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

Answer: Option D. -> Rs.18,50,000
Answer: (d)Using Rule 1,Let the income of company in 2010 be Rs.PAccording to the question,A = P$(1 + R/100)^T$2664000 = P$(1 + 20/100)^2$2664000 = P$(1 + 1/5)^2$2664000 = P × $(6/5)^2$P = ${2664000 × 5 × 5}/{6 × 6}$ = Rs.1850000

Answer: Option B. -> 64 = $(1.01)^{12y}$
Answer: (b)Rate of interest = 12% p.a.= 1% per monthTime = 12y monthsA = P$(1 + R/100)^T$64 = 1$(1 + 1/100)^{12y}$64 = $1(1.01)^{12y}$

Answer: Option B. -> 10%
Answer: (b)Let the rate of interest be r% per annum,According to the question,4840 = P$(1 + r/100)^2$ ..... (i)and 5324 = P$(1 + r/100)^3$....(ii)On dividing equation (ii) by equation (i), we have,$1 + r/100 = 5324/4840 = 1 + 484/4840$$r/100 = 484/4840$ ⇒ r = 10%Using Rule 7,If on compound interest, a sum becomes Rs.A in 'a' years and Rs.B in 'b' years then,(i) If b - a = 1, then, R% = $(B/A - 1)$ × 100%(ii) If b - a = 2, then, R% = $(√{B/A} - 1)$ × 100%(iii) If b - a = n then, R% = $[(B/A)^{1/n} - 1]$ × 100%where n is a whole number.

Answer: Option A. -> Rs.625
Answer: (a)Principal = Rs.P (let)Rate = R% per annumA = P$(1 + R/100)^T$650 = P$(1 + R/100)$$650/P = (1 + R/100)$ ...(i)Again, 676 = P$(1 + R/100)^2$676 = P$(650/P)^2$= ${P × 650 × 650}/P^2$P = ${650 × 650}/676$ = Rs.625

Answer: Option C. -> Rs.500
Answer: (c)C.I. = P$[(1 + R/100)^T - 1]$525 = P$[(1 + 10/100)^2 - 1]$525 = P$(121/100 - 1)$525 = ${P × 21}/100$P = ${525 × 100}/21$ = Rs.2500Again, new rate = 5% per annumS.I. = $\text"Principal × Time × Rate"/100$= ${2500 × 5 × 4}/100$ = Rs.500

Answer: Option A. -> 6%
Answer: (a)Difference = 238.50 - 225 = Rs.13.50= S.I. on Rs.225 for 1 yearRate = $\text"S.I. × 100"/\text"Principal × Time"$= ${13.50 × 100}/{225 × 1}$ = 6% per annumUsing Rule 7(i),Here, b - a = 1B = Rs.238.50, A = Rs.225R% = $(B/A - 1)$ × 100%= $({238.50}/225 - 1) × 100%$= $({238.50 - 225}/225) × 100%$= $({13.5}/225) × 100%$ = 6%

Answer: Option C. -> Rs.625
Answer: (c)Interest on Rs.650 for 1 year= 676 - 650 = Rs.26So, r = $26/650 × 100$r = 4% per annumP = $A/[1 + r/100]^t = 650/[1 + 4/100]^1$= $650/{26/25} = 650 × 25/26$ = Rs.625Using Rule 7(i),Here, b - a = 1B = Rs.676, A = Rs.650R% = $(B/A - 1)$ × 100%= $[676/650 - 1] × 100%$= $[{676 - 650}/650] × 100%$= $26/650 × 100% = 100/25$ = 4%Amount= P$(1 + R/100)^1$650 = P$(1 + 4/100)$P = ${650 × 100}/104$ = Rs.625Note : A sum at a rate of interest compounded yearly becomes Rs.$A_1$, in n years and Rs. $A_2$ in (n + 1) years, thenP = $A_1(A_1/A_2)^n$

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