 6800$$x = {6800 × 81}/136$ = Rs.4050Using Rule 9(i),Here, P = Rs.6800, R = $25/2$%, n = 2Each instalment= $p/{(100/{100 + r}) + (100/{100 + r})^2$= $6800/{(100/{100 +{25/2}}) + (100/{100 + {25/2}})^2$= $6800/{200/225 + (200/225)^2}$= $6800/{200/225(1 + {200/225})}$= ${6800 × 225 × 225}/{200 × 425}$ = Rs.4050

Answer: Option D. -> Rs.121
Answer: (d)Using Rule 9(i),Let the value of each instalment be Rs.xPrincipal = Present worth of Rs.x due 1 year hence, present worth of Rs. x due 2 years hence210 = $x/(1 + R/100) + x/(1 + R/100)^2$210 = $x/(1 + 10/100) + x/(1 + 10/100)^2$210 = $x/{1 + 1/10} + x/(1 + 1/10)^2$210 = $x/{11/10} + x/(11/10)^2$210 = ${10x}/11 + {100x}/121$210 = ${110x + 100x}/121$210 × 121 = 210 x$x = {210 × 121}/210$ = Rs.121